🧠:[<think>Okay, let's tackle this problem step by step. The question is asking how many irrational numbers are in the given list: √2, 1/4, π, ∛8, -22/7, and 0.̄32̄. The options are A through D, with the numbers 1 to 4. Alright, first, I need to recall what makes a number irrational. An irrational number is a number that cannot be expressed as a fraction a/b, where a and b are integers and b is not zero. So, basically, numbers that can't be written as simple fractions. Their decimal expansions are non-repeating and non-terminating. Now, let me go through each number one by one:1. √2: I remember that the square root of any non-perfect square is irrational. Since 2 isn't a perfect square, √2 should be irrational. Yeah, that's right. So √2 is irrational.2. 1/4: This is a fraction. 1 divided by 4 is 0.25, which is a terminating decimal. Terminating decimals can always be expressed as fractions, so 1/4 is rational. So that's not irrational.3. π: Pi is a well-known irrational number. It's the ratio of a circle's circumference to its diameter, and its decimal expansion never repeats or terminates. So π is definitely irrational.4. ∛8: Hmm, the cube root of 8. Let's calculate that. 2 cubed is 8, right? So ∛8 is 2. 2 is an integer, and integers are rational because they can be expressed as a fraction over 1. So ∛8 is rational.5. -22/7: This is a negative fraction. Since it's already written as a fraction of two integers, -22 and 7, this is rational. Even though 22/7 is an approximation of π, it's not π itself; it's a fraction, so it's rational.6. 0.̄32̄: The notation here is a bit confusing. The overline is over both 3 and 2, right? So that means the decimals repeat every two digits? So it's 0.32323232..., repeating "32" indefinitely. A repeating decimal is a rational number because it can be expressed as a fraction. For example, 0.323232... can be written as 32/99. Let me confirm that. Let x = 0.323232..., then 100x = 32.323232..., subtract x from 100x: 99x = 32, so x = 32/99. Yep, that's a fraction. So 0.̄32̄ is rational.So, going through each one:- √2: Irrational (1)- 1/4: Rational- π: Irrational (2)- ∛8: Rational- -22/7: Rational- 0.̄32̄: RationalSo that gives us two irrational numbers: √2 and π. Therefore, the number of irrationals is 2, which corresponds to option B.Wait, let me double-check each one to make sure I didn't make a mistake. Sometimes cube roots can be tricky. For ∛8, since 8 is 2³, cube root of 8 is definitely 2. So that's rational. Pi is definitely irrational. √2 is irrational. The fractions and repeating decimals are rational. Yeah, seems correct. So the answer should be B: 2.Final Answerboxed{B}</think>To determine the number of irrational numbers among the given values, we analyze each one:1. (sqrt{2}): The square root of 2 is irrational because 2 is not a perfect square.2. (frac{1}{4}): This is a fraction, hence rational.3. (pi): Pi is a well-known irrational number with a non-repeating, non-terminating decimal expansion.4. (sqrt[3]{8}): The cube root of 8 is 2, which is an integer and therefore rational.5. (-frac{22}{7}): This is a fraction, hence rational.6. (0.overline{3}overline{2}): This repeating decimal is rational, as it can be expressed as (frac{32}{99}).The irrational numbers are (sqrt{2}) and (pi). Therefore, there are 2 irrational numbers.(boxed{B}),<think>Okay, so I need to figure out how many irrational numbers are in this list: √2, 1/4, π, cube root of 8, -22/7, and 0.̅3̅2. Then choose the correct answer from A to D. Let me start by recalling what an irrational number is. An irrational number is a number that cannot be expressed as a fraction a/b where a and b are integers and b is not zero. So, they can't be written as simple fractions, and their decimal expansions are non-repeating and non-terminating. Got it.Let me go through each number one by one.First is √2. Hmm, I remember that the square root of 2 is a classic example of an irrational number. Because if you try to write it as a fraction, it can't be done. The decimal goes on forever without repeating. So √2 is irrational. That's one.Next is 1/4. Well, that's a fraction. 1 divided by 4 is 0.25, which terminates. So 1/4 is a rational number. So that's not irrational.Third is π. Pi is another famous irrational number. It's the ratio of a circle's circumference to its diameter, but it can't be expressed as a fraction. Its decimal goes on forever without repeating. So π is irrational. That's two so far.Fourth is the cube root of 8. Let me calculate that. The cube root of 8 is 2 because 2*2*2=8. So cube root of 8 is 2, which is an integer. Integers are rational because they can be expressed as a fraction over 1. So that's rational. Not irrational.Fifth is -22/7. That's already a fraction. Negative numbers can be rational too, as long as they can be expressed as a fraction. So -22 divided by 7 is a rational number. Even though 22/7 is an approximation of π, here it's exactly written as a fraction, so it's rational. Not irrational.Last one is 0.̅3̅2. The overline notation here means that the digits 32 are repeating. So 0.32323232... and so on. Any repeating decimal is a rational number because it can be expressed as a fraction. For example, let's say x = 0.323232..., then 100x = 32.323232..., subtract x from 100x: 99x = 32, so x = 32/99. Therefore, 0.̅3̅2 is 32/99, which is a fraction. So that's rational too.So summarizing:- √2: irrational (1)- 1/4: rational- π: irrational (2)- cube root of 8: rational- -22/7: rational- 0.̅3̅2: rationalSo out of the six numbers given, two of them are irrational: √2 and π. Therefore, the answer should be B: 2. Let me check again to make sure I didn't make a mistake.√2: Definitely irrational. Yep.1/4: 0.25, which is rational. Correct.π: Yes, irrational. No doubt.Cube root of 8: 2, which is rational. Right.-22/7: That's a fraction. Rational. Correct.0.323232...: Repeating decimal, which converts to 32/99. Rational. Right.So yes, only two irrational numbers. The answer is B.Final Answerboxed{B}</think>To determine the number of irrational numbers among the given values, we analyze each one:1. (sqrt{2}): This is a well-known irrational number as it cannot be expressed as a fraction of integers and its decimal expansion is non-repeating and non-terminating.2. (frac{1}{4}): This is a rational number since it is a fraction of integers and its decimal expansion (0.25) terminates.3. (pi): This is an irrational number because it cannot be expressed as a fraction of integers and its decimal expansion is non-repeating and non-terminating.4. (sqrt[3]{8}): The cube root of 8 is 2, which is an integer and thus a rational number.5. (-frac{22}{7}): This is a rational number as it is a fraction of integers.6. (0.overline{3}overline{2}): This is a repeating decimal, which can be expressed as the fraction (frac{32}{99}), making it a rational number.Summarizing the results:- Irrational numbers: (sqrt{2}) and (pi)- Rational numbers: (frac{1}{4}), (sqrt[3]{8}), (-frac{22}{7}), and (0.overline{3}overline{2})Thus, there are 2 irrational numbers among the given values.(boxed{B})]

🔑:To determine the number of irrational numbers among the given list, let's examine each number individually:1. sqrt{2}: This is a classic example of an irrational number because it cannot be expressed as a fraction of two integers. The decimal expansion of sqrt{2} is non-terminating and non-repeating.2. frac{1}{4}: This is a rational number because it can be expressed as a fraction of two integers (in this case, 1/4).3. pi: This is well-known to be an irrational number. It represents the ratio of the circumference of a circle to its diameter, and its decimal expansion is non-terminating and non-repeating.4. sqrt[3]{8}: This is equal to 2, which is a rational number because it can be expressed as a fraction (2/1).5. -frac{22}{7}: This is a rational number because it is expressed as a fraction of two integers.6. 0.overline{3}overline{2}: This represents a repeating decimal, which means it is a rational number because repeating decimals can be expressed as fractions.Therefore, among the given numbers, there are 2 irrational numbers: sqrt{2} and pi.So, the correct answer is boxed{text{B}}.

🧠:[<think>Okay, let's see. I need to find the sum of all the numbers in a 2014x2014 square table where each cell at position (i, j) has the number a_{i,j} = (-1)^i * (2015 - i - j)^2. Hmm, that seems a bit complicated, but maybe I can break it down step by step.First, let me understand the formula for a_{i,j}. The value in each cell depends on the row number i and the column number j. The term (-1)^i means that the sign of the number alternates depending on the row: even rows (i even) will be positive, and odd rows (i odd) will be negative. Then, the magnitude is given by (2015 - i - j)^2. So each cell's value is the square of (2015 minus the row number minus the column number), multiplied by (-1) raised to the row number.The problem is to sum all these numbers over all i and j from 1 to 2014. So, we need to compute the double sum S = sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i * (2015 - i - j)^2.Let me see. Since the square is 2014x2014, both i and j range from 1 to 2014. Maybe changing variables could help here. Let me consider substituting k = 2015 - i - j. Let's check if that substitution simplifies things. Wait, k would be 2015 - i - j, which is equivalent to (2014 - i + 1) - j. Not sure if that helps immediately.Alternatively, maybe I can reverse the order of summation. Let's note that (2015 - i - j) is equal to (2015 - (i + j)). So, perhaps if I fix a value for s = i + j, then for each s, I can count how many times each (2015 - s)^2 term appears, multiplied by (-1)^i. Then, the total sum would be sum_{s} [ (2015 - s)^2 * sum_{i,j such that i + j = s} (-1)^i ].But how many pairs (i, j) have i + j = s? For s from 2 to 4028 (since i and j each go up to 2014, so the maximum i + j is 2014 + 2014 = 4028). For each s, the number of pairs is s - 1 when s is between 2 and 2015, and 4029 - s when s is between 2015 and 4028. Wait, that's the standard count for the number of integer solutions (i, j) to i + j = s with 1 ≤ i, j ≤ n. In our case, n = 2014. So, for s from 2 to 2015, the number of solutions is s - 1, and for s from 2016 to 4028, it's 4028 - s + 1 = 4029 - s. But since our maximum s is 4028, which is 2014 + 2014.But perhaps instead of splitting into two cases, maybe there's a more symmetric way. However, since we also have the (-1)^i factor, which depends on i, the sum over i and j with i + j = s would be sum_{i=1}^{s-1} (-1)^i, but only for i ≤ 2014 and j = s - i ≤ 2014. Hmm, maybe this complicates things. Let's try to formalize it.For each s, the sum over all i and j such that i + j = s of (-1)^i is equal to sum_{i=max(1, s - 2014)}^{min(2014, s - 1)} (-1)^i. Then, multiply that by (2015 - s)^2 and sum over s. But this might be complex because s ranges over a lot of values, and the inner sum varies depending on s.Alternatively, maybe we can separate the variables i and j. Let's see. The term (2015 - i - j)^2 can be expanded. Let me compute that:(2015 - i - j)^2 = (2015)^2 - 2*2015*(i + j) + (i + j)^2Therefore, a_{i,j} = (-1)^i * [2015² - 2*2015*(i + j) + (i + j)^2]Therefore, the total sum S can be written as sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i * [2015² - 2*2015*(i + j) + (i + j)^2]Since this is a linear combination, we can split the sum into three separate sums:S = 2015² * sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i - 2*2015 * sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i*(i + j) + sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i*(i + j)^2Let me compute each of these three terms separately.First term: T1 = 2015² * sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^iSecond term: T2 = -2*2015 * sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i*(i + j)Third term: T3 = sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i*(i + j)^2Starting with T1. Let's compute sum_{j=1}^{2014} (-1)^i. Since (-1)^i doesn't depend on j, sum_{j=1}^{2014} (-1)^i = 2014*(-1)^i. Therefore, T1 = 2015² * sum_{i=1}^{2014} 2014*(-1)^i = 2015² * 2014 * sum_{i=1}^{2014} (-1)^iNow, sum_{i=1}^{2014} (-1)^i. Since (-1)^i alternates between -1 and 1. Since 2014 is even, the number of terms is even. So, pairing terms:sum_{i=1}^{2014} (-1)^i = (-1 + 1) + (-1 + 1) + ... + (-1 + 1) = 0 (1007 pairs). Therefore, T1 = 2015² * 2014 * 0 = 0.So the first term cancels out. Nice.Now, moving on to the second term, T2 = -2*2015 * sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i*(i + j). Let's split the sum into two parts: sum_{i,j} (-1)^i*i + sum_{i,j} (-1)^i*jTherefore, T2 = -2*2015 * [ sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i*i + sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i*j ]Let's compute these two sums separately. Let's denote S1 = sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i*i and S2 = sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i*jFor S1: sum_{j=1}^{2014} (-1)^i*i doesn't depend on j, so S1 = sum_{i=1}^{2014} [ (-1)^i*i * sum_{j=1}^{2014} 1 ] = sum_{i=1}^{2014} [ (-1)^i*i * 2014 ] = 2014 * sum_{i=1}^{2014} (-1)^i*iSimilarly, S2: sum_{i=1}^{2014} [ (-1)^i * sum_{j=1}^{2014} j ] = sum_{i=1}^{2014} [ (-1)^i * (sum_{j=1}^{2014} j) ] = sum_{i=1}^{2014} [ (-1)^i * (2014*(2014 + 1)/2) ] = (2014*2015/2) * sum_{i=1}^{2014} (-1)^iBut we already saw that sum_{i=1}^{2014} (-1)^i = 0. Therefore, S2 = 0.Therefore, T2 = -2*2015*(S1 + S2) = -2*2015*S1 = -2*2015*(2014 * sum_{i=1}^{2014} (-1)^i*i )So we need to compute sum_{i=1}^{2014} (-1)^i*i. Let's compute this sum.Let’s denote S = sum_{i=1}^{n} (-1)^i*i, where n = 2014.Since n is even, let's group terms in pairs:For i = 1: (-1)^1*1 = -1i = 2: (-1)^2*2 = 2Sum for i=1 and 2: (-1 + 2) = 1Similarly, i=3: -3, i=4: 4, sum: (-3 + 4) = 1Each pair (odd, even) gives 1. Since n is 2014, the number of such pairs is 2014 / 2 = 1007.Therefore, total sum S = 1007 * 1 = 1007.Wait, let me verify with a small n. Let's take n=2: sum is -1 + 2 = 1. n=4: -1 +2 -3 +4 = (-1 +2) + (-3 +4) = 1 +1 = 2. So for n=4, number of pairs 2, sum 2. So yes, each pair contributes 1, so for n even, sum = n/2 * 1 = n/2. Wait, but in the example with n=4, sum is 2 which is 4/2. For n=2, sum is 1 which is 2/2. Wait, so in general, sum_{i=1}^{n} (-1)^i*i = n/2 when n is even.But wait in our case, n=2014. Then sum would be 2014 / 2 = 1007. Yes, so that's correct. Therefore, sum_{i=1}^{2014} (-1)^i*i = 1007.Therefore, S1 = 2014 * 1007.Therefore, T2 = -2*2015*(2014 * 1007 )Compute 2014 * 1007: Let's note that 2014 = 2 * 1007. Therefore, 2014 * 1007 = 2 * 1007 * 1007 = 2 * (1007)^2.Therefore, T2 = -2 * 2015 * 2 * (1007)^2 = -4 * 2015 * (1007)^2.Wait, hold on. Let me check again:T2 = -2*2015*(2014 * 1007 )But 2014 = 2*1007, so 2014*1007 = 2*1007*1007 = 2*(1007)^2Therefore, T2 = -2*2015*2*(1007)^2 = -4*2015*(1007)^2.Okay, so T2 is equal to -4*2015*(1007)^2.Now, moving on to T3 = sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i*(i + j)^2.This looks more complicated. Let's try to expand (i + j)^2:(i + j)^2 = i² + 2ij + j²Therefore, T3 = sum_{i,j} (-1)^i*(i² + 2ij + j²) = sum_{i,j} (-1)^i*i² + 2*sum_{i,j} (-1)^i*ij + sum_{i,j} (-1)^i*j²Let’s split T3 into three parts: T3a = sum_{i,j} (-1)^i*i², T3b = 2*sum_{i,j} (-1)^i*ij, T3c = sum_{i,j} (-1)^i*j²Compute each part.Starting with T3a: sum_{i,j} (-1)^i*i². Since this doesn't depend on j, sum_{j=1}^{2014} (-1)^i*i² = 2014*(-1)^i*i². Therefore, T3a = 2014*sum_{i=1}^{2014} (-1)^i*i².Similarly, T3c: sum_{i,j} (-1)^i*j² = sum_{i=1}^{2014} (-1)^i * sum_{j=1}^{2014} j². Since sum_{j=1}^{2014} j² is a constant with respect to i, so T3c = [sum_{j=1}^{2014} j²] * sum_{i=1}^{2014} (-1)^i. But we already know sum_{i=1}^{2014} (-1)^i = 0. Therefore, T3c = 0.Therefore, T3 = T3a + T3b + T3c = 2014*sum_{i=1}^{2014} (-1)^i*i² + 2*sum_{i,j} (-1)^i*ij + 0So, we need to compute T3a and T3b.First, let's compute T3a: 2014*sum_{i=1}^{2014} (-1)^i*i².Let me denote S3a = sum_{i=1}^{2014} (-1)^i*i². Let's compute S3a.Again, since 2014 is even, we can pair the terms:sum_{i=1}^{2014} (-1)^i*i² = sum_{k=1}^{1007} [ (-1)^{2k-1}*(2k -1)^2 + (-1)^{2k}*(2k)^2 ]Simplify each pair:For each k from 1 to 1007:Term1: (-1)^{2k-1}*(2k -1)^2 = - (2k -1)^2Term2: (-1)^{2k}*(2k)^2 = + (2k)^2So each pair is: - (2k -1)^2 + (2k)^2 = (2k)^2 - (2k -1)^2Compute this difference:(2k)^2 - (2k -1)^2 = [4k²] - [4k² -4k +1] = 4k² -4k² +4k -1 = 4k -1Therefore, each pair contributes (4k -1). Therefore, the total sum S3a = sum_{k=1}^{1007} (4k -1)Compute this sum:sum_{k=1}^{1007} (4k -1) = 4*sum_{k=1}^{1007} k - sum_{k=1}^{1007} 1 = 4*(1007*1008/2) - 1007*1Compute 1007*1008/2 = (1007*504) = Let's compute 1007*500 = 503,500 and 1007*4 = 4,028, so total 503,500 + 4,028 = 507,528. Therefore, 4*507,528 = 2,030,112Then subtract 1007: 2,030,112 - 1007 = 2,029,105Therefore, S3a = 2,029,105Therefore, T3a = 2014 * 2,029,105Wait, let me compute that. 2014 multiplied by 2,029,105.But maybe we can hold off on computing the exact number yet, and see if there is cancellation in T3.Now, compute T3b = 2*sum_{i,j} (-1)^i*ijAgain, let's split the sum:sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i*ij = sum_{i=1}^{2014} [ (-1)^i*i * sum_{j=1}^{2014} j ]Because ij = i * j, and sum over j is sum_{j=1}^{2014} j = (2014)(2015)/2Therefore, sum_{i=1}^{2014} [ (-1)^i*i * (2014*2015/2) ] = (2014*2015/2) * sum_{i=1}^{2014} (-1)^i*iWe already computed sum_{i=1}^{2014} (-1)^i*i = 1007. Therefore, T3b = 2 * [ (2014*2015/2) * 1007 ] = 2014*2015*1007So T3b = 2014*2015*1007Therefore, T3 = T3a + T3b = 2014*2,029,105 + 2014*2015*1007Factor out 2014:T3 = 2014*(2,029,105 + 2015*1007)Let me compute the term inside the parentheses:First compute 2015*1007. Let's compute 2000*1007 = 2,014,000 and 15*1007 = 15,105. So total 2,014,000 +15,105 = 2,029,105.Wait, that's interesting. 2015*1007 = 2,029,105. So the term inside becomes 2,029,105 + 2,029,105 = 2*2,029,105 = 4,058,210.Therefore, T3 = 2014 * 4,058,210But wait, 2014 * 4,058,210. Let me compute that. Let's note that 2014 * 4,058,210. Let's break it down:2014 * 4,000,000 = 8,056,000,0002014 * 58,210 = ?Compute 2014 * 50,000 = 100,700,0002014 * 8,210 = ?Compute 2014 * 8,000 = 16,112,0002014 * 210 = 422, 940Wait:Wait, 2014 * 8,000 = 2014*8*1,000 = 16,112*1,000 = 16,112,0002014 * 210 = 2014*200 + 2014*10 = 402,800 + 20,140 = 422,940Therefore, 2014*8,210 = 16,112,000 + 422,940 = 16,534,940Therefore, 2014 * 58,210 = 100,700,000 + 16,534,940 = 117,234,940Therefore, total T3 = 8,056,000,000 + 117,234,940 = 8,173,234,940Wait, that's 8,173,234,940. But let me check with another method:Alternatively, 2014*4,058,210. Let's note that 4,058,210 * 2000 = 8,116,420,000Then 4,058,210 *14 = 56,815, 940So total 8,116,420,000 +56,815,940 = 8,173,235,940. Wait, but earlier I had 8,173,234,940. There is a discrepancy here. Let's check again.Wait, 4,058,210 * 14:4,058,210 *10 = 40,582,1004,058,210 *4 = 16,232,840Add them: 40,582,100 +16,232,840 = 56,814,940Therefore, total T3 = 8,116,420,000 +56,814,940 =8,173,234,940. So previous calculation was correct. So T3 =8,173,234,940.But let's check if this is necessary. Wait, but perhaps instead of expanding T3, maybe there is a smarter way. Let me see.Wait, but let's recall that T2 was -4*2015*(1007)^2, and T3 is 2014*(2,029,105 + 2015*1007). Wait, but we saw that 2015*1007 = 2,029,105, so T3 became 2014*(2*2,029,105). Then 2*2,029,105 = 4,058,210, so T3 =2014*4,058,210. Which is 2014 multiplied by that number. But 2014 is 2*1007. So T3 = 2*1007*4,058,210 = 1007*8,116,420. Hmm. Not sure if that helps.Alternatively, let's note that T3 = 2014 * 4,058,210 = let's compute 2014 *4,058,210.But maybe this is necessary, but let's note that in the total sum S = T1 + T2 + T3 = 0 + (-4*2015*(1007)^2) + 2014*4,058,210. Let's compute S.But we need to compute S = T2 + T3.So S = -4*2015*(1007)^2 + 2014*4,058,210.Wait, but 2014*4,058,210 = let me check:Compute 2014 * 4,058,210:Let me write 4,058,210 as 4,000,000 + 58,210So, 2014*4,000,000 = 8,056,000,0002014*58,210:Compute 2014*50,000=100,700,0002014*8,210=16,534,940 (as before)Total 100,700,000 +16,534,940=117,234,940Therefore, total 8,056,000,000 +117,234,940=8,173,234,940So T3=8,173,234,940T2= -4*2015*(1007)^2. Let me compute that:First compute (1007)^2. 1007*1007. Let's compute:1000^2 =1,000,0002*1000*7=14,0007^2=49So (1000 +7)^2=1,000,000 +14,000 +49=1,014,049Therefore, (1007)^2=1,014,049Then, 4*2015=8,060Therefore, T2= -8,060*1,014,049Compute 8,060*1,014,049.First compute 8,000*1,014,049=8,112,392,000Then compute 60*1,014,049=60,842, 940Wait, 1,014,049*60=60,842,940Therefore, total T2= - (8,112,392,000 +60,842,940) = -8,173,234,940Therefore, T2= -8,173,234,940Therefore, S= T2 + T3= -8,173,234,940 +8,173,234,940=0Wait, really? So the total sum S=0?But that seems surprising. Let me verify again.We had T3 =8,173,234,940 and T2= -8,173,234,940, so they cancel each other. Therefore, S=0.But is that correct? Let's check again.Wait, let's recap:T2 = -4*2015*(1007)^2T3 =2014*4,058,210But 2015*1007=2,029,105, which is equal to 4,058,210 /2. Wait, 4,058,210=2*2,029,105. And since 2014=2*1007, then T3=2*1007*2*2,029,105=4*1007*2,029,105. Wait, but 2015*1007=2,029,105, so 2,029,105=2015*1007. Therefore, T3=4*1007*(2015*1007)=4*2015*(1007)^2. Therefore, T3=4*2015*(1007)^2. Therefore, T3 = -T2, so indeed T3 + T2=0.Therefore, the total sum S=0+0=0.Wait, but that's really unexpected. The entire sum cancels out? Let me check again.Given that T1=0, T2= -4*2015*(1007)^2, T3=4*2015*(1007)^2, so T2 + T3=0, hence S=0.But maybe there's an error in the calculation. Let's check T3.From earlier steps, T3 = sum (-1)^i*(i + j)^2 = T3a + T3b =2014*S3a +2*S3b.We had S3a = sum (-1)^i*i² =2,029,105. Then T3a=2014*2,029,105. Then T3b=2014*2015*1007=2014*2,029,105. Therefore, T3=2014*2,029,105 +2*2014*2,029,105= (1 +2)*2014*2,029,105=3*2014*2,029,105. Wait, no, that can't be. Wait, wait:Wait, T3a=2014*2,029,105T3b=2014*2015*1007=2014*(2015*1007)=2014*2,029,105Therefore, T3=2014*2,029,105 + 2*(2014*2,029,105)= (1 +2)*2014*2,029,105=3*2014*2,029,105. Wait, this contradicts previous conclusion. Wait, where is the error?Wait, original T3a and T3b:T3a =2014*sum (-1)^i*i²=2014*S3a=2014*2,029,105T3b=2*sum (-1)^i*ij=2*(2014*2015/2)*sum (-1)^i*i=2*(2014*2015/2)*1007=2014*2015*1007=2014*(2015*1007)=2014*2,029,105Therefore, T3 = T3a + T3b=2014*2,029,105 +2014*2,029,105=2*2014*2,029,105Which is equal to 2*2014*2,029,105. However, 2015*1007=2,029,105, so 2*2014*2,029,105=2*2014*2015*1007=2*2015*1007*2014. But 2014=2*1007. Therefore, 2*2015*1007*2*1007=4*2015*(1007)^2. Therefore, T3=4*2015*(1007)^2.Which is exactly equal to -T2=4*2015*(1007)^2. Therefore, T2= -4*2015*(1007)^2, T3=4*2015*(1007)^2. So T2 + T3=0. Therefore, S=0.Therefore, the total sum of all numbers in the table is zero. That seems surprising, but according to the calculations, the terms cancel each other out. Let me just verify with a smaller table to see if this makes sense.Let's take a smaller case, say 2x2 table. Compute the sum and see if it's zero.For a 2x2 table, rows i=1,2 and columns j=1,2.Compute each cell:i=1:a_{1,1}=(-1)^1*(2015 -1 -1)^2= -1*(2013)^2a_{1,2}=(-1)^1*(2015 -1 -2)^2= -1*(2012)^2i=2:a_{2,1}=(-1)^2*(2015 -2 -1)^2=1*(2012)^2a_{2,2}=(-1)^2*(2015 -2 -2)^2=1*(2011)^2So sum:-2013² -2012² +2012² +2011² = (-2013²) + ( -2012² +2012² ) +2011²= -2013² +0 +2011²= (2011² -2013²)= (2011 -2013)(2011 +2013)= (-2)(4024)= -8048 ≠0Wait, but according to our formula, S should be zero. But in this 2x2 case, it's not zero. Therefore, our conclusion for the general case must be wrong. Hmm. So something is wrong here.This inconsistency suggests that there's a mistake in the earlier reasoning. Therefore, our approach must have an error.Wait, in the 2x2 case, the sum is -2013² -2012² +2012² +2011²= -2013² +2011²= (2011 -2013)(2011 +2013)= (-2)(4024)= -8048. But according to our general formula, S=0. But in reality, it's not zero, so the general formula must be incorrect.Therefore, there must be a mistake in the previous steps.Let me check again.In the general case, we broke down the sum into T1, T2, T3, and found that T1=0, T2 and T3 cancel each other. But in the 2x2 case, this is not true, so the error must be in how we computed T3.Wait, let's recalculate T3 for the 2x2 case using the formula.For the 2x2 case:T1: 2015² * sum_{i=1}^2 sum_{j=1}^2 (-1)^i. sum_{j=1}^2 (-1)^i = 2*(-1)^i. sum_{i=1}^2 2*(-1)^i=2*(-1 +1)=0. So T1=0. Correct.T2= -2*2015*(sum_{i,j} (-1)^i*(i + j)). Compute sum_{i,j} (-1)^i*(i + j).sum_{i=1}^2 sum_{j=1}^2 (-1)^i*(i + j).For i=1: (-1)^1*(1 +1) + (-1)^1*(1 +2)= -2 + -3= -5For i=2: (-1)^2*(2 +1) + (-1)^2*(2 +2)= 3 +4=7Total sum= -5 +7=2Therefore, T2= -2*2015*2= -8060T3= sum_{i,j} (-1)^i*(i + j)^2.Compute each term:For i=1:j=1: (-1)^1*(1+1)^2= -4j=2: (-1)^1*(1+2)^2= -9For i=2:j=1: (-1)^2*(2+1)^2=9j=2: (-1)^2*(2+2)^2=16Total T3= -4 -9 +9 +16=12Therefore, total sum S= T1 + T2 + T3= 0 -8060 +12= -8048, which matches the manual calculation. But according to our general formula, T3=4*2015*(1007)^2 when n=2014, but in the case n=2, substituting n=2:Wait, substituting n=2 into our general formula:T2= -4*2015*(1007)^2. But n=2, so 1007 should be n/2=1. So T2= -4*2015*(1)^2= -8060. Which matches.T3=4*2015*(1)^2=8060. But in reality, T3=12, not 8060. Therefore, our general formula for T3 is incorrect when n=2. Therefore, the error must be in how we generalized the formula.Wait, where did we get T3=4*2015*(1007)^2?In the case of n=2014, we had T3=4*2015*(1007)^2, but in n=2, this would imply T3=4*2015*(1)^2=8060, but actual T3=12. So clearly, the formula is not generalizable. Therefore, our approach must have a mistake.Looking back, when we expanded (i + j)^2 and split into three sums, perhaps there was an error in the calculation of S3a and S3b.Wait, for the 2x2 case:T3a=2014*sum_{i=1}^2 (-1)^i*i². But n=2, so 2014 should be 2. Wait, in the general case, T3a= n * sum_{i=1}^n (-1)^i*i². For n=2, T3a=2*sum_{i=1}^2 (-1)^i*i²=2*(-1^1*1 + (-1)^2*4)=2*(-1 +4)=2*3=6.T3b=2*sum_{i,j} (-1)^i*ij. For n=2, sum_{i,j} (-1)^i*ij= sum_{i=1}^2 (-1)^i*i*sum_{j=1}^2 j. sum_{j=1}^2 j=3. Therefore, sum_{i=1}^2 (-1)^i*i*3=3*(-1*1 +1*2)=3*(-1 +2)=3*1=3. Therefore, T3b=2*3=6.Therefore, T3= T3a + T3b=6 +6=12, which matches the actual T3. Therefore, in the n=2 case, our formula works. But according to our general calculation for n=2014, we said that T3=4*2015*(1007)^2. However, in the n=2 case, T3=12. Let's check with the formula: 4*2015*(n/2)^2. But n=2, so 4*2015*(1)^2=8060, which does not equal 12. Therefore, the mistake is that in the general case, we expressed T3 as 4*2015*(1007)^2, but actually, that is not the case.Wait, let's re-examine the steps for T3. For the general case:T3=2014*S3a +2*S3b, where S3a=sum_{i=1}^{2014} (-1)^i*i²=2,029,105 and S3b=2014*2015*1007/2=2014*2015*1007/2.Wait, no. Wait, in the n=2 case, S3a=sum_{i=1}^2 (-1)^i*i²= -1 +4=3. Then T3a=2*3=6.In the general case, S3a=sum_{i=1}^{n} (-1)^i*i². For n even, this sum is equal to n(n+1)/2. Wait, no. For n=2, sum is -1 +4=3. For n=4, sum is -1 +4 -9 +16=10. Hmm, which is different.Wait, in the case of n=2014, we calculated S3a=2,029,105. Which is equal to (4*1007 -1)*1007= (4028 -1)*1007=4027*1007. Wait, let's verify 4027*1007. 4000*1007=4,028,000, 27*1007=27,189. So total 4,028,000 +27,189=4,055,189. But 2,029,105*2=4,058,210. Hmm, no. So maybe my earlier calculation was wrong. Wait, no. When we did the pairing for S3a=sum_{i=1}^{2014} (-1)^i*i², we said each pair contributes (4k -1), and there are 1007 terms. Therefore, S3a=sum_{k=1}^{1007} (4k -1)=4*sum_{k=1}^{1007}k - sum_{k=1}^{1007}1=4*(1007*1008)/2 -1007=2*1007*1008 -1007=1007*(2016 -1)=1007*2015=2,029,105. Yes, that's correct. So S3a=1007*2015=2,029,105.Therefore, T3a=2014*S3a=2014*1007*2015. Since 2014*1007=2014*1007=2*1007*1007=2*(1007)^2. Wait, 2014=2*1007. Therefore, 2014*1007=2*1007*1007. Therefore, T3a=2*1007*1007*2015.Then T3b=2014*2015*1007=2*1007*2015*1007= same as above.Wait, T3a=2*1007*2015*1007T3b=2*1007*2015*1007Therefore, T3= T3a + T3b=4*1007*2015*1007=4*2015*(1007)^2. Therefore, T3=4*2015*(1007)^2. But in the n=2 case, according to this formula, substituting n=2:1007=n/2=1, 2015=2015. So T3=4*2015*(1)^2=8060. But in reality, for n=2, T3=12. So clearly, this formula is not applicable for n=2. Therefore, the error is in the assumption that T3=4*2015*(n/2)^2 for any n. But in reality, that's derived specifically for n=2014.But in the n=2 case, T3=12=4*3*1^2, where 3=2015 - i - j when n=2. Wait, no. Wait, 2015 in the problem is fixed. Wait, hold on. Wait, in the original problem, 2015 is part of the formula a_{i,j}=(-1)^i*(2015 -i -j)^2, regardless of the size of the table. Wait, but in the n=2 case, the table is 2x2, but the formula still uses 2015. So in the n=2 case, the numbers are like (2015 -i -j)^2, even though i and j are up to 2, so 2015 -i -j can be as large as 2015 -1 -1=2013. Therefore, the numbers can be very large even for small tables, leading to large sums. Therefore, in the general problem, the size of the table is 2014x2014, but the formula uses 2015 -i -j. So the n in the problem is 2014, but the formula uses 2015. Therefore, in the original problem, even if we check with n=2014, but in the 2x2 example, even though the table is smaller, the formula still references 2015. Therefore, the 2x2 example is not a scaled-down version of the original problem, because the 2015 is fixed. Therefore, perhaps my test case is not appropriate. Let me instead create a different test case where the formula is scaled down.Suppose instead the problem were: a square table with n rows and columns, each cell a_{i,j}=(-1)^i*(C -i -j)^2, where C is some constant. Then, perhaps if we set C =n +1, so that the terms become (n +1 -i -j)^2. Then, in this case, when n=2, C=3, and the cells would be similar. Let me compute that.Test case: n=2, C=3. So a_{i,j}=(-1)^i*(3 -i -j)^2.Compute all cells:i=1:j=1: (-1)^1*(3 -1 -1)^2= -1*(1)^2= -1j=2: (-1)^1*(3 -1 -2)^2= -1*(0)^2=0i=2:j=1: (-1)^2*(3 -2 -1)^2=1*(0)^2=0j=2: (-1)^2*(3 -2 -2)^2=1*(-1)^2=1Total sum: -1 +0 +0 +1=0. So in this case, the sum is zero.Therefore, when C =n +1, the sum is zero for n=2.Similarly, let's check n=1. But n=1 is trivial.Another test case: n=3, C=4.i ranges 1-3, j ranges 1-3.Compute a_{i,j}=(-1)^i*(4 -i -j)^2.Compute each cell:i=1:j=1: -1*(4 -1 -1)^2= -1*(2)^2= -4j=2: -1*(4 -1 -2)^2= -1*(1)^2= -1j=3: -1*(4 -1 -3)^2= -1*(0)^2=0i=2:j=1:1*(4 -2 -1)^2=1*(1)^2=1j=2:1*(4 -2 -2)^2=1*(0)^2=0j=3:1*(4 -2 -3)^2=1*(-1)^2=1i=3:j=1: -1*(4 -3 -1)^2= -1*(0)^2=0j=2: -1*(4 -3 -2)^2= -1*(-1)^2= -1j=3: -1*(4 -3 -3)^2= -1*(-2)^2= -4Sum all cells:-4 -1 +0 +1 +0 +1 +0 -1 -4= (-4 -1 -4) + (-1 +1) + (0 +0 +0)= -9 +0 +0= -9 ≠0Hmm, so in this case, the sum is not zero. But according to the formula, if C= n +1=4, sum might not be zero. Wait, maybe my previous assumption is wrong.Wait, in the original problem, C=2015, and n=2014. So C= n +1. For n=2, C=3. In that case, the sum was zero. For n=3, C=4, sum was -9.Wait, inconsistency. Wait, why is the sum zero for n=2 but not for n=3. Maybe the parity of n matters?Wait, n=2 is even, n=3 is odd. Maybe when n is even, the sum is zero, and when n is odd, it's not.Wait, let's check n=1, C=2. Table 1x1:a_{1,1}= (-1)^1*(2 -1 -1)^2= -1*0=0. Sum=0.n=1, sum=0.n=3, sum=-9. Let's try n=4, C=5.But this might take too long. Alternatively, maybe there's a pattern when n is even.Wait, in the original problem, n=2014 (even), and according to the previous calculation, sum=0, which matched the n=2 case. For n=3 (odd), the sum was -9. Let me check n=4 to see.Test case n=4, C=5.i=1 to 4, j=1 to4.Compute a_{i,j}=(-1)^i*(5 -i -j)^2.This would be tedious, but let's try to find a pattern.Alternatively, notice that when C=n +1 and n is even, maybe the sum is zero, but when n is odd, it's not. But for n=3, C=4, sum was -9. For n=1, C=2, sum=0. Hmm, not sure.But in the original problem, n=2014 (even), and we derived sum=0. But when we took n=2 (even), sum=0. So perhaps when n is even, the sum is zero. For odd n, sum might not be. Therefore, the answer for the original problem is zero.But the test case with n=3 gave sum=-9. However, in the original problem, with n=2014 even, sum=0. Therefore, perhaps the answer is indeed zero.But in the initial 2x2 example with the original formula (using 2015), the sum was -8048, not zero, because the formula uses 2015 regardless of n. But in the modified 2x2 example where C=n +1=3, the sum was zero. So the difference is that in the original problem, C=2015, which is n +1 when n=2014. Therefore, in the original problem, C=2015=2014 +1, so it's following the C=n +1 pattern. Therefore, in that case, the sum is zero. However, when I took n=2 but kept C=2015, then it's not C=n +1, so the sum is not zero. But in the modified test case where C=n +1=3, the sum was zero. Therefore, when C=n +1 and n is even, the sum is zero. Therefore, in the original problem, since C=2015=2014 +1, and n=2014 is even, the sum is zero. Therefore, the answer is zero.But the initial test case with n=2 and C=3 (i.e., n +1) gave sum zero. So it seems the formula holds. Therefore, the answer for the original problem is zero.Therefore, despite the initial miscalculation with an irrelevant test case (using C=2015 for n=2), the correct answer is zero.Final Answerboxed{0}</think>Given a square table with 2014 rows and 2014 columns, where each cell at position ((i, j)) contains the number (a_{i,j} = (-1)^i (2015 - i - j)^2), we need to find the sum of all the numbers in the table.First, we express the total sum (S) as a double sum:[S = sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i (2015 - i - j)^2]We expand ((2015 - i - j)^2) as follows:[(2015 - i - j)^2 = 2015^2 - 2 cdot 2015 (i + j) + (i + j)^2]This allows us to split the sum into three parts:[S = 2015^2 sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i - 2 cdot 2015 sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i (i + j) + sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i (i + j)^2]1. First Term: [ 2015^2 sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i ] Since ((-1)^i) alternates between -1 and 1, and there are 2014 terms (even number), the sum over (i) is zero. Therefore, the first term is zero.2. Second Term: [ -2 cdot 2015 sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i (i + j) ] This can be split into two sums: [ sum_{i=1}^{2014} (-1)^i i sum_{j=1}^{2014} 1 + sum_{i=1}^{2014} (-1)^i sum_{j=1}^{2014} j ] The second part of this sum is zero because (sum_{i=1}^{2014} (-1)^i = 0). The first part involves (sum_{i=1}^{2014} (-1)^i i), which for even (2014) is (1007). Thus, the second term simplifies to (-4 cdot 2015 cdot 1007^2).3. Third Term: [ sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i (i + j)^2 ] Expanding ((i + j)^2) and splitting into three sums, we find that the terms involving (j^2) sum to zero. The remaining terms involve (i^2) and (ij). After detailed calculations, we find that this term simplifies to (4 cdot 2015 cdot 1007^2).Combining the second and third terms, we see they cancel each other out:[-4 cdot 2015 cdot 1007^2 + 4 cdot 2015 cdot 1007^2 = 0]Thus, the total sum of all the numbers in the table is:[boxed{0}],<think>Okay, so I need to find the sum of all the numbers in a 2014x2014 square table where each cell at position (i, j) has the number a_{i,j} = (-1)^i * (2015 - i - j)^2. Hmm, let's see. First, let me make sure I understand the problem correctly. The table has 2014 rows and 2014 columns, so i and j both range from 1 to 2014. Each entry is computed by taking (-1) raised to the power of the row number i, multiplied by (2015 - i - j) squared. Then, I need to sum all these entries. Alright, so the total sum S would be the double sum over all rows i and columns j:S = Σ_{i=1}^{2014} Σ_{j=1}^{2014} (-1)^i * (2015 - i - j)^2Hmm. That's a lot of terms. Since it's a double summation, maybe there's a way to simplify this by changing variables or recognizing a pattern. Let's see.First, notice that (2015 - i - j) is the same as (2015 - (i + j)). Let me denote k = i + j. Then, (2015 - k) is the term inside the square. But k would range from 2 (when i=1, j=1) to 4028 (when i=2014, j=2014). But since we have i and j each up to 2014, the maximum i + j is 2014 + 2014 = 4028, and the minimum is 2. But 2015 - k would then range from 2013 down to -2013. Wait, but squaring it makes it positive, so maybe that's okay.But changing variables from (i, j) to k and something else might complicate things. Alternatively, maybe fix the value of k = i + j and see how many times each (2015 - k)^2 appears in the sum, multiplied by (-1)^i. Wait, but for each k, how many pairs (i, j) have i + j = k? That's a standard combinatorial problem. For a given k, the number of pairs (i, j) with i ≥ 1, j ≥ 1, and i + j = k is (k - 1). But here, since i and j are both bounded by 2014, for k from 2 to 2015, the number of pairs is (k - 1), but for k from 2015 to 4028, the number of pairs decreases. Specifically, for k ≤ 2015, the number of pairs is (k - 1), and for k > 2015, the number of pairs is (4028 - k + 1) = (4029 - k). Wait, is that right?Wait, let's think again. For i and j each from 1 to n (here, n = 2014), the number of pairs (i, j) with i + j = k is:- If k is between 2 and n + 1 (i.e., k ≤ 2015), then the number is k - 1.- If k is between n + 2 and 2n (i.e., 2016 ≤ k ≤ 4028), then the number is 2n - k + 1 = 4028 - k + 1 = 4029 - k.So in general, the number of pairs (i, j) with i + j = k is:N(k) = {k - 1, if 2 ≤ k ≤ 2015,4029 - k, if 2016 ≤ k ≤ 4028}But in our case, n = 2014, so n + 1 = 2015, and 2n = 4028. So that works out.Therefore, if we can express the original sum in terms of k, then for each k from 2 to 4028, we have N(k) terms of (2015 - k)^2 multiplied by (-1)^i. But here's the problem: (-1)^i depends on i, not just k. So even if i + j = k, the sign alternates depending on i. So we can't directly factor that out unless we can relate it to k.Alternatively, maybe express (-1)^i in terms of (-1)^k and (-1)^j. Since i + j = k, then (-1)^i = (-1)^{k - j} = (-1)^k * (-1)^{-j} = (-1)^k * (-1)^j, since (-1)^{-j} = [(-1)^{-1}]^j = (-1)^j. Therefore, (-1)^i = (-1)^k * (-1)^j.Therefore, the sum S can be rewritten as:S = Σ_{k=2}^{4028} Σ_{(i,j) such that i + j = k} [(-1)^i * (2015 - k)^2]But since (2015 - k)^2 is constant for a given k, and (-1)^i = (-1)^k * (-1)^j, then:S = Σ_{k=2}^{4028} (2015 - k)^2 * (-1)^k * Σ_{j=1}^{N(k)} (-1)^jWait, because for each k, the pairs (i, j) with i + j = k correspond to j ranging from 1 to k - 1 (if k ≤ 2015), but actually, when i ranges from 1 to k - 1, j = k - i. But since i and j must be ≤ 2014, the exact range depends on k. However, the previous expression is using that for each k, we can write j as varying, and correspondingly, since i = k - j, the term (-1)^i = (-1)^{k - j} = (-1)^k (-1)^{-j} = (-1)^k (-1)^j. Therefore, for each k, we can factor out (-1)^k, and then sum over j (with appropriate limits) the term (-1)^j.But the number of terms N(k) is as we had before, and for each k, the sum over j (with i = k - j) of (-1)^j, but j is determined by i. Wait, maybe another approach.Alternatively, fix k = i + j, then for each k, the possible i's go from max(1, k - 2014) to min(2014, k - 1). Then, for each k, the terms in the inner sum would be:Σ_{i = max(1, k - 2014)}^{min(2014, k - 1)} (-1)^i * (2015 - k)^2But (2015 - k)^2 is constant with respect to i, so we can factor that out:(2015 - k)^2 * Σ_{i = max(1, k - 2014)}^{min(2014, k - 1)} (-1)^iTherefore, the total sum S is:Σ_{k=2}^{4028} (2015 - k)^2 * [Σ_{i = a}^{b} (-1)^i]Where a = max(1, k - 2014) and b = min(2014, k - 1).Therefore, the key is to compute the inner sum Σ_{i = a}^{b} (-1)^i for each k, multiply it by (2015 - k)^2, and then sum over all k from 2 to 4028.So now, let's analyze the inner sum. Let's denote:Sum_{i = a}^{b} (-1)^iThis is a finite geometric series with ratio -1. The formula for the sum of a geometric series from m to n is:Σ_{k=m}^n r^k = (r^{n+1} - r^m)/(r - 1)Here, r = -1, so:Σ_{i=a}^b (-1)^i = [ (-1)^{b + 1} - (-1)^a ] / (-1 - 1) = [ (-1)^{b + 1} - (-1)^a ] / (-2)Simplify:= [ (-1)^a - (-1)^{b + 1} ] / 2Alternatively, factor out (-1)^a:= (-1)^a [1 - (-1)^{b - a + 1}] / 2But maybe better to leave it as:[ (-1)^a - (-1)^{b + 1} ] / 2But let's see. The sum can be either 0 or ±1 depending on the parity of the number of terms. Wait, since the terms alternate between -1 and 1. If the number of terms is even, they cancel out to 0; if it's odd, the sum is either 1 or -1 depending on the starting term.Wait, actually, the sum from i = a to b of (-1)^i is:If a and b have the same parity, then the number of terms is (b - a)/2 + 1. If there are an even number of terms, the sum is 0; if odd, then (-1)^a.Wait, no. Let's take an example. Suppose a is even and b is even. Then each pair of terms is (-1)^{even} + (-1)^{even + 1} = 1 - 1 = 0. So if the number of terms is even, the sum is 0. If odd, there's a leftover term. Wait, but actually, let's be precise.Suppose we start at i = a and end at i = b. The number of terms is N = b - a + 1.If N is even, then we can pair them up: (a, a+1), (a+2, a+3), ..., (b-1, b). Each pair sums to (-1)^a + (-1)^{a+1} = (-1)^a (1 - 1) = 0. So the total sum is 0.If N is odd, then there are (N - 1)/2 pairs summing to 0, and the last term is (-1)^b. So the total sum is (-1)^a * (N mod 2). Wait, no. Wait, let me take specific examples.Case 1: a = 1, b = 1. Sum = (-1)^1 = -1. Number of terms = 1 (odd).Case 2: a = 1, b = 2. Sum = -1 + 1 = 0. Number of terms = 2 (even).Case 3: a = 1, b = 3. Sum = -1 + 1 -1 = -1. Number of terms = 3 (odd).Case 4: a = 2, b = 4. Sum = 1 -1 +1 = 1. Number of terms = 3 (odd).So in general, if the number of terms N is even, sum is 0. If N is odd, sum is (-1)^a. Wait, in Case 3, a=1 (odd), N=3 (odd), sum is -1 = (-1)^1.In Case 4, a=2 (even), N=3 (odd), sum is 1 = (-1)^2.Wait, yes, if N is odd, the sum is (-1)^a. If N is even, sum is 0. Wait, is that correct?Wait, let's check.If N is the number of terms, which is b - a + 1. If N is even, then the sum is 0. If N is odd, then the sum is (-1)^a. Let me verify:Take a = 1, b = 3. N = 3, which is odd. Sum is (-1) + 1 + (-1) = -1. Which is (-1)^1 = -1. Correct.Take a = 2, b = 4. N = 3. Sum is 1 + (-1) + 1 = 1. Which is (-1)^2 = 1. Correct.Take a = 1, b = 4. N = 4. Sum is (-1) + 1 + (-1) + 1 = 0. Correct.Take a = 2, b = 5. N = 4. Sum is 1 + (-1) + 1 + (-1) = 0. Correct.Okay, so indeed, if the number of terms is even, sum is 0. If odd, sum is (-1)^a.So in our case, for each k, the inner sum Σ_{i = a}^b (-1)^i is equal to:If (b - a + 1) is even: 0If (b - a + 1) is odd: (-1)^aTherefore, to compute the inner sum, we need to determine whether the number of terms (i.e., the number of valid i's for each k) is even or odd.But first, let's recall that a = max(1, k - 2014) and b = min(2014, k - 1). So, for each k:- When 2 ≤ k ≤ 2015: i ranges from 1 to k - 1 (since k - 2014 ≤ 0, so a = 1, and b = k - 1). Therefore, the number of terms is (k - 1) - 1 + 1 = k - 1. Wait, wait. Wait, if a = 1 and b = k - 1, then the number of terms is (k - 1 - 1) + 1 = k - 1. Wait, no, number of terms from a to b inclusive is b - a + 1. So for a=1, b=k-1: number of terms is (k - 1) - 1 + 1 = k - 1. Wait, yes. For example, k=2: a=1, b=1, number of terms=1. k=3: a=1, b=2, number of terms=2. Correct.Similarly, when 2016 ≤ k ≤ 4028: a = k - 2014 (since k - 2014 ≥ 2, as k ≥ 2016), and b = 2014. So the number of terms is 2014 - (k - 2014) + 1 = 2014 - k + 2014 + 1 = 4029 - k.Therefore, the number of terms for each k is:- For 2 ≤ k ≤ 2015: N(k) = k - 1- For 2016 ≤ k ≤ 4028: N(k) = 4029 - kSo, the inner sum Σ_{i=a}^b (-1)^i is:If N(k) is even: 0If N(k) is odd: (-1)^aSo we need to check for each k whether N(k) is even or odd.First, let's consider k from 2 to 2015:For 2 ≤ k ≤ 2015:N(k) = k - 1. So when is k - 1 even or odd? That's equivalent to k being odd or even. So:If k is even, then N(k) = even - 1 = odd. Wait, k is even: k - 1 is odd. If k is odd, k - 1 is even. Therefore, for k from 2 to 2015:- If k is even (i.e., k = 2m), then N(k) = 2m - 1, which is odd. Therefore, sum is (-1)^a = (-1)^1 = -1 (since a = 1 for these k).- If k is odd (i.e., k = 2m + 1), then N(k) = 2m, which is even. Therefore, sum is 0.Wait, but hold on. For k from 2 to 2015, a is always 1. So when N(k) is odd (which is when k is even), the sum is (-1)^1 = -1. When N(k) is even (k odd), the sum is 0. So for k even: sum is -1; for k odd: sum is 0.Similarly, for k from 2016 to 4028:N(k) = 4029 - k. So when is 4029 - k even or odd? 4029 - k is even when k is odd, and odd when k is even.So:If k is even (2016 ≤ k ≤ 4028), then N(k) = 4029 - k is odd. Therefore, sum is (-1)^a, but a here is k - 2014. So a = k - 2014. Since k is even (as per the case), a is even - 2014. Since 2014 is even, even - even = even. Therefore, (-1)^a = 1.If k is odd (2016 ≤ k ≤ 4028), then N(k) = 4029 - k is even. Therefore, sum is 0.Therefore, for k from 2016 to 4028:- If k is even, sum is 1- If k is odd, sum is 0Putting this all together, the total sum S can be split into two parts:S = Σ_{k=2}^{2015} (2015 - k)^2 * [ -1 if k even else 0 ] + Σ_{k=2016}^{4028} (2015 - k)^2 * [ 1 if k even else 0 ]But let's make a substitution in the second sum. Let me denote m = 2015 - k. When k = 2016, m = 2015 - 2016 = -1. When k = 4028, m = 2015 - 4028 = -2013. So the second sum can be rewritten in terms of m from -1 to -2013. But squaring m makes it positive, so (2015 - k)^2 = m^2.But maybe instead, let's reverse the index. Let’s let m = 4029 - k (since k ranges from 2016 to 4028, m ranges from 1 to 2013). Then 2015 - k = 2015 - (4029 - m) = 2015 - 4029 + m = m - 2014. So (2015 - k)^2 = (m - 2014)^2. But maybe that complicates things. Alternatively, note that (2015 - k)^2 is the same as (k - 2015)^2, so it's symmetric around k = 2015. So the second sum is from k=2016 to 4028 of (k - 2015)^2 multiplied by 1 if k is even. Similarly, the first sum is from k=2 to 2015 of (2015 - k)^2 multiplied by -1 if k is even.But notice that if we let m = 2015 - k in the first sum, when k ranges from 2 to 2015, m ranges from 2013 down to 0. So (2015 - k) = m. So the first sum becomes Σ_{m=0}^{2013} m^2 * [ -1 if (2015 - m) is even else 0 ]Similarly, for the second sum, let m = k - 2015, which ranges from 1 to 2013 as k goes from 2016 to 4028. Then (2015 - k)^2 = m^2. Therefore, the second sum becomes Σ_{m=1}^{2013} m^2 * [ 1 if (2015 + m) is even else 0 ]But note that (2015 - m) in the first sum corresponds to k = 2015 - m, so whether k is even is equivalent to whether 2015 - m is even. Since 2015 is odd, 2015 - m is even if and only if m is odd. Similarly, in the second sum, k = 2015 + m, and 2015 is odd, so 2015 + m is even if and only if m is odd.Therefore, the first sum is -Σ_{m=1,3,5,...,2013} m^2 (since only odd m contribute, as m must be odd for 2015 - m to be even, i.e., k even). Similarly, the second sum is Σ_{m=1,3,5,...,2013} m^2 (since m must be odd for 2015 + m to be even). Therefore, the first sum is negative the sum over odd m from 1 to 2013 of m^2, and the second sum is positive the same sum. Therefore, they cancel each other out!Wait, hold on. Let's verify that.First sum: when k ranges from 2 to 2015. If k is even, then m = 2015 - k must be odd (since 2015 is odd, odd minus even is odd). Therefore, m is odd, so the first sum becomes -Σ_{m odd, m=1}^{2013} m^2.Second sum: when k ranges from 2016 to 4028, and k is even, then m = k - 2015 must be odd (since k is even and 2015 is odd, even minus odd is odd). Therefore, m is odd, so the second sum becomes Σ_{m odd, m=1}^{2013} m^2.Therefore, S = (-Σ_{m odd} m^2) + (Σ_{m odd} m^2) = 0.Wait, so does that mean the total sum S is zero? But that seems surprising. Let me check my reasoning again.First sum: For k even from 2 to 2015, m = 2015 - k is odd (since 2015 is odd and k is even), so m ranges from 2013 (when k=2) down to 1 (when k=2014). Wait, but k=2014 is even, m=2015 - 2014=1. So m takes odd values from 2013 to 1. Therefore, the first sum is -Σ_{m odd, m=1}^{2013} m^2.Second sum: For k even from 2016 to 4028, m = k - 2015 is odd (since k is even, 2015 is odd, so even - odd = odd). m ranges from 1 (when k=2016) up to 2013 (when k=4028). So the second sum is Σ_{m odd, m=1}^{2013} m^2.Thus, S = (-Σ_{m odd} m^2) + (Σ_{m odd} m^2) = 0. Therefore, the total sum is zero? That's unexpected. But let's check with a smaller example.Let me take n=2 (2x2 table) and see if the sum is zero.Wait, original problem is for 2014x2014 table, but let's test n=2. For n=2, rows and columns are 1 to 2.Compute each entry:a_{1,1} = (-1)^1*(2015 -1 -1)^2 = -1*(2013)^2a_{1,2} = -1*(2015 -1 -2)^2 = -1*(2012)^2a_{2,1} = (-1)^2*(2015 -2 -1)^2 = 1*(2012)^2a_{2,2} = 1*(2015 -2 -2)^2 = 1*(2011)^2Sum: -2013² -2012² +2012² +2011² = (-2013² + 2011²) + (-2012² +2012²) = (-2013² +2011²)Compute that: (2011 -2013)(2011 +2013) = (-2)(4024) = -8048So sum is -8048, which is not zero. Hmm, so my previous conclusion that the sum is zero is contradicted by this small example. Therefore, my reasoning must be flawed.Wait, so what's wrong here?In the n=2 case, following the same steps: the first sum is for k from 2 to 2015 (which in the original problem is up to 2015, but here with n=2, we need to adjust. Wait, actually, in the original problem, n=2014, so the k ranges from 2 to 4028. But in the n=2 case, k ranges from 2 to 4.But my mistake was that in the smaller example, when I set n=2, the value inside the square is 2015 - i - j, which for n=2 would be 2015 - i - j, but that's a very large number even for small i and j. That seems to be an issue because in the original problem, the numbers inside are (2015 - i -j)^2, but when n=2, that term is still 2015 - i - j, which is not related to n. So perhaps testing with n=2 isn't a good idea because the numbers are not scaled down.Wait, perhaps the problem is that when n=2014, the value 2015 - i -j can be positive or negative, but when squared, it's always positive. However, when n is smaller, like n=2, 2015 - i -j is still a large positive number. Therefore, testing with n=2 is not helpful because the actual numbers are different.Alternatively, maybe create a similar problem with smaller numbers. Let's say instead of 2015, use a smaller number. Let's adjust the problem to be a 2x2 table with numbers a_{i,j} = (-1)^i*(5 - i - j)^2. Then 5 is like the original 2015 (since n=2, 5 = 2 + 2 +1). Let's compute this.Cells:a_{1,1} = (-1)^1*(5 -1 -1)^2 = -1*(3)^2 = -9a_{1,2} = -1*(5 -1 -2)^2 = -1*(2)^2 = -4a_{2,1} = (-1)^2*(5 -2 -1)^2 = 1*(2)^2 = 4a_{2,2} = 1*(5 -2 -2)^2 = 1*(1)^2 = 1Sum: -9 -4 +4 +1 = (-9 -4) + (4 +1) = -13 +5 = -8If we apply the same reasoning as before, would we get zero?For this adjusted problem, let's see. Compute S = Σ_{k=2}^{4} (5 -k)^2 * [sum factors]For k=2 to 4:For k=2: a=1, b=1 (i=1, j=1). Sum of (-1)^i = -1. So term: (5 -2)^2 * (-1) = 9*(-1) = -9For k=3: a=1, b=2. Number of terms=2 (even), sum=0. So term: (5 -3)^2 *0 =0For k=4: a=4 -2=2, b=2. Number of terms=1 (odd). Sum=(-1)^2=1. So term: (5 -4)^2 *1=1*1=1Wait, but in the actual sum, the total is -9 +0 +1 = -8, which matches the manual calculation. So according to the original approach, the sum is -9 +0 +1 = -8. But according to the previous reasoning, where we thought the sum cancels out, it's not. So where is the mistake?Wait, in this adjusted example, k ranges from 2 to 4. For k=2 (even): sum is -1. For k=3 (odd): sum is 0. For k=4 (even): sum is 1. So the total sum is (5 -2)^2*(-1) + (5 -4)^2*(1) = 9*(-1) +1*1= -9 +1= -8. Correct. But according to the reasoning for the original problem, splitting into two parts:First sum for k=2 to n+1=3 (since n=2, k=2 to 3). Wait, but in the adjusted problem, n=2, so original k would go up to 4. Wait, perhaps my previous substitution is getting confused with different n.Wait, maybe the key is that in the original problem, the split is at k=2015 (n +1). For n=2, split at k=3. Then:First sum: k=2 to 3 (n=2). For k=2: N(k)=1 (odd), sum=-1. For k=3: N(k)=2 (even), sum=0. So first sum: (5 -2)^2*(-1) + (5 -3)^2*0 =9*(-1) +4*0= -9.Second sum: k=4 (since n=2, 2016 becomes 4 in the adjusted problem). For k=4: N(k)=4029 -k=4029 -4=4025 (wait, no, in the adjusted problem, n=2, so 4029 -k is not relevant. Wait, this is where the confusion arises. The substitution in the original problem for the second sum was k from n +2 to 2n. In the adjusted problem, n=2, so k=4. Then N(k)=2n -k +1=4 -4 +1=1. Then, a=k -n=4 -2=2. So the sum over i=2 to 2: (-1)^2=1. So term: (5 -4)^2*1=1*1=1.Therefore, total sum is -9 +1= -8, which matches. So according to this, the sum is not zero, but in the original problem with n=2014, why did we think it cancels? Because in the adjusted problem, the first sum (k=2 to 3) had contributions from k even (k=2) and the second sum (k=4) also had a contribution from k even (k=4). However, the terms (5 -k)^2 are different for each k. In the original problem, when we did the substitution for m, we saw that terms could be mirrored, but in reality, the (2015 -k)^2 is mirrored around k=2015, but the coefficients (sum of (-1)^i) for k < 2015 and k >2015 are negatives of each other? Wait, in the adjusted problem, k=2: term=-9, k=4: term=+1. These are not symmetric. So in the original problem, perhaps the terms don't cancel out.Wait, but in the original problem, when we made the substitution m = 2015 - k for the first sum and m = k - 2015 for the second sum, we saw that the terms would be negatives of each other if the coefficients were the same. But in reality, the coefficients for the first sum (k from 2 to 2015) have a factor of -1 for even k, and the coefficients for the second sum (k from 2016 to 4028) have a factor of +1 for even k. However, when you make the substitution m = 2015 -k and m = k -2015, the terms (2015 -k)^2 and (k -2015)^2 are the same, but the coefficients are -1 and +1 respectively. So for each m, the term -m^2 and +m^2 would cancel each other. But in the adjusted problem, for m=1 (k=2014 in original, m=1 corresponds to k=2014, but in the adjusted problem, m=1 corresponds to k=4), but in the adjusted problem, m=1 in the second sum gives a +1, but m=3 in the first sum gives a -9. Wait, this seems inconsistent.Wait, maybe the problem is that in the adjusted problem, n=2, but the numbers inside are not scaled accordingly. Let's adjust the problem to use a different base number. Let me define a_{i,j} = (-1)^i*(n +1 -i -j)^2. For example, if n=2, then a_{i,j} = (-1)^i*(3 -i -j)^2. Let's compute that.For n=2:a_{1,1} = (-1)^1*(3 -1 -1)^2 = -1*(1)^2 = -1a_{1,2} = -1*(3 -1 -2)^2 = -1*(0)^2 = 0a_{2,1} = 1*(3 -2 -1)^2 = 1*(0)^2 = 0a_{2,2} =1*(3 -2 -2)^2 =1*(-1)^2=1Sum: -1 +0 +0 +1 =0. So sum is zero here.Hmm, interesting. So in this adjusted problem where the number inside the square is (n +1 -i -j), the sum is zero. However, in the original problem, the number is (2015 -i -j). Note that 2015 = 2014 +1, so actually, the original problem is similar to this adjusted problem with n=2014. Therefore, in the original problem, the sum should also be zero.But in my first adjusted example where I used 5 instead of n +1 (for n=2), the sum was -8, but when I used (n +1 -i -j), i.e., 3 -i -j, the sum was zero. Therefore, perhaps the key is whether the number inside the square is (n +1 -i -j) or a different number. In the original problem, it's 2015 -i -j, and 2015 =2014 +1, which is exactly n +1. Therefore, the original problem is equivalent to the adjusted problem where the number inside the square is (n +1 -i -j). Therefore, the sum should be zero.But in my first test with n=2 and (5 -i -j), which is not (n +1 -i -j), the sum was non-zero. So the key seems to be whether the constant term is n +1. Since in the original problem, it is n +1 (2015 =2014 +1), then the sum is zero. But how?Wait, let me re-examine the adjusted problem with a_{i,j} = (-1)^i*(3 -i -j)^2 for n=2. The sum was indeed zero. Let's see:For i=1, j=1: -1*(1)^2=-1For i=1, j=2: -1*(0)^2=0For i=2, j=1:1*(0)^2=0For i=2, j=2:1*(-1)^2=1Total sum: -1 +0 +0 +1=0So in this case, the sum is zero. Similarly, if we take another n, say n=1 (1x1 table). Then a_{1,1} = (-1)^1*(2 -1 -1)^2 = -1*(0)^2=0. Sum is 0.Wait, n=3. Let's try that. For n=3, a_{i,j}=(-1)^i*(4 -i -j)^2. Compute the sum.i from 1 to 3, j from 1 to 3.Compute each cell:i=1:j=1: -1*(4 -1 -1)^2 = -1*(2)^2=-4j=2: -1*(4 -1 -2)^2 = -1*(1)^2=-1j=3: -1*(4 -1 -3)^2 = -1*(0)^2=0i=2:j=1:1*(4 -2 -1)^2=1*(1)^2=1j=2:1*(4 -2 -2)^2=1*(0)^2=0j=3:1*(4 -2 -3)^2=1*(-1)^2=1i=3:j=1:(-1)^3*(4 -3 -1)^2=-1*(0)^2=0j=2: -1*(4 -3 -2)^2=-1*(-1)^2=-1j=3: -1*(4 -3 -3)^2=-1*(-2)^2=-4Now, sum all these:Row i=1: -4 -1 +0 =-5Row i=2:1 +0 +1=2Row i=3:0 -1 -4= -5Total sum: -5 +2 -5= -8. Wait, that's not zero. But according to the previous reasoning, it should be zero. Hmm, so there's a contradiction here. Wait, but in this case, the a_{i,j} = (-1)^i*(4 -i -j)^2 for n=3 (since 4=3+1). But the sum is -8, not zero. So why is that?Wait, maybe my previous conclusion was wrong. Let me check again.Wait, in the first adjusted problem with n=2 and using (n +1 -i -j), the sum was zero, but with n=3, it's -8. So perhaps the conclusion that the sum is zero only holds for even n? But n=2 is even, and sum is zero; n=3 is odd, sum is non-zero. Hmm.Wait, original problem is n=2014, which is even, so maybe in that case, the sum is zero. Let's check with n=4.Take n=4, so a_{i,j}=(-1)^i*(5 -i -j)^2. Compute the sum.This would be a 4x4 grid. Calculating all 16 terms would be tedious, but maybe we can use symmetry.Note that the terms where i + j = k will have certain symmetries. Let's see.Alternatively, note that for even n, perhaps the sum cancels out, while for odd n, it doesn't. Let's see.For n=2 (even), sum is zero. For n=3 (odd), sum is non-zero. For n=4 (even), maybe sum is zero. Let's test n=1 (odd? Well, n=1 is odd, but sum is zero). Wait, n=1: sum is zero. Hmm. So maybe my previous assumption is incorrect.Alternatively, perhaps the key lies in the pairing of terms. For each (i, j), there is a corresponding term (j, i) when we swap i and j. Let's see:a_{i,j} = (-1)^i*(C -i -j)^2a_{j,i} = (-1)^j*(C -j -i)^2 = (-1)^j*(C -i -j)^2Since (C -i -j)^2 is symmetric in i and j, so a_{i,j} and a_{j,i} have the same magnitude but different signs if i and j have different parity.Therefore, if i and j are both even or both odd, then (-1)^i and (-1)^j are the same, so a_{i,j} and a_{j,i} are equal. If i and j have different parity, then a_{i,j} and a_{j,i} are negatives of each other.Wait, let's see:Case 1: i and j have the same parity. Then (-1)^i = (-1)^j, so a_{i,j} = a_{j,i}. Therefore, these pairs contribute 2*a_{i,j} if i ≠ j, and a_{i,i} if i = j.Case 2: i and j have different parity. Then (-1)^i = -(-1)^j, so a_{i,j} = -a_{j,i}. Therefore, these pairs contribute a_{i,j} + a_{j,i} =0.Therefore, in the entire sum, all off-diagonal terms where i and j have different parity cancel out. The remaining terms are the diagonal terms (i=j) and the terms where i and j have the same parity but i ≠ j.But the diagonal terms are a_{i,i} = (-1)^i*(C -2i)^2. For the original problem, C=2015, so a_{i,i}= (-1)^i*(2015 -2i)^2.The terms where i and j have the same parity and i ≠ j contribute 2*a_{i,j} for i < j.But unless the contributions from these terms also cancel out, the sum won't necessarily be zero.However, in the n=2 case, let's see:i=1, j=1: -1*(3 -2)^2=-1i=1, j=2: different parity, cancels with i=2, j=1: -1*(3-3)^2=0 and 1*(3-3)^2=0. So 0.i=2, j=2:1*(3 -4)^2=1*1=1i=1, j=3: doesn't exist in n=2.So sum is -1 +1=0.For n=3:Wait, let's see for n=3, 3x3 grid.Pairs with same parity:i=1, j=1: (-1)^1*(4 -2)^2=-4i=1, j=3: (-1)^1*(4 -4)^2=0i=3, j=1: (-1)^3*(4 -4)^2=0i=3, j=3: (-1)^3*(4 -6)^2=-4i=2, j=2:1*(4 -4)^2=0Pairs with different parity:i=1, j=2: cancels with i=2, j=1i=1, j=4: doesn't existi=2, j=3: cancels with i=3, j=2i=3, j=4: doesn't existThus, the sum would be -4 (i=1,j=1) +0 +0 -4 (i=3,j=3) +0 (i=2,j=2) + cancellation of other terms. But in reality, when we computed manually, the sum was -8. So where's the discrepancy? Because in the manual calculation, the i=1,j=3 and i=3,j=1 are zero, but other terms like i=2,j=1 and i=1,j=2 also involve cancellation. Wait, the sum for n=3 was -5 +2 -5=-8. Let's check again:Row 1: i=1, j=1 to 3: -4, -1, 0. Sum: -5Row 2: i=2, j=1 to3:1,0,1. Sum:2Row3: i=3, j=1 to3:0, -1, -4. Sum:-5Total sum: -5 +2 -5= -8But according to the pairing argument, we have:Diagonal terms: i=1,j=1: -4; i=2,j=2:0; i=3,j=3:-4. Total: -8Other same parity terms: i=1,j=3:0; i=3,j=1:0. Total:0Different parity terms cancel out. So total sum is -8, which matches. Therefore, the sum is accounted for by the diagonal terms. Therefore, in this case, the sum is -8, not zero.So why in n=2 it was zero? Because the diagonal terms canceled each other. For n=2:Diagonal terms:i=1,j=1: -1*(3-2)^2=-1i=2,j=2:1*(3 -4)^2=1*1=1Sum: -1 +1=0Other same parity terms: i=1,j=3 (doesn't exist), etc. So sum is zero.For n=3, the diagonal terms give -4 -4 = -8, and other same parity terms are zero. So the sum is -8.For n=4, if we follow this pattern, maybe the diagonal terms would cancel out again?Let me check n=4:Diagonal terms:i=1,j=1: (-1)^1*(5 -2)^2 = -9i=2,j=2: (-1)^2*(5 -4)^2=1*1=1i=3,j=3: (-1)^3*(5 -6)^2=-1*1=-1i=4,j=4: (-1)^4*(5 -8)^2=1*9=9Sum of diagonals: -9 +1 -1 +9=0Other same parity terms:For example, i=1,j=3: (-1)^1*(5 -4)^2=-1*1=-1i=3,j=1: (-1)^3*(5 -4)^2=-1*1=-1Sum: -1 +(-1)=-2. But since they are same parity, they are not canceled but doubled. Wait, but if i < j, then i=1,j=3 contributes -1, and i=3,j=1 contributes -1, but they are same parity, so actually, they don't cancel. Wait, no. Wait, if i and j have the same parity, they are both odd or both even. For i=1,j=3: both odd, so same parity. Then a_{i,j} and a_{j,i} are both calculated as (-1)^i*(5 -i -j)^2 and (-1)^j*(5 -j -i)^2. Since i and j are both odd, (-1)^i = -1, (-1)^j = -1. So a_{i,j}= -1*(5 -4)^2=-1*1=-1, a_{j,i}= -1*(5 -4)^2=-1*1=-1. Therefore, sum is -1 + (-1) = -2.Similarly, i=1,j=1: already considered.i=2,j=4: both even. a_{2,4}=1*(5 -6)^2=1*1=1; a_{4,2}=1*(5 -6)^2=1*1=1. Sum:1 +1=2.i=4,j=2: same as above.i=2,j=2: already considered.So total sum would be diagonal terms 0, plus other same parity terms. For example, in a 4x4 grid, same parity pairs (i,j) where i ≠ j:- i=1,j=3: -2- i=2,j=4: +2- i=1,j=1: -9 +1 -1 +9=0- i=1,j=3, i=3,j=1: -2- i=2,j=4, i=4,j=2: +2- Other same parity pairs:i=1,j=5: doesn't existi=3,j=5: nope.So total sum would be diagonals 0 plus (-2 +2)=0. So total sum is zero.But let's compute manually to verify.For n=4, a_{i,j}=(-1)^i*(5 -i -j)^2:i=1:j=1: -1*(5-2)^2=-9j=2: -1*(5-3)^2=-4j=3: -1*(5-4)^2=-1j=4: -1*(5-5)^2=0Sum: -9 -4 -1 +0= -14i=2:j=1:1*(5-3)^2=4j=2:1*(5-4)^2=1j=3:1*(5-5)^2=0j=4:1*(5-6)^2=1Sum:4 +1 +0 +1=6i=3:j=1: -1*(5-4)^2=-1j=2: -1*(5-5)^2=0j=3: -1*(5-6)^2=-1j=4: -1*(5-7)^2=-4Sum:-1 +0 -1 -4= -6i=4:j=1:1*(5-5)^2=0j=2:1*(5-6)^2=1j=3:1*(5-7)^2=4j=4:1*(5-8)^2=9Sum:0 +1 +4 +9=14Total sum: (-14) +6 + (-6) +14=0. Okay, so sum is zero. Therefore, for n=4 (even), the sum is zero, but for n=3 (odd), it's non-zero. Therefore, this suggests that when n is even, the sum is zero, and when n is odd, it's non-zero. Therefore, in the original problem with n=2014 (even), the sum should be zero. That aligns with the first reasoning where the terms from k=2 to2015 and k=2016 to4028 canceled each other out due to the symmetric nature and the coefficients being negatives. Therefore, the answer should be zero.But let's make sure with one more example. Take n=4 as above, sum is zero. So for even n, sum is zero.Therefore, concluding that for even n, the sum is zero. Therefore, the original problem with n=2014 (even) should have a sum of zero.But why in the first adjusted problem with n=2, where the number inside the square wasn't (n +1 -i -j) but (5 -i -j), the sum wasn't zero? Because in that case, the number inside the square wasn't symmetric around the center. In the problem where we have a_{i,j}=(-1)^i*(n +1 -i -j)^2, for even n, the sum is zero due to symmetric cancellation. In the original problem, 2015 =2014 +1, so it fits into this category. Therefore, the sum is zero.Therefore, the answer should be boxed{0}Final Answerboxed{0}</think>To find the sum of all the numbers in the 2014x2014 table where each cell at position ((i, j)) contains the number (a_{i,j} = (-1)^i cdot (2015 - i - j)^2), we start by considering the double sum:[S = sum_{i=1}^{2014} sum_{j=1}^{2014} (-1)^i cdot (2015 - i - j)^2]We analyze the sum by changing variables and considering symmetry. Let (k = i + j). For each (k), the term ((2015 - k)^2) is constant, and we need to consider the number of pairs ((i, j)) such that (i + j = k) and the sign factor ((-1)^i).For (2 leq k leq 2015), the number of pairs ((i, j)) is (k - 1). For (2016 leq k leq 4028), the number of pairs is (4029 - k). We need to determine the sum of ((-1)^i) for these pairs.By examining the geometric series and parity, we find that the sum of ((-1)^i) for pairs ((i, j)) with (i + j = k) is zero if the number of terms is even, and (pm 1) if odd. However, due to the alternating signs and symmetry around (k = 2015), we find that the contributions from terms for (k leq 2015) and (k geq 2016) cancel each other out.This cancellation occurs because the terms ((2015 - k)^2) are symmetric around (k = 2015), and the alternating signs result in equal and opposite contributions. For even (n = 2014), this symmetry leads to the total sum being zero.Thus, the sum of all the numbers in the table is:[boxed{0}]]

🔑:Problem 3. Answer: mathbf{0 .}

🧠:[<think>Okay, let's see. The problem is about arranging three numbers x, y = x^x, and z = x^{x^x} in order of increasing magnitude when x is between 0.9 and 1.0. The options are given as (A) through (E). Hmm, so first, I need to figure out the relationship between x, x^x, and x^{x^x} when x is a number slightly less than 1, like 0.95 or something. Let me start by recalling some properties of exponents and logarithms.Since x is between 0.9 and 1.0, it's a fractional number less than 1 but greater than 0.9. Let me note that for 0 < x < 1, the function x^a decreases as a increases. Wait, actually, no. For 0 < x < 1, the function x^a is a decreasing function if a is in the exponent. But when both the base and the exponent are variables, things get trickier. Maybe I should consider taking logarithms to compare the sizes.Let me pick a specific value of x in the given interval to test. Let's choose x = 0.9, which is at the lower end. Then compute y = 0.9^{0.9} and z = 0.9^{0.9^{0.9}}. Let me calculate these values step by step.First, x = 0.9.Compute y = x^x = 0.9^0.9. Since 0.9 is less than 1, raising it to a power also less than 1 will result in a number greater than 0.9. Wait, how does that work? Let me recall: for numbers between 0 and 1, when you raise them to a power between 0 and 1, the result is actually larger than the original number. For example, 0.25^0.5 = 0.5, which is larger than 0.25. Similarly, 0.9^0.9 should be slightly larger than 0.9. Let me verify with a calculator. 0.9^0.9 ≈ e^{0.9 ln 0.9} ≈ e^{0.9*(-0.10536)} ≈ e^{-0.09482} ≈ 0.9097. So y ≈ 0.91, which is indeed larger than x = 0.9.Then z = x^{x^x} = 0.9^{0.9^{0.9}} ≈ 0.9^{0.9097}. Let's compute that. Again, 0.9 raised to a power slightly less than 1. Since 0.9 < 1, raising it to a power less than 1 will make it larger, but how much? Let's compute 0.9^{0.9097} ≈ e^{0.9097 * ln 0.9} ≈ e^{0.9097*(-0.10536)} ≈ e^{-0.0958} ≈ 0.9087. So z ≈ 0.9087. Wait, so z is approximately 0.9087, which is slightly less than y ≈ 0.9097 but more than x = 0.9. Hmm, so in this case, the order would be x < z < y. But let me check again because that might not be the case. Wait, 0.9^0.9 is about 0.91, and then 0.9^{0.91} is about 0.9087. So, x=0.9, z≈0.9087, y≈0.9097. So increasing order is x < z < y. But this is for x=0.9. But the problem states that x is between 0.9 and 1.0. So maybe I should check another value, say x = 0.95, to see if the relationship holds.Let's try x = 0.95.First, compute y = 0.95^0.95. Let's calculate ln(0.95) ≈ -0.051293, so 0.95*ln(0.95) ≈ -0.051293*0.95 ≈ -0.04873. Then y ≈ e^{-0.04873} ≈ 0.9531. So y ≈ 0.9531.Now compute z = 0.95^{y} = 0.95^{0.9531}. Let's compute ln(0.95) again ≈ -0.051293. Multiply by 0.9531: -0.051293 * 0.9531 ≈ -0.04889. Then z ≈ e^{-0.04889} ≈ 0.9530. So here, z ≈ 0.9530, which is very slightly less than y ≈ 0.9531. So again, x = 0.95, z ≈0.9530, y≈0.9531. So same order: x < z < y.Wait a minute, so in both x=0.9 and x=0.95, the order is x < z < y. But the answer options don't include that. Let's check the options again.Options:(A) x, z, y(B) x, y, z(C) y, x, z(D) y, z, x(E) z, x, ySo according to my calculations, the order is x < z < y, which would be option A. But hold on, when x=0.9, z≈0.9087 and y≈0.9097, so z < y. Similarly, for x=0.95, z≈0.9530 < y≈0.9531. So x < z < y. Therefore, the order is x, z, y, which is option A. But the answer options have (A) as x, z, y. But let me check the problem statement again.Wait, the problem says "arranged in order of increasing magnitude", so from smallest to largest. So if x < z < y, then the order is x, z, y, which is option A. However, the original answer choices for the problem might have a different answer. Wait, but according to my calculations, that's the case. Hmm. But let me think again. Wait, perhaps there is a mistake here.Wait, when x is between 0 and 1, x^x is greater than x, which I saw in my calculations. Then, x^{x^x} would be x raised to something greater than x, but since the base x is less than 1, increasing the exponent would make the result smaller. Wait, hold on. Let me clarify:For 0 < x < 1, a^b is a decreasing function of b if a is fixed and less than 1. So, if you have x < 1, then if you increase the exponent, the result decreases. Therefore, since x^x > x (as we saw in examples), then z = x^{x^x} would be x raised to a higher exponent than x, so z < x^x. But since x < 1, x^x > x, but z = x^{x^x} < x^x. However, how does z compare to x?So, we have x < 1, x^x > x, and z = x^{x^x}. Since x^x > x, and because for 0 < a < 1, a^b is decreasing in b, then z = x^{x^x} < x^x. But how does z compare to x? Is z > x or z < x?Wait, in the examples above, for x=0.9, z≈0.9087, which is greater than x=0.9. Wait, 0.9087 is actually greater than 0.9. Wait, but 0.9^{0.9} ≈0.91, then 0.9^{0.91}≈0.9087. So 0.9087 is less than 0.91 but greater than 0.9.Similarly, for x=0.95, z≈0.9530, which is slightly less than y≈0.9531 but greater than x=0.95. Wait, 0.95 is x, then z=0.9530 and y=0.9531. So in this case, x < z < y.Therefore, in both examples, z is between x and y, closer to y. So order is x < z < y, which is option A.But wait, the answer choices given in the problem are:(A) x, z, y(B) x, y, z(C) y, x, z(D) y, z, x(E) z, x, ySo according to this, if it's x < z < y, then the answer is (A). However, I need to make sure that this holds for all x in (0.9, 1.0). Let me check another value, say x=0.99.Compute y = 0.99^0.99. Let's calculate ln(0.99) ≈ -0.01005, so 0.99*ln(0.99) ≈ -0.01005*0.99 ≈ -0.00995. Then y ≈ e^{-0.00995} ≈ 0.9901. So y ≈0.9901.Then z = 0.99^{y} = 0.99^{0.9901}. So compute ln(0.99) ≈ -0.01005. Multiply by 0.9901: -0.01005*0.9901 ≈ -0.00995. Then z ≈ e^{-0.00995} ≈0.9901. Wait, so z ≈0.9901, which is the same as y? Hmm, but that can't be. Wait, maybe my approximation is too rough. Let me compute more accurately.First, compute y = 0.99^0.99. Let me use natural logarithm:ln(y) = 0.99 * ln(0.99) ≈0.99*(-0.01005034) ≈-0.0099498.Therefore, y ≈ e^{-0.0099498} ≈1 -0.0099498 + (0.0099498)^2/2 ≈approximately 0.9901. Similarly, z =0.99^{0.9901}. Then ln(z) =0.9901*ln(0.99)≈0.9901*(-0.01005034)≈-0.009955. Therefore, z≈e^{-0.009955}≈0.9901. So y and z are almost the same here, but maybe z is slightly less than y. Because the exponent in z is 0.9901, which is very slightly more than the exponent in y, which is 0.99. Since the base 0.99 is less than 1, increasing the exponent makes the result smaller. So z would be slightly smaller than y. Therefore, z ≈0.990099, and y≈0.9901. So again, x=0.99, z≈0.990099, y≈0.9901. So x=0.99 < z≈0.990099 < y≈0.9901. The difference is minuscule, but the order still holds.Therefore, in all tested cases (x=0.9, 0.95, 0.99), the order is x < z < y, which would be option A. But wait, let me think again. Wait, when x approaches 1 from below, what happens?Let x approach 1. Let’s take x = 0.999.Compute y = 0.999^0.999. ln(y) =0.999*ln(0.999) ≈0.999*(-0.0010015)≈-0.0010005. Thus y ≈ e^{-0.0010005} ≈0.9990005. Then z =0.999^{y}≈0.999^{0.9990005}. Then ln(z) ≈0.9990005*ln(0.999)≈0.9990005*(-0.0010015)≈-0.0010005. Thus z≈e^{-0.0010005}≈0.9990005. So again, z is equal to y in this approximation. Wait, but since x is approaching 1, both y and z approach 1 as well, but the difference between them becomes negligible. However, strictly speaking, when x is just below 1, x < z < y still holds? Let me think.Suppose x =1 - ε, where ε is a small positive number, like 0.001.Compute y = x^x = (1 - ε)^{1 - ε} ≈1 - ε + ε^2/2 (using the expansion (1 - ε)^{1 - ε} ≈ e^{(1 - ε)(-ε)} ≈ e^{-ε + ε^2} ≈1 - ε + ε^2/2 + ...).Similarly, z =x^{x^x} ≈(1 - ε)^{y} ≈(1 - ε)^{1 - ε + ε^2/2} ≈e^{(1 - ε + ε^2/2)(-ε)} ≈e^{-ε + ε^2 - ε^3/2} ≈1 - ε + ε^2 - ε^3/2 - ε^2/2 + ...≈1 - ε + ε^2/2 - ε^3/2 + ...Compare y and z. So y ≈1 - ε + ε^2/2, and z≈1 - ε + ε^2/2 - ε^3/2. Therefore, z is slightly smaller than y, as the term -ε^3/2 makes it smaller. Therefore, even as x approaches 1, the order remains x < z < y. Because x=1 - ε, z≈1 - ε + ε^2/2 - ..., which is larger than x=1 - ε. Wait, but wait, in the expansion, x=1 - ε, z≈1 - ε + (ε^2/2 - ε^3/2). But x=1 - ε, so z is larger than x. Therefore, x < z < y even as x approaches 1.Therefore, in all cases where 0.9 < x < 1.0, the order is x < z < y, which is option A. But wait, the answer choices have (A) as x, z, y, which is increasing order, so that's correct.Wait, but hold on. Let me check again with another example. Maybe x=0.5? Although the problem states x is between 0.9 and 1.0, but just to test the intuition. Let me take x=0.5. Then y=0.5^0.5=√0.5≈0.707. Then z=0.5^{0.707}≈e^{0.707*ln0.5}≈e^{0.707*(-0.693)}≈e^{-0.490}≈0.612. So here, x=0.5, z≈0.612, y≈0.707. So order is x < z < y, same as before. Wait, even when x=0.5, which is far from 1, the order is still x < z < y. So perhaps for all x in (0,1), the order is x < z < y. Is that true?Let me check with x=0.2. y=0.2^0.2. Let's compute that. ln(y)=0.2*ln0.2≈0.2*(-1.609)≈-0.3218. So y≈e^{-0.3218}≈0.724. Then z=0.2^{0.724}≈e^{0.724*ln0.2}≈e^{0.724*(-1.609)}≈e^{-1.165}≈0.311. So here, x=0.2, z≈0.311, y≈0.724. So order is x < z < y. So even at x=0.2, the order holds.Wait, so perhaps in general, for 0 < x < 1, x < x^{x^x} < x^x. Therefore, the order is x < z < y, which is option A. But the answer given in the problem's options is (A) x, z, y, so that must be the answer. But wait, the original problem statement says "arranged in order of increasing magnitude", so the correct order from smallest to largest is x, z, y. Therefore, the answer is (A).But wait, I need to confirm this with another method. Let's try to analyze the functions without plugging numbers.Given x is between 0.9 and 1.0. Let me consider the three numbers: x, y = x^x, z = x^{x^x}.First, compare x and y = x^x. For x in (0,1), x^x is greater than x. Let me verify this.Take the function f(x) = x^x. Let's compare f(x) with x. For x in (0,1):At x=1, f(x)=1^1=1, which equals x. At x approaching 0+, f(x) approaches 1 (since x^x = e^{x ln x} and x ln x approaches 0). So f(x) = x^x is greater than x for x in (0,1). For example, at x=0.5, 0.5^0.5≈0.707>0.5. So y = x^x > x.Now compare z = x^{x^x} with x and y.First, z vs x: Since y =x^x >x, and since for 0 <a<1, a^b is a decreasing function of b, then z =x^{y} <x^x = y. But how does z compare to x?Since y =x^x >x, and z =x^{y}. Since y >x, and a =x <1, then z =x^{y} <x^x = y. But z compared to x: Is z >x?Yes. Let me think. If we take logarithms, we can compare ln(z) = y ln x, and ln(x) = ln x. Since y =x^x, ln(z) =x^x ln x. Compare ln(z) and ln(x):ln(z) - ln(x) = x^x ln x - ln x = ln x (x^x -1). Since x <1, ln x <0. Also, x^x <1 because x <1 and x^x = e^{x ln x}, and x ln x <0. Wait, but earlier we saw that x^x >x. Wait, x^x is between x and 1 when x is between 0 and1. Wait, let's check:If x is in (0,1), then x^x is in (0,1). But is x^x greater than x?Yes. For x in (0,1), x^x >x. For example, x=0.5, 0.5^0.5≈0.707>0.5. x=0.9, 0.9^0.9≈0.91>0.9. So x^x is greater than x but less than 1. Therefore, x^x is in (x,1). Then, z =x^{x^x} since x <1 and exponent x^x >x, so z =x^{x^x} <x^x. But z compared to x:z =x^{x^x} >x because x^{x^x} = e^{x^x ln x}. Since x <1, ln x <0, and x^x is between x and1. So x^x ln x is between x ln x and ln x. Since ln x is negative, and x^x >x, so x^x ln x <x ln x (because multiplying a negative number by a larger positive number makes it smaller). Therefore, ln z =x^x ln x <x ln x. Therefore, z = e^{x^x ln x} < e^{x ln x} =x^x. But how does z compare to x?Wait, to check if z >x, we can consider the difference z -x =x^{x^x} -x. Let's analyze the sign of this difference.Since x^{x^x} >x ?Is x^{x^x} >x for x in (0.9,1.0)?Take natural logs: x^x ln x > ln x.But since ln x <0 for x in (0,1), we can divide both sides by ln x (inequality reverses):x^x <1.But x^x is always <1 for x in (0,1). Therefore, the inequality x^x ln x > ln x is equivalent to x^x <1, which is true. Therefore, ln z =x^x ln x > ln x (since ln x is negative and x^x ln x is less negative than ln x). Wait, no:Wait, x^x is <1, so multiplying ln x (negative) by x^x (less than 1 but positive) gives a number that is less negative than ln x. Therefore, x^x ln x > ln x (because it's less negative). Therefore, ln z =x^x ln x > ln x, which implies z >x. Therefore, z >x.Therefore, combining these results: x < z < y.Therefore, in general, for x in (0,1), we have x < x^{x^x} <x^x.Therefore, the order is x, z, y, which is option A. Therefore, the answer should be (A).But let me verify once again with x approaching 1 from below. Let x approach 1, so x =1 - ε where ε is small.Then, x^x = (1 - ε)^{1 - ε} ≈1 - ε + ε^2/2.Then, z =x^{x^x} ≈(1 - ε)^{1 - ε + ε^2/2} ≈1 - ε*(1 - ε + ε^2/2) + [ε*(1 - ε + ε^2/2)]^2/2.Wait, expanding using (1 - ε)^a ≈1 - aε + (a^2 ε^2)/2.So, a =1 - ε + ε^2/2.Thus, z≈1 - a ε + (a ε)^2 /2.Plugging a:z≈1 - [1 - ε + ε^2/2] ε + ([1 - ε + ε^2/2] ε)^2 /2.≈1 - ε + ε^2 - ε^3/2 + (ε^2 - 2 ε^3 + ... ) /2≈1 - ε + ε^2 - ε^3/2 + ε^2/2 - ε^3 + ...≈1 - ε + (3/2) ε^2 - (3/2) ε^3 + ...Comparing this to y =x^x≈1 - ε + ε^2/2.So, z ≈1 - ε + (3/2) ε^2 - ..., which is greater than x=1 - ε, but compared to y=1 - ε + ε^2/2, z is larger than y when ε is small?Wait, no. Wait, z ≈1 - ε + (3/2) ε^2, and y≈1 - ε + ε^2/2. Therefore, z is larger than y. But earlier, when we computed x=0.9, 0.95, 0.99, z was less than y. Wait, there must be a mistake here.Wait, maybe the expansion is not accurate enough. Let me check with x=0.99, which is close to 1.x=0.99, so ε=0.01.Compute y=0.99^0.99≈1 -0.01 +0.00005≈0.99005 (approximate expansion).But the actual value calculated earlier was y≈0.9901. Similarly, z=0.99^{0.9901}≈1 -0.01*0.9901 + ...≈0.9901 -0.000099≈0.9900.Wait, but actual calculation showed z≈0.990099, which is very slightly less than y≈0.9901. Therefore, even as x approaches 1, z remains slightly less than y, hence the order x < z < y holds.Therefore, the expansion might not capture this because higher-order terms make z less than y. Therefore, in reality, even as x approaches 1, z remains less than y, maintaining the order x < z < y.Therefore, the correct order is x < z < y, which is option A.But wait, the original problem is from a past exam, and sometimes answer choices might be different. Let me check the problem again.The problem states that x is between 0.9 and 1.0, and we need to arrange x, y=x^x, z=x^{x^x} in increasing order. The options are (A) x,z,y; (B) x,y,z; (C) y,x,z; (D) y,z,x; (E) z,x,y.Given that in all test cases and analytical approaches, the order is x < z < y, which is option A. Therefore, the answer is (A).But I recall that in some similar problems, the order is different. Let me think if there's a different way to approach this.Alternatively, take logarithms to compare the numbers.Let’s define x, y =x^x, z =x^{x^x}.We know that x <1, so ln x <0.Compare x and z:We need to check if z >x, which is equivalent to x^{x^x} >x.Take natural log: x^x ln x > ln x.Since ln x <0, divide both sides by ln x (inequality reverses):x^x <1.But x^x <1 because x <1. Therefore, the inequality holds, so x^{x^x} >x. Therefore, z >x.Compare z and y:z =x^{x^x}, y =x^x.Check if z <y:x^{x^x} <x^x.Take natural log: x^x ln x < ln x.Again, ln x <0, so divide both sides (inequality reverses):x^x >1.But x^x <1 because x <1. Therefore, the inequality x^{x^x} <x^x holds. Hence, z <y.Therefore, combining both results: x < z <y. Therefore, the order is (A).Hence, the answer is option A.Final Answerboxed{A}</think>Given ( x ) in the interval ( (0.9, 1.0) ), we need to arrange ( x ), ( y = x^x ), and ( z = x^{x^x} ) in increasing order.1. Compare ( x ) and ( y ): - For ( 0 < x < 1 ), ( x^x ) is greater than ( x ). This is because raising a number between 0 and 1 to a power less than 1 results in a number greater than the original. For example, ( 0.9^{0.9} approx 0.9097 ), which is greater than 0.9.2. Compare ( z ) and ( y ): - Since ( y = x^x ) is greater than ( x ), and the base ( x ) is less than 1, raising ( x ) to a higher power (i.e., ( y )) results in a smaller number. Therefore, ( z = x^{x^x} ) is less than ( y ). For example, ( 0.9^{0.9097} approx 0.9087 ), which is less than 0.9097.3. Compare ( z ) and ( x ): - Since ( y = x^x ) is greater than ( x ), and ( z = x^{x^x} ) is ( x ) raised to a power greater than ( x ), ( z ) is greater than ( x ). For example, ( 0.9^{0.9097} approx 0.9087 ), which is greater than 0.9.Thus, the order from smallest to largest is ( x < z < y ).[boxed{A}],<think>Okay, so I have this problem here where I need to arrange three numbers in increasing order: x, y = x^x, and z = x^{x^x}, with the condition that x is between 0.9 and 1.0. The options are given from (A) to (E), and I need to figure out which one is correct. Let me start by understanding the problem step by step.First, let me note that x is a number slightly less than 1 but greater than 0.9. So x is like 0.95 or something. Since x is less than 1, raising it to a higher power should make it smaller, right? Wait, but exponents can sometimes be tricky when the base is between 0 and 1. Let me recall: if you have a number between 0 and 1, say a, then a^b decreases as b increases, right? For example, (1/2)^2 = 1/4, which is smaller than 1/2. So, if x is between 0.9 and 1, which is close to 1 but less than 1, then x^x should be greater than x, because you're raising a number less than 1 to a power less than 1? Wait, hold on. Let's test with x = 0.5. Then x^x = 0.5^0.5 = sqrt(0.5) ≈ 0.707, which is greater than 0.5. Similarly, if x is 0.9, x^x would be 0.9^0.9. Let me calculate that. Hmm, 0.9^0.9 is approximately e^{0.9 ln 0.9} ≈ e^{0.9*(-0.105)} ≈ e^{-0.0945} ≈ 0.91. So 0.9^0.9 is about 0.91, which is greater than 0.9. So in this case, x^x is actually greater than x when x is between 0 and 1. Wait, that's interesting. So for x between 0 and 1, x^x is greater than x? Let me check with x approaching 1. If x is 1, then x^x is 1^1 = 1. So as x approaches 1 from below, x^x approaches 1 from below? Wait, but when x is 0.9, x^x is approximately 0.91, which is greater than x. So maybe there's a maximum somewhere? Wait, the function x^x when x is between 0 and 1: Let's think about its derivative. Let me recall that the derivative of x^x is x^x (1 + ln x). So setting derivative to zero, 1 + ln x = 0 => ln x = -1 => x = e^{-1} ≈ 0.3678. So the function x^x has a minimum at x = 1/e, but wait, between 0 and 1, the function x^x first decreases from x=0 to x=1/e, reaching a minimum at 1/e, then increases towards 1 as x approaches 1. Wait, that seems conflicting with my previous thought. Wait, when x approaches 0, what does x^x approach? Let's see: x^x = e^{x ln x}, and x ln x approaches 0 as x approaches 0 from the right, because ln x approaches -infty, but x approaches 0, so the product x ln x approaches 0. Therefore, x^x approaches e^0 = 1. Wait, that's interesting. So as x approaches 0 from the right, x^x approaches 1. Then, at x = 1/e, x^x is e^{-1/e} ≈ 0.6922, and at x=1, it's 1. So the function x^x on (0,1) starts at approaching 1 as x approaches 0, decreases to a minimum at x=1/e, then increases back to 1 at x=1. Therefore, for x between 0 and 1/e, x^x is decreasing, and between 1/e and 1, it's increasing. So, in our case, x is between 0.9 and 1.0, so 0.9 is greater than 1/e ≈ 0.367, so we are in the increasing portion of x^x. Therefore, as x increases from 0.9 to 1, x^x increases from ~0.91 to 1. But since x is between 0.9 and 1.0, which is less than 1, then x^x is still greater than x? Wait, when x is 0.9, x^x ≈ 0.91, which is greater than 0.9. When x is 0.95, x^x ≈ e^{0.95 ln 0.95} ≈ e^{0.95*(-0.0513)} ≈ e^{-0.0487} ≈ 0.953, which is still greater than 0.95. Wait, so x^x is still greater than x even when x is approaching 1? Wait, but when x approaches 1, x^x approaches 1 as well. Let me check when x is 0.99. Then x^x ≈ e^{0.99 ln 0.99} ≈ e^{0.99*(-0.01005)} ≈ e^{-0.00995} ≈ 0.9901. So 0.99^0.99 ≈ 0.9901, which is slightly greater than 0.99. So even at x = 0.99, x^x is still slightly greater than x. So in general, for x in (0,1), x^x > x? Wait, but when x approaches 0, x^x approaches 1, so x^x would be greater than x. At x = 1/e, x^x ≈ 0.6922, which is greater than 1/e ≈ 0.3678. So yes, x^x is always greater than x on (0,1). Therefore, for our problem, x is between 0.9 and 1.0, so x^x is greater than x. Therefore, x < x^x. So that would mean y = x^x is greater than x. So in terms of order, x < y. So far, so good.Now, what about z = x^{x^x}. Let's analyze z. Since x is between 0.9 and 1, and x^x is between ~0.9 and 1 (but actually, since x is between 0.9 and 1, x^x is between ~0.91 and 1, as we saw). So z is x raised to the power of x^x. So, since x is less than 1, and the exponent x^x is greater than x (since x^x > x as we established), then z = x^{x^x} would be less than x^x, because when you raise a number between 0 and 1 to a higher power, it gets smaller. For example, if x = 0.5, x^x = 0.707, then z = 0.5^{0.707} ≈ 0.612, which is less than 0.707. Similarly, take x = 0.9: x^x ≈ 0.91, so z = 0.9^{0.91} ≈ e^{0.91 ln 0.9} ≈ e^{0.91*(-0.105)} ≈ e^{-0.0955} ≈ 0.909. So z ≈ 0.909, which is less than x^x ≈ 0.91, but greater than x = 0.9. Wait, so in this case, z is between x and y. Let's verify with x = 0.9: x = 0.9, y ≈ 0.91, z ≈ 0.909. So order would be x < z < y. Wait, but z is approximately 0.909, which is greater than x = 0.9, but less than y ≈ 0.91. So in this case, x < z < y. But in another example, take x = 0.95. Then, x^x ≈ e^{0.95 ln 0.95} ≈ e^{-0.95*0.0513} ≈ e^{-0.0487} ≈ 0.953. Then z = x^{x^x} = 0.95^{0.953} ≈ e^{0.953 ln 0.95} ≈ e^{-0.953*0.0513} ≈ e^{-0.0489} ≈ 0.953. Wait, so z ≈ 0.953, which is almost equal to y ≈ 0.953. Hmm, but is it slightly less? Let me compute more accurately.Take x = 0.95. Compute x^x:First, ln(0.95) ≈ -0.051293.So x^x = e^{0.95 * (-0.051293)} ≈ e^{-0.048728} ≈ 1 - 0.048728 + 0.048728²/2 ≈ 0.953 (approximate Taylor series). Let's use calculator-like steps:Compute exponent: 0.95 * ln 0.95 ≈ 0.95 * (-0.051293) ≈ -0.048728.Then e^{-0.048728} ≈ 1 - 0.048728 + (0.048728)^2 / 2 - (0.048728)^3 / 6 ≈First term: 1Second term: -0.048728Third term: (0.002374) / 2 ≈ 0.001187Fourth term: -(0.0001156) / 6 ≈ -0.00001926Adding them up: 1 - 0.048728 = 0.951272; 0.951272 + 0.001187 ≈ 0.952459; 0.952459 - 0.00001926 ≈ 0.952439. So approximately 0.9524.So y ≈ 0.9524.Then z = x^{x^x} = 0.95^{0.9524}.Compute ln(0.95) ≈ -0.051293.So exponent is 0.9524 * (-0.051293) ≈ -0.04885.Then e^{-0.04885} ≈ similar to before: 1 - 0.04885 + (0.04885)^2 / 2 - (0.04885)^3 / 6 ≈1 - 0.04885 = 0.95115(0.04885)^2 ≈ 0.002386, divided by 2: ≈ 0.001193(0.04885)^3 ≈ 0.0001166, divided by 6: ≈ 0.00001943So 0.95115 + 0.001193 = 0.952343; 0.952343 - 0.00001943 ≈ 0.952323.Therefore, z ≈ 0.9523, which is slightly less than y ≈ 0.9524. So in this case, z is very slightly less than y, but still greater than x = 0.95. Therefore, the order is x < z < y.Wait, but when x approaches 1, what happens? Let's take x = 0.99.Compute x^x: 0.99^0.99 ≈ e^{0.99 ln 0.99} ≈ e^{-0.99*0.01005} ≈ e^{-0.0099495} ≈ 0.9901. Then z = x^{x^x} = 0.99^{0.9901} ≈ e^{0.9901 ln 0.99} ≈ e^{-0.9901*0.01005} ≈ e^{-0.009956} ≈ 0.9901. So here, z ≈ 0.9901, which is almost equal to y ≈ 0.9901. But technically, since the exponent for z is 0.9901, which is slightly less than 1, but x is 0.99. Wait, no, x^x is 0.9901, so z = x^{x^x} = 0.99^{0.9901} ≈ 0.99^{0.99 + 0.0001} ≈ 0.99^{0.99} * 0.99^{0.0001} ≈ 0.9901 * (1 - 0.0001*0.01) ≈ 0.9901 - 0.0000001 ≈ approximately 0.9901. So z is almost equal to y. But perhaps slightly less? Let's compute more precisely.Compute ln(0.99) ≈ -0.01005034.x^x exponent: 0.99 * (-0.01005034) ≈ -0.00995.So x^x ≈ e^{-0.00995} ≈ 1 - 0.00995 + 0.00995²/2 - ... ≈ 0.9901.Then z = x^{x^x} = 0.99^{0.9901} ≈ e^{0.9901 * (-0.01005034)} ≈ e^{-0.009956} ≈ same as before, 0.9901.So in this case, z is approximately equal to y. However, because x is less than 1, and the exponent x^x is slightly less than 1 (since x is 0.99, x^x is 0.9901), then z = x^{x^x} = 0.99^{0.9901} is slightly greater than x^x, because when you raise a number less than 1 to a power less than 1, the result is greater than the original number. Wait, hold on. Wait, if you have a number a between 0 and 1, then a^b for b < 1 is greater than a. Wait, is that true? Let me check with a = 0.5, b = 0.5. Then 0.5^0.5 ≈ 0.707, which is greater than 0.5. If a = 0.8, b = 0.9, then 0.8^0.9 ≈ e^{0.9 ln 0.8} ≈ e^{-0.9*0.223} ≈ e^{-0.2007} ≈ 0.818, which is greater than 0.8. So yes, if a is between 0 and 1 and b is between 0 and 1, then a^b > a. Therefore, in our case, z = x^{x^x} where x is between 0.9 and 1, so x^x is between ~0.9 and 1. If x^x is less than 1, then z = x^{x^x} is greater than x. Wait, but we saw earlier that when x is 0.9, z ≈ 0.909, which is greater than x = 0.9 but less than y ≈ 0.91. When x is 0.95, z ≈ 0.9523, which is slightly less than y ≈ 0.9524. Wait, that contradicts the previous statement. Let me check that again.Wait, if x is 0.95, x^x is approximately 0.9524. Then z = 0.95^{0.9524}. Since 0.95 is less than 1 and 0.9524 is less than 1, then 0.95^{0.9524} should be greater than 0.95. But we computed it as approximately 0.9523, which is indeed greater than 0.95 but less than x^x = 0.9524. So in this case, z is between x and y. So even though a^b > a when a is between 0 and 1 and b is between 0 and 1, z can be less than y. Wait, but why?Because x^x is greater than x, so z = x^{x^x} is x raised to a higher power than x^x, which is already greater than x. Wait, but x is between 0.9 and 1. So when you raise x to a higher power (since exponents x^x are greater than x), the result should be smaller. Wait, that seems conflicting. Wait, if x is between 0 and 1, then the higher the exponent, the smaller the result. So if you have x, then x^x is larger (since exponent x is less than 1, so x^x > x). Then z = x^{x^x} is x raised to a higher exponent than x^x, so since the exponent is higher (x^x > x), then z = x^{x^x} < x^x. But x^{x^x} is greater than x, because x^{x^x} = (x^x)^x. Wait, no, that's not the case. Wait, exponentiation is not associative. z = x^{x^x} is different from (x^x)^x. Actually, (x^x)^x = x^{x*x} = x^{x^2}, which is different. So z = x^{x^x} is a tower of exponents: x^(x^x). So it's x raised to the power of x^x. Since x is between 0.9 and 1, and x^x is between ~0.9 and 1, let's see:If we have x = 0.9, then x^x ≈ 0.91, so z = 0.9^0.91 ≈ 0.909, which is less than x^x but greater than x. So z is between x and y. If x is 0.95, z ≈ 0.9523, which is just a bit less than y ≈ 0.9524, but still greater than x = 0.95. When x approaches 1, say x = 0.99, then z ≈ 0.9901, which is almost equal to y ≈ 0.9901, but perhaps very slightly less. Therefore, in all cases in (0.9, 1), z is between x and y, but closer to y. So the order is x < z < y. Therefore, the correct arrangement is x, z, y, which corresponds to option (A) x, z, y? Wait, the options are (A) x,z,y; (B) x,y,z; (C) y,x,z; (D) y,z,x; (E) z,x,y.Wait, according to my analysis, x < z < y, so increasing order is x, z, y, which is option (A). But wait, when x is 0.95, z ≈ 0.9523 and y ≈ 0.9524, so z is slightly less than y. So the order is x < z < y. Therefore, the answer should be (A). But wait, let's check again with x approaching 1. For example, x = 0.999.Compute x^x: 0.999^0.999 ≈ e^{0.999 * ln 0.999} ≈ e^{-0.999*0.001001} ≈ e^{-0.001000} ≈ 0.9990. Then z = x^{x^x} = 0.999^{0.9990} ≈ e^{0.9990 * ln 0.999} ≈ e^{-0.9990*0.001001} ≈ e^{-0.001000} ≈ 0.9990. So here, z ≈ 0.9990, y = x^x ≈ 0.9990, and x = 0.999. So as x approaches 1, z approaches y, and both approach 1. So in the limit as x approaches 1 from below, x, y, z all approach 1, but the order remains x < z < y. Because even at x = 0.999, x is 0.999, y and z are both approximately 0.9990, but actually, let's compute more precisely.Let me take x = 0.999.Compute ln(0.999) ≈ -0.0010015.Then x^x = e^{0.999 * (-0.0010015)} ≈ e^{-0.0010005} ≈ 0.9990005.Then z = x^{x^x} = 0.999^{0.9990005} ≈ e^{0.9990005 * (-0.0010015)} ≈ e^{-0.0010015} ≈ 0.999000 approximately.Wait, so in this case, z ≈ 0.999000, which is slightly less than y ≈ 0.9990005. So even in this case, z is very slightly less than y, but still greater than x. Therefore, x < z < y still holds.So in all cases where 0.9 < x < 1.0, we have x < z < y. Therefore, the order is x, z, y, which is option (A). But let's check again the answer options:(A) x, z, y(B) x, y, z(C) y, x, z(D) y, z, x(E) z, x, yWait, according to the analysis, the order is x < z < y, which is (A). But let me cross-verify with another example. Let's take x = 0.92.Compute x^x: 0.92^0.92.ln(0.92) ≈ -0.083375.So x^x ≈ e^{0.92*(-0.083375)} ≈ e^{-0.0767} ≈ 0.926.Then z = x^{x^x} = 0.92^{0.926}.ln(0.92) ≈ -0.083375.Exponent: 0.926*(-0.083375) ≈ -0.0773.So z ≈ e^{-0.0773} ≈ 0.9256.So x = 0.92, z ≈ 0.9256, y ≈ 0.926. So order is x < z < y, which is 0.92 < 0.9256 < 0.926. So still, (A) is correct.Alternatively, maybe I made a mistake in assuming that z is always less than y. Let me check with x approaching 0.9 from above.Take x = 0.91.x^x = 0.91^0.91 ≈ e^{0.91*ln 0.91} ≈ e^{0.91*(-0.09431)} ≈ e^{-0.0857} ≈ 0.918.Then z = 0.91^{0.918} ≈ e^{0.918*ln 0.91} ≈ e^{0.918*(-0.09431)} ≈ e^{-0.0866} ≈ 0.917.Wait, so x = 0.91, z ≈ 0.917, y ≈ 0.918. So order is x < z < y. Therefore, (A) still holds.Wait, but here z ≈ 0.917 and y ≈ 0.918, so z is less than y. Therefore, even in this case, the order is x < z < y.Therefore, based on all these examples, it seems that the order is always x < z < y, so the answer should be (A). But wait, the options given are:(A) x, z, y(B) x, y, z(C) y, x, z(D) y, z, x(E) z, x, yTherefore, the correct answer is (A). However, let me check the original problem statement again to ensure I didn't misinterpret anything.The problem states: "Given the three numbers x, y = x^x, z = x^{x^x} with 0.9 < x < 1.0. Arranged in order of increasing magnitude, they are: (A) x,z,y etc."Yes, that's correct. So according to my analysis, (A) is correct. But wait, let me check with x = 0.99 again. If x = 0.99, then:x = 0.99y = x^x ≈ 0.99^0.99 ≈ 0.990099z = x^{x^x} ≈ 0.99^{0.990099} ≈ 0.990099005Wait, actually, in reality, these values are so close that maybe z is slightly greater than y? Wait, no. Let's compute with more precision.Take x = 0.99.Compute y = x^x:Let me compute ln(0.99) = -0.01005034.So ln(y) = x * ln(x) = 0.99 * (-0.01005034) ≈ -0.00995.Therefore, y ≈ e^{-0.00995} ≈ 1 - 0.00995 + 0.00995²/2 - 0.00995³/6 ≈ approximately 0.990099.Now compute z = x^{y} = 0.99^{0.990099}.Compute ln(z) = y * ln(x) ≈ 0.990099 * (-0.01005034) ≈ -0.009956.Thus, z ≈ e^{-0.009956} ≈ 1 - 0.009956 + (0.009956)^2/2 - (0.009956)^3/6 ≈ 0.990099.Wait, but with more precise calculation, perhaps?Let me compute e^{-0.009956}:The exponent is -0.009956.Using Taylor series around 0:e^a ≈ 1 + a + a²/2 + a³/6 + a^4/24.So:1 + (-0.009956) + (0.009956)^2/2 + (-0.009956)^3/6 + (0.009956)^4/24Compute each term:1st term: 12nd term: -0.009956 ≈ -0.0099563rd term: (0.00009912)/2 ≈ 0.000049564th term: (-0.000000986)/6 ≈ -0.0000001645th term: (0.00000000982)/24 ≈ 0.00000000041Adding up:1 - 0.009956 = 0.9900440.990044 + 0.00004956 ≈ 0.990093560.99009356 - 0.000000164 ≈ 0.99009340.9900934 + 0.00000000041 ≈ 0.9900934So z ≈ 0.9900934.Compare with y ≈ 0.990099.Therefore, z ≈ 0.9900934 < y ≈ 0.990099. So even here, z is slightly less than y, so order is x < z < y. Therefore, even at x = 0.99, the order holds.Therefore, in all cases in (0.9, 1), x < z < y, so answer (A). But wait, the original options: (A) x, z, y; (B) x, y, z; (C) y, x, z; (D) y, z, x; (E) z, x, y. So (A) is x < z < y.But wait, looking back at the problem statement: "arranged in order of increasing magnitude". So if it's increasing magnitude, the order is from smallest to largest. So if x < z < y, then the arrangement is x, z, y, which is option (A). So that should be the correct answer. However, let me check once more with another value of x, say x = 0.93.Compute x = 0.93.Compute y = x^x:ln(0.93) ≈ -0.07257Thus, ln(y) = 0.93 * (-0.07257) ≈ -0.0675Thus, y ≈ e^{-0.0675} ≈ 1 - 0.0675 + 0.0675²/2 ≈ 0.9348.Then z = x^{y} = 0.93^{0.9348}ln(z) = 0.9348 * ln(0.93) ≈ 0.9348 * (-0.07257) ≈ -0.0679Thus, z ≈ e^{-0.0679} ≈ 1 - 0.0679 + (0.0679)^2 / 2 ≈ 0.9343.Therefore, x = 0.93, z ≈ 0.9343, y ≈ 0.9348. So order is x < z < y, again. So (A) is the correct answer.But wait, according to the options, (A) is x, z, y. So that's correct. Therefore, the answer should be (A). However, let me check if there is any possibility that for some x in (0.9, 1), z could be less than x, which would change the order. For example, take x = 0.95: z ≈ 0.9523, which is greater than x = 0.95. So no, z is always greater than x. Therefore, x is the smallest, then z, then y. So the order is x < z < y, which is option (A). Therefore, I think the correct answer is (A).But wait, the original problem lists the options as (A) x,z,y; (B) x,y,z; etc. So if the answer is (A), why is the option (B) present? Maybe I made a mistake. Let me think again.Wait, perhaps I misapplied the exponentiation. Let me check z = x^{x^x}.Wait, is z = x^(x^x) or (x^x)^x? The problem says z = x^{x^x}, which is x^(x^x). Exponentiation is right-associative, so x^{x^x} = x^(x^x). So yes, that's correct. So z is x raised to the x^x power. So with x in (0.9,1), and x^x in (0.9,1). Therefore, since x < 1, raising it to a higher power (x^x) which is greater than x, gives a smaller number. Wait, but x^x is greater than x, so the exponent is higher, hence z = x^{x^x} < x^x. However, since x is between 0.9 and 1, and the exponent x^x is between 0.9 and 1, which is a number less than 1. Therefore, z = x^{x^x} is greater than x^{1} = x. Wait, this is a bit confusing.Wait, another approach: Let me set t = x^x, so y = t, and z = x^t. So since x is between 0.9 and 1, t = x^x is between ~0.9 and 1. Then z = x^t. Since x is between 0.9 and 1, and t is between ~0.9 and 1, then z = x^t is between x^1 = x and x^{0.9}. Since x^{0.9} is greater than x (because x < 1 and exponent < 1), so z is between x and x^{0.9}, but x^{0.9} > x. Wait, but in reality, z = x^t where t = x^x. Since t > x (as we saw earlier), then because x < 1, z = x^t < x^x = y. Also, since t = x^x < 1, z = x^t > x^1 = x. Therefore, x < z < y. So yes, that reasoning also shows that z is between x and y.Therefore, the order is x < z < y, which is option (A). Therefore, the answer is (A). But wait, initially, I thought that raising x to a higher power (t = x^x) would make it smaller, so z < y, but since t is still less than 1, then x^t is greater than x. Therefore, z is between x and y.Hence, after all this detailed analysis and multiple examples, I confirm that the correct order is x < z < y, which is option (A).Final Answerboxed{A}</think>Given ( x ) in the interval ( 0.9 < x < 1.0 ), we need to arrange ( x ), ( y = x^x ), and ( z = x^{x^x} ) in increasing order.1. Understanding ( y = x^x ): - For ( x ) in ( (0, 1) ), ( x^x ) is greater than ( x ). This is because raising a number between 0 and 1 to a power less than 1 results in a value greater than the original number. For example, ( 0.9^{0.9} approx 0.91 ), which is greater than 0.9.2. Understanding ( z = x^{x^x} ): - Since ( x ) is between 0.9 and 1, and ( x^x ) is slightly less than 1 (but greater than ( x )), raising ( x ) to the power ( x^x ) (which is a number slightly less than 1) results in a value greater than ( x ) but less than ( x^x ). For example, ( 0.9^{0.91} approx 0.909 ), which is between 0.9 and 0.91.3. Order Analysis: - Through various examples and calculations, it is shown that ( x < z < y ). For instance: - If ( x = 0.95 ), ( y approx 0.9524 ) and ( z approx 0.9523 ), so ( x < z < y ). - If ( x = 0.99 ), ( y approx 0.9901 ) and ( z approx 0.990099 ), so ( x < z < y ).4. Conclusion: - The order of increasing magnitude is ( x < z < y ).Thus, the correct answer is (boxed{A}).]

🔑:fbox{A}

🧠:[<think>Okay, let's tackle this geometry problem step by step. So, we have a pyramid ABCDE with a square base ABCD. Point E is directly above A, and the edges AE and AB are given as 1024 and 640 units long, respectively. The pyramid is cut by a horizontal plane parallel to the base, creating a smaller pyramid on top. We need to find how many integer values of h (the distance from the base to the cutting plane) result in the volume of the new pyramid being an integer.First, let's visualize the pyramid. Since ABCD is a square base, all sides AB, BC, CD, and DA are equal, each 640 units. Point E is directly above A, meaning that the height of the pyramid is AE = 1024 units. So, the apex E is vertically above point A, making this a square-based pyramid but not a regular pyramid since the apex isn't over the center of the base.Now, when the pyramid is cut by a horizontal plane parallel to the base, the resulting cross-section is a smaller square. The new pyramid formed on top will be similar to the original pyramid because the cut is parallel, so all corresponding dimensions are scaled by the same factor.The key idea here is that the volume of similar solids scales with the cube of the linear dimensions. So, if the height of the new pyramid is k times the height of the original pyramid, its volume will be k³ times the original volume. However, in this case, the height from the cutting plane to the apex is not h but rather the original height minus h. Wait, let's be careful here.Wait, the original pyramid has height AE = 1024. But since E is directly above A, the height of the pyramid is 1024 units. However, when we make a horizontal cut at a distance h above the base, the height from the cutting plane to the apex E is 1024 - h. But since the cutting plane is parallel to the base, the height of the new pyramid (from the cutting plane to E) will be (1024 - h). But the original pyramid's height is 1024, so the scaling factor k is (1024 - h)/1024. Therefore, the volume of the new pyramid is k³ times the original volume.But wait, hold on. Let me confirm this. If the original pyramid has height H, and we cut it at height h above the base, the remaining height from the cut to the apex is H - h. Since the pyramids are similar, the ratio of their heights is (H - h)/H, so the volume scales by [(H - h)/H]^3. Therefore, the volume of the new pyramid is [(1024 - h)/1024]^3 times the original volume. But the problem states that this volume must be an integer. So, we need [(1024 - h)/1024]^3 * V_original is an integer.But first, let's compute the original volume V_original. Since ABCD is a square with side length 640, the area of the base is 640². The height of the pyramid is 1024. The volume of a pyramid is (1/3)*base area*height. Therefore, V_original = (1/3)*(640^2)*1024. Let me compute that:V_original = (1/3)*(640*640)*(1024). Let's compute 640*640 first: 640^2 = 409600. Then, 409600*1024 = 409600*1024. Let me compute that: 409600*1024 = 409600*1000 + 409600*24 = 409,600,000 + 9,830,400 = 419,430,400. Then, divide by 3: V_original = 419,430,400 / 3 ≈ 139,810,133.333... So, it's not an integer. Wait, but the problem says the new pyramid's volume must be an integer. So, [(1024 - h)/1024]^3 * (419,430,400 / 3) must be an integer.Let me write that out:Volume_new = [(1024 - h)/1024]^3 * (419,430,400 / 3)We need Volume_new to be an integer. Let's denote k = (1024 - h)/1024. Then, Volume_new = k³ * (419,430,400 / 3). For this to be an integer, k³ must be such that when multiplied by 419,430,400 / 3, the result is an integer.Alternatively, we can express k as a fraction. Since k = (1024 - h)/1024, let's let k = t/1024, where t = 1024 - h. Then, Volume_new = (t/1024)^3 * (419,430,400 / 3) = (t³ / 1024³) * (419,430,400 / 3). Let's compute this expression:First, 1024³ is (2^10)^3 = 2^30. 419,430,400 is 640² * 1024. Let's factor these numbers:640 = 64*10 = 2^6 * 5 * 2 = 2^7 * 5. Therefore, 640² = (2^7 *5)^2 = 2^14 *5².1024 = 2^10. Therefore, 640² *1024 = 2^14 *5² *2^10 = 2^24 *5². Then, 419,430,400 = 2^24 *5². Therefore, V_original = (2^24 *5²)/3. Therefore, Volume_new = (t³ / 2^30) * (2^24 *5² /3) = (t³ * 2^24 *5²) / (3 *2^30) ) = (t³ *5²) / (3 *2^6) ) = (25 t³)/(3*64) = (25 t³)/(192). So, Volume_new = (25 t³)/(192). For this to be an integer, 192 must divide 25 t³. Since 25 and 192 are coprime (their GCD is 1), 192 must divide t³. Therefore, t³ must be a multiple of 192.But 192 factors as 2^6 *3. So, t³ must have at least 2^6 *3 in its prime factors. Let's factor t³. If t is an integer, then t³'s prime factors will have exponents that are multiples of 3. So, for t³ to have 2^6 *3, t must have at least 2^(2) *3^(1) in its prime factors. Because when you cube t, 2^2 becomes 2^6 and 3^1 becomes 3^3. Wait, but 2^6 is required, so to get 2^6 in t³, t needs to have at least 2^(6/3) = 2^2. Similarly, 3 is required in t³, so t needs to have at least 3^(1/3). But since t must be an integer, exponents must be integers, so t must have at least 3^1.Wait, no. If t³ must be divisible by 2^6 *3, then t must have prime factors such that:For the prime 2: exponent in t must be at least ceiling(6/3) = 2. So, 2^2.For the prime 3: exponent in t must be at least ceiling(1/3) = 1. So, 3^1.Therefore, t must be a multiple of 2^2 *3 = 12. Therefore, t = 12k, where k is an integer. Therefore, t is a multiple of 12.But t = 1024 - h, and h is a distance above the base, so h must be between 0 and 1024 (since the plane is cutting the pyramid, h can't be more than the height of the pyramid). Therefore, t must be between 0 and 1024. Since h is an integer, t = 1024 - h is also an integer between 0 and 1024.But since t must be a multiple of 12, let's find all integers t such that t = 12k, where k is an integer, and 0 < t ≤ 1024. The upper limit is t ≤ 1024 because h ≥ 0. However, when h = 0, t = 1024, which would correspond to the original pyramid. But since the problem says the portion above the plane is a new pyramid, h must be less than 1024. So, h can be from 0 up to 1023, but since t = 1024 - h, t ranges from 1 to 1024. But t must be a multiple of 12, so t can be 12, 24, ..., up to the largest multiple of 12 less than or equal to 1024.Wait, but t must be at least 12*1 = 12. But if t is allowed to be 12, then h = 1024 - 12 = 1012. But is t allowed to be 0? If t=0, then h=1024, but h=1024 would mean the cutting plane is at the apex, so the "new pyramid" would have zero volume. However, the problem states "the portion above the plane is a new pyramid", which probably implies that h must be such that the cutting plane is strictly below the apex, so t must be at least 1. Therefore, t ranges from 1 to 1024. However, since t must be a multiple of 12, let's find the number of multiples of 12 in that range.Wait, but earlier conclusion was that t must be a multiple of 12 (i.e., t=12k) for t³ to be divisible by 192. Is that correct?Wait, let's check again. We had Volume_new = (25 t³)/192 must be integer. So 192 divides 25 t³. Since 25 and 192 are coprime, 192 divides t³. So, t³ must be divisible by 192. Let's factor 192: 192 = 2^6 * 3. Therefore, in t³, the exponent of 2 must be at least 6 and exponent of 3 must be at least 1. Therefore, in t, the exponent of 2 must be at least 2 (since 2^2 in t gives 2^6 in t³) and exponent of 3 must be at least 1 (since 3^1 in t gives 3^3 in t³). Therefore, t must be divisible by 2^2 *3 = 12. Therefore, t must be a multiple of 12. Therefore, t can be 12, 24, 36, ..., up to the largest multiple of 12 less than or equal to 1024.Now, how many multiples of 12 are there between 1 and 1024 inclusive?Wait, t starts at 12 (since t=12k, k=1,2,...) up to the maximum t such that t ≤ 1024. So the maximum k is floor(1024/12). Let's compute 1024 divided by 12: 12*85 = 1020, so 12*85 = 1020, which is less than 1024. The next multiple is 12*86=1032, which is more than 1024. Therefore, the largest multiple of 12 less than or equal to 1024 is 1020. Therefore, the number of multiples of 12 from 12 to 1020 inclusive is 85. So, there are 85 possible values of t. Therefore, h = 1024 - t, so each t corresponds to an h. Therefore, h = 1024 -12k where k=1,2,...,85. Then, h would range from 1024 -12*1=1012 down to 1024 -12*85=1024 -1020=4. So, h would take values 4, 16, 28, ..., 1012. Wait, 1024 -12*85=1024-1020=4. But wait, when k=85, t=1020, so h=1024-1020=4. So, h starts at 4 and goes up to 1012 in steps of 12. But the problem asks for how many integer h's, so how many h's are there? Since t starts at 12 (k=1) and ends at 1020 (k=85), that gives 85 values. Therefore, there are 85 integer h's such that Volume_new is an integer.Wait, but hold on. Let's check if this is correct. Suppose t is a multiple of 12, so t=12k. Then, Volume_new = (25*(12k)^3)/192 = (25*1728k³)/192 = (25*1728)/192 *k³. Let's compute 1728/192: 1728 ÷ 192 = 9. Therefore, Volume_new = 25*9*k³ = 225k³. So, Volume_new is 225k³, which is an integer since k is an integer. Therefore, yes, if t is a multiple of 12, then Volume_new is an integer. Therefore, the number of such h is equal to the number of multiples of 12 in t=12k where t ≤1024. As we found, t can be from 12 to 1020 in steps of 12, which gives 85 terms.But wait, h=1024 - t. So, h would be from 1024 -12=1012 down to 1024 -1020=4. So, h=4,16,...,1012. But these h values are spaced by 12 units apart. So, starting at 4 and adding 12 each time. Let's check how many terms that is.The sequence is 4,16,28,...,1012. The difference between consecutive terms is 12. The formula for the nth term is a_n = 4 + (n-1)*12. We set a_n =1012: 4 +12(n-1)=1012 ⇒12(n-1)=1008⇒n-1=84⇒n=85. So, yes, 85 terms. Therefore, the answer should be 85.But wait, before concluding, let's verify with an example. Let's take k=1, t=12, h=1024-12=1012. Then Volume_new=225*1=225, which is an integer. If k=2, t=24, h=1024-24=1000, Volume_new=225*8=1800, which is integer. Similarly, for k=85, t=1020, h=4, Volume_new=225*(85)^3. Since 85 is an integer, 85³ is an integer, so Volume_new is integer. So, all these h's result in integer volumes.But wait, the problem statement says "the portion of ABCDE that is above the plane is a new pyramid". So, h must be strictly less than 1024, as h=1024 would result in the plane passing through E, making the "new pyramid" have zero volume. Therefore, h can be from 0 up to 1023. However, in our solution, the smallest h is 4 and the largest is 1012. Wait, but h=0 would correspond to t=1024, but t=1024 is a multiple of 12? 1024 ÷12 ≈85.333... So 12*85=1020, so t=1024 is not a multiple of 12. Therefore, h=0 is not included in our count. So, the possible h's are from 4 to 1012, multiples of 12 subtracted from 1024, which gives 85 values.Wait, but the problem says "For how many integers h is the volume of the new pyramid an integer?" So, h must be an integer. Our answer is 85. However, let's check if there are other possible h's where Volume_new is integer, even if t is not a multiple of 12. But according to our earlier reasoning, Volume_new =25 t³ /192. So, for this to be integer, 192 must divide 25 t³. Since GCD(25,192)=1, 192 must divide t³. Which requires t³ divisible by 192=2^6 *3. Therefore, t must be divisible by 12, as we concluded. Therefore, only multiples of 12 for t. Hence, 85 values. Therefore, the answer is 85.But wait, let's check the edge case when h=0. At h=0, the cutting plane is the base itself, so the "new pyramid" would be the entire original pyramid, whose volume is (1/3)*640²*1024, which we computed earlier as approximately 139,810,133.333..., which is not an integer. Therefore, h=0 is excluded because the volume is not an integer, which aligns with our previous result.Similarly, h=4 gives t=1020, which is a multiple of 12, and Volume_new=225*85³. Let's compute 85³: 85*85=7225, 7225*85=614,125. Then, 225*614,125=138,178,125, which is an integer. So, that's correct.Therefore, the number of integer h's is 85. But let me just verify once more. The multiples of 12 for t from 12 to 1020 inclusive. The number of terms is (1020 -12)/12 +1 = (1008)/12 +1=84 +1=85. Correct. Therefore, the answer is 85.Final Answerboxed{85}</think>Given a pyramid (ABCDE) with a square base (ABCD) and apex (E) directly above (A), where (AE = 1024) and (AB = 640). The pyramid is cut by a horizontal plane parallel to the base, creating a smaller pyramid. We need to find the number of integer values (h) such that the volume of the new pyramid is an integer.1. Volume of the Original Pyramid: - The base area is (640^2). - The height of the pyramid is (1024). - Volume (V_{text{original}} = frac{1}{3} times 640^2 times 1024 = frac{419430400}{3}).2. Scaling Factor: - The new pyramid's height is (1024 - h). - The scaling factor (k = frac{1024 - h}{1024}). - Volume of the new pyramid (V_{text{new}} = k^3 times V_{text{original}}).3. Integer Volume Requirement: - The volume (V_{text{new}} = left(frac{1024 - h}{1024}right)^3 times frac{419430400}{3}) must be an integer. - Simplifying, (V_{text{new}} = frac{25(1024 - h)^3}{192}). - For (V_{text{new}}) to be an integer, ((1024 - h)^3) must be divisible by (192).4. Prime Factorization: - (192 = 2^6 times 3). - ((1024 - h)) must be a multiple of (12) (since (12 = 2^2 times 3) ensures divisibility by (192) when cubed).5. Range of (h): - (1024 - h) must be a multiple of (12) and between (1) and (1024). - The smallest multiple of (12) is (12) and the largest is (1020). - The number of multiples of (12) between (12) and (1020) is (frac{1020 - 12}{12} + 1 = 85).Thus, the number of integer values (h) such that the volume of the new pyramid is an integer is (boxed{85}).,<think>Okay, so I need to solve this geometry problem about a pyramid with a square base. Let me first visualize the pyramid. It's called ABCDE, and the base is ABCD, which is a square. Point E is directly above A, so the apex of the pyramid is right above vertex A. The height of the pyramid, which is the distance from E to the base ABCD, is given as AE = 1024. The side length of the square base AB is 640 units.The problem says that the pyramid is cut by a horizontal plane parallel to the base ABCD. This plane is at a distance h above the base. The part of the pyramid above this plane is a new pyramid. We need to find how many integer values of h make the volume of this new pyramid an integer.Hmm. So, let me recall that when you cut a pyramid with a plane parallel to its base, the resulting smaller pyramid is similar to the original one. That means all the linear dimensions are scaled by the same factor. The volume of similar solids scales with the cube of the linear scaling factor.So, if the original pyramid has a volume V, then the new pyramid's volume will be V' = V * (k)^3, where k is the scaling factor. But in this problem, the height of the original pyramid is 1024, and the height from the cutting plane to the apex is (1024 - h). Wait, but the problem states the plane is a distance h above the base, so the height from the plane to the apex would be 1024 - h. Therefore, the scaling factor k would be (1024 - h)/1024, since the heights in similar pyramids are proportional.But wait, actually, since the original pyramid's height is 1024, and the new pyramid's height is (1024 - h), the scaling factor is (1024 - h)/1024. Therefore, the volume of the new pyramid would be the original volume times [(1024 - h)/1024]^3. But we need this volume to be an integer. However, the problem says "the portion of ABCDE that is above the plane is a new pyramid," so maybe the new pyramid's volume is [(1024 - h)/1024]^3 times the original volume, right?Wait, but we also need to compute the original volume. The original pyramid has a square base with side length 640, so the area of the base is 640^2. The height is 1024, so the volume V is (1/3) * 640^2 * 1024. Then the new pyramid's volume would be (1/3) * (640 * k)^2 * (1024 * k) = (1/3) * 640^2 * 1024 * k^3, which is indeed V * k^3. So, that checks out.Therefore, the volume of the new pyramid is V * [(1024 - h)/1024]^3. We need this to be an integer. But the original volume V is (1/3)*640^2*1024. Let me compute that first.First, let's compute the original volume:V = (1/3) * (640)^2 * 1024640^2 is 409600. Then, multiply by 1024: 409600 * 1024 = 409600 * 1024. Let me compute that step by step.409600 * 1000 = 409,600,000409600 * 24 = 409600 * 20 + 409600 * 4 = 8,192,000 + 1,638,400 = 9,830,400So total V = (1/3)*(409,600,000 + 9,830,400) = (1/3)*(419,430,400) = 139,810,133.333...Wait, so V is not an integer? Wait, 640 is 64*10, so 640^2 is 409600, which is 64^2 * 10^2 = 4096 * 100. Then 409600 * 1024 is 4096 * 100 * 1024. 4096 * 1024 is 4,194,304, so 4,194,304 * 100 = 419,430,400. Then, divide by 3: 419,430,400 / 3 ≈ 139,810,133.333... So yes, V is not an integer. Therefore, the volume of the new pyramid is (419,430,400 / 3) * [(1024 - h)/1024]^3. We need this to be an integer.Therefore, (419,430,400 / 3) * [(1024 - h)/1024]^3 must be an integer. Let's denote k = (1024 - h)/1024, so k is a scaling factor between 0 and 1, because h is between 0 and 1024 (since the plane is above the base and below the apex). Then, the volume is (419,430,400 / 3) * k^3. For this to be an integer, 419,430,400 * k^3 must be divisible by 3. Since 419,430,400 is equal to 419,430,400. Let's check if this number is divisible by 3.Sum of its digits: 4+1+9+4+3+0+4+0+0 = 4+1=5, 5+9=14, 14+4=18, 18+3=21, 21+0=21, 21+4=25, 25+0=25, 25+0=25. 25 is not divisible by 3, so 419,430,400 is not divisible by 3. Therefore, (419,430,400 / 3) is not an integer, which we already saw. So, in order for the entire expression (419,430,400 / 3) * k^3 to be an integer, k^3 must be a multiple of 3. Wait, but k is a rational number since h is an integer, so k = (1024 - h)/1024. Therefore, k^3 = [(1024 - h)/1024]^3. Let's write this as [(1024 - h)^3]/1024^3. Then, multiplying by 419,430,400 / 3 gives:(419,430,400 / 3) * [(1024 - h)^3]/1024^3Let me compute 419,430,400 / 1024^3. Let's compute 1024^3 first.1024^3 = (2^10)^3 = 2^30. Let's confirm: 2^10=1024, so 1024^3 = 2^30 = 1,073,741,824.But 419,430,400 divided by 1,073,741,824 is equal to:419,430,400 ÷ 1,073,741,824Let me compute this. Let's note that 419,430,400 is 409,600 * 1024. 409,600 is 640^2. Alternatively, 419,430,400 divided by 1,073,741,824: both numbers are divisible by 1024. Let's divide numerator and denominator by 1024:419,430,400 / 1024 = 409,6001,073,741,824 / 1024 = 1,048,576So now, 409,600 / 1,048,576Simplify further. Let's divide numerator and denominator by 4096:409,600 / 4096 = 1001,048,576 / 4096 = 256So now it's 100 / 256 = 25 / 64.Therefore, 419,430,400 / 1,073,741,824 = 25/64.Therefore, the volume expression simplifies to:(25/64) / 3 * (1024 - h)^3 = (25/192) * (1024 - h)^3So, Volume = (25/192) * (1024 - h)^3We need this volume to be an integer. Therefore, (25/192) * (1024 - h)^3 must be integer. Let's denote x = 1024 - h. Then, x ranges from 1 to 1023 (since h is an integer distance above the base, so h >= 1, and h < 1024, since the plane is below the apex). Wait, but h is the distance above the base, so h can be 0, but if h=0, the plane is the base itself, so the "new" pyramid would be the entire pyramid. But the problem says "the portion above the plane", so h must be from 1 to 1023 inclusive. So x = 1024 - h, which would be from 1 to 1023. Therefore, x is an integer between 1 and 1023.Therefore, the volume is (25/192) * x^3, and we need this to be an integer. So (25/192) * x^3 is integer. Let's write this as (25 x^3)/192 must be integer. Therefore, 192 must divide 25 x^3. Since 25 and 192 are coprime (25 is 5^2, 192 is 2^6 * 3), so 192 divides x^3. Therefore, x^3 must be divisible by 192, which is 2^6 * 3. Therefore, x^3 must be divisible by 2^6 * 3. Let's factor x^3. For x^3 to be divisible by 2^6 * 3, x must be divisible by 2^(6/3) * 3^(1/3) = 2^2 * 3^(1/3). But since x must be an integer, 3^(1/3) is not an integer, so x must be divisible by 2^2 * 3 = 12. Wait, let me check:If x is divisible by 12, then x = 12k, so x^3 = (12k)^3 = 12^3 k^3 = 1728 k^3. Then 1728 is 2^6 * 3^3. Therefore, 1728 is divisible by 192 (since 192 = 2^6 * 3, 1728 = 2^6 * 3^3). Therefore, x^3 must be divisible by 2^6 * 3. Since 3 is a prime, x must be divisible by 2^2 and 3. Because in x^3, the exponent of 2 in x must be at least 2, since (2^2)^3 = 2^6, which is sufficient. For the prime 3, the exponent of 3 in x must be at least 1, since 3^1 in x would give 3^3 in x^3, but we only need 3^1. Wait, but x^3 must have at least 3^1. Therefore, x must be divisible by 3, because if x is not divisible by 3, then x^3 is not divisible by 3. Similarly, x must be divisible by 2^2 = 4. Therefore, x must be divisible by lcm(4,3) = 12. Therefore, x must be a multiple of 12.Therefore, x ∈ {12, 24, ..., 1023}. But wait, 1024 - h = x, so h = 1024 - x. Since h must be at least 1, x must be at most 1023. The maximum multiple of 12 less than or equal to 1023 is floor(1023/12)*12. Let's compute 1023 divided by 12: 12*85 = 1020, so 85*12 = 1020, which is less than 1023. The next multiple would be 86*12 = 1032, which is too big. So the maximum x is 1020. Therefore, x ranges from 12 to 1020 in steps of 12. So the number of possible x is (1020 - 12)/12 + 1 = (1008)/12 + 1 = 84 + 1 = 85. Wait, but hold on. Let me check.Wait, if x starts at 12 and goes up to 1020, inclusive, stepping by 12 each time, then the number of terms is ((1020 - 12)/12) + 1 = (1008 /12) + 1 = 84 +1 =85. So 85 possible x's. Therefore, h = 1024 - x would be h = 1024 - 12k, where k ranges from 1 to 85. Wait, but when x=12, h=1024-12=1012, and when x=1020, h=1024-1020=4. Wait, but h must be at least 1. When x=1023, h=1, but x=1023 is not a multiple of 12. The minimal x that is a multiple of 12 is 12, which gives h=1024-12=1012. But then, if x=1020, h=4. So h would go from 4 to 1012 in steps of 12? Wait, but 1020 is x, so h=1024 - 1020=4. Then x=12, h=1012. So the h values would be 4, 16, 28, ..., 1012? Wait, stepping by 12. Wait, no. Wait, if x is multiples of 12, then h = 1024 - x, so each h is 1024 minus a multiple of 12. So h would be 1024 - 12, 1024 -24, ..., 1024 -1020. So 1024 -12=1012, 1024-24=1000, ..., 1024 -1020=4. So h=1012, 1000, ..., 4. So these are h values decreasing by 12 each time. So starting at 1012, down to 4, step of -12. How many terms?Let me compute the number of terms. From 4 to 1012 inclusive, step of 12. The formula for the number of terms is ((last - first)/step) +1. So ((1012 -4)/12)+1 = (1008/12)+1=84 +1=85. So 85 terms. Therefore, there are 85 integer values of h where the volume is an integer. But wait, but the problem says "the portion of ABCDE that is above the plane is a new pyramid". So h is the distance above the base, so h must be less than 1024. So h can be 1 up to 1023. But our calculation gives h from 4 to 1012 in steps of 12, which are 85 values. However, hold on. The problem states h is an integer, but does it specify that h has to be such that the new pyramid is non-degenerate? Well, if h=0, the plane is the base, but the problem says "the portion above the plane", so h must be at least 1. But if x=1023, then h=1. But x=1023 is not a multiple of 12, so h=1 is not included in our 85 values. Similarly, h=2,3, etc., up to h=3, are not included. So the minimal h in our case is 4. Therefore, the answer would be 85.But wait, hold on. Let me check the logic again. The key point is that (25 x^3)/192 must be integer. So 192 divides 25 x^3. Since 25 and 192 are coprime (GCD(25,192)=1), this implies that 192 divides x^3. Therefore, x^3 must be a multiple of 192=2^6*3. Therefore, x must be divisible by 2^2 and 3, as I thought before. So x must be divisible by 12. Therefore, x=12k, where k is an integer. Then, x can be 12, 24,..., up to the maximum multiple of 12 less than or equal to 1023. Since 1023 divided by 12 is 85.25, so the maximum k is 85, so x=12*85=1020. Therefore, x can be 12*1, 12*2,...,12*85=1020. So 85 values. Therefore, h=1024 -x, so h=1024 -12*1=1012, h=1024 -12*2=1000, ..., h=1024 -12*85=1024 -1020=4. So h=4,16,28,...,1012. Wait, no. Wait, 1024 -12*1=1012, 1024 -12*2=1000, 1024 -12*3=988, ..., 1024 -12*85=1024 -1020=4. So h decreases by 12 each time. Therefore, the h values are 1012,1000,...,4. How many terms?As computed earlier, from 4 to 1012 inclusive, step of 12. The number of terms is ((1012 -4)/12)+1=(1008/12)+1=84+1=85. Therefore, 85 integer values. Therefore, the answer is 85. But wait, the problem is in the translation from x to h. So x must be divisible by 12, hence h=1024 -x must be congruent to 1024 mod 12. Let me check 1024 divided by 12: 12*85=1020, so 1024=12*85 +4. Therefore, 1024 ≡4 mod12. Therefore, h=1024 -x ≡4 -0 mod12, since x is divisible by12. Therefore, h ≡4 mod12. Therefore, h=4,16,28,...,1012. So starting at 4, each subsequent h is 12 more. Wait, but 1012 -4=1008, which is 84*12, so the number of terms is 84+1=85. So 85 terms. Therefore, the answer is 85.But let me confirm with an example. Let's take h=4. Then x=1020. Then volume is (25/192)*(1020)^3. Let's check if this is integer.First, compute 1020^3. 1020=1000+20. 1020^3 = (1000+20)^3 = 1000^3 +3*1000^2*20 +3*1000*20^2 +20^3 = 1,000,000,000 + 60,000,000 + 1,200,000 + 8,000 = 1,061,208,000.Then, (25/192)*1,061,208,000. Let's compute 1,061,208,000 /192 first. 1,061,208,000 ÷192.Divide numerator and denominator by 16: 1,061,208,000 ÷16=66,325,500; 192 ÷16=12.So now, 66,325,500 /12=5,527,125.Multiply by25: 5,527,125 *25=138,178,125. Which is an integer. So yes, for h=4, the volume is 138,178,125, which is integer. Similarly, take h=16, then x=1024 -16=1008. x=1008, which is divisible by12 (1008=12*84). Then x^3=1008^3. 1008=1000+8. 1008^3=1,000,000,000 +3*1,000,000*8 +3*1,000*8^2 +8^3=1,000,000,000 +24,000,000 + 192,000 + 512=1,024,192,512. Then (25/192)*1,024,192,512. Let's divide 1,024,192,512 by192. 1,024,192,512 ÷192. Again, divide numerator and denominator by16: 1,024,192,512 ÷16=64,012,032; 192 ÷16=12. Then 64,012,032 ÷12=5,334,336. Multiply by25: 5,334,336*25=133,358,400, which is integer. So this works.Therefore, the logic holds. So each h that is congruent to4 mod12, from4 to1012 inclusive, gives an integer volume. So the number of such h is85. Therefore, the answer is85.Wait, but let me check if x must be divisible by12. Suppose x is divisible by 12, then x=12k. So x^3=12^3k^3=1728k^3. Then 25x^3/192=25*1728k^3/192=25*9k^3=225k^3, which is integer. Therefore, yes, if x is divisible by12, then 25x^3/192=225k^3, which is integer. So that works. But is this the only possibility?Wait, 192 divides25x^3. Since GCD(25,192)=1, then 192 dividesx^3. So x^3 must be divisible by192=2^6*3. Therefore, x must be divisible by2^2 and3, which is equivalent tox divisible by12. Therefore, yes, only x divisible by12 would satisfy this. Therefore, the answer is85.Wait, but let me check if x=24. x=24, then h=1024-24=1000. Then volume is (25/192)*(24)^3. 24^3=13,824. 13,824*25=345,600. 345,600/192=1,800. Which is integer. So yes, works. So the logic is correct.Therefore, the number of integers h is85. Therefore, the answer is boxed{85}.Final Answerboxed{85}</think>Given a square-based pyramid (ABCDE) with (ABCD) as the base and (E) directly above (A). The height (AE = 1024) and the side length (AB = 640). The pyramid is cut by a horizontal plane parallel to the base at a distance (h) above the base, forming a new pyramid. We need to determine how many integer values of (h) result in an integer volume for the new pyramid.1. Volume of the Original Pyramid: - The volume (V) of the original pyramid is calculated as: [ V = frac{1}{3} times 640^2 times 1024 = frac{1}{3} times 409600 times 1024 = frac{419430400}{3} ]2. Volume of the New Pyramid: - The new pyramid is similar to the original pyramid with a scaling factor (k = frac{1024 - h}{1024}). - The volume of the new pyramid is: [ V' = V times k^3 = frac{419430400}{3} times left(frac{1024 - h}{1024}right)^3 ]3. Condition for Integer Volume: - The volume (V') must be an integer. This requires: [ frac{419430400}{3} times left(frac{(1024 - h)^3}{1024^3}right) text{ is an integer} ] - Simplifying, we find: [ frac{25 times (1024 - h)^3}{192} text{ must be an integer} ] - This implies (192) must divide (25 times (1024 - h)^3). Since (25) and (192) are coprime, (192) must divide ((1024 - h)^3).4. Factorization and Divisibility: - (192 = 2^6 times 3). Therefore, ((1024 - h)^3) must be divisible by (2^6 times 3). - For ((1024 - h)^3) to be divisible by (2^6 times 3), (1024 - h) must be divisible by (2^2 times 3 = 12).5. Range and Count of Valid (h): - Let (x = 1024 - h). Then (x) must be a multiple of (12) and (1 leq x leq 1023). - The valid values of (x) are (12, 24, 36, ldots, 1020). - The number of terms in this sequence is (leftlfloor frac{1023}{12} rightrfloor = 85).Thus, the number of integer values of (h) for which the volume of the new pyramid is an integer is (boxed{85}).]

🔑:We first find an expression for the volume of the top pyramid in terms of h.The volume of any pyramid is equal to frac{1}{3} times (area of base) times (height).The height of the upper pyramid is 1024-h.Suppose the plane intersects E A at F and E B at G.Since the plane is parallel to the base, we have that F G is parallel to A B.This means triangle E F G is similar to triangle E A B, which means frac{F G}{A B}=frac{E F}{E A}. We know that A B=640, E A=1024, and E F=1024-h, so substitution and rearranging leads toF G=frac{(1024-h)(640)}{1024}=frac{5(1024-h)}{8} .Next, we show that the base of the new pyramid is a square.Label by H the point at which the plane intersects E C and I the point at which the plane intersects E D.We have already observed that F G is parallel to A B. Similarly, G H is parallel to B C, H I is parallel to C D, and I F is parallel to D A.This means F G is parallel to H I and perpendicular to I F, which is parallel to G H, so F G H I is a rectangle.To see that F G H I is a square, we will show that I F=F G.We have already observed that triangle E F G is similar to triangle E A B. For similar reasons, triangle E F I is similar to triangle E A D.Since E is directly above A, we have that angle E A B=angle E A D=90^{circ}.As well, A B=A D, so triangle E A B is congruent to triangle E A D by side-angle-side congruence.It follows that triangle E F I is similar to triangle E F G, and since they share the side E F, triangle E F I is congruent to triangle E F G.Therefore, F I=F G, so F G H I is a square.We can now compute the volume of the square-based pyramid FGHIE.The area of the base isF G^{2}=left(frac{5(1024-h)}{8}right)^{2}and the height is 1024-h, so the volume isfrac{1}{3}(1024-h)left(frac{5(1024-h)}{8}right)^{2}=frac{(1024-h)^{3} 5^{2}}{3 times 2^{6}}We now need to count the positive integers h with the property that frac{(1024-h)^{3} 5^{2}}{3 times 2^{6}} is an integer.This is the same as counting the integers h with the property that 3 times 2^{6} is a divisor of (1024-h)^{3} 5^{2}.Since 5 is prime, 5^{2} has no factor of 2 and no factor of 3 . Therefore, we need to count integers h with the property that (1024-h)^{3} is a multiple of 3 times 2^{6}.Since 3 is prime, (1024-h)^{3} is a multiple of 3 exactly when 1024-h is a multiple of 3 .Suppose 1024-h is a multiple of 2^{2}=4. Then (1024-h)^{3} is a multiple 2^{6}.If (1024-h)^{3} is a multiple of 2^{6}, then 1024-h must be even.However, if 1024-h is a multiple of 2 but not a multiple of 4 , then (1024-h)^{3} cannot have 6 copies of 2 in its prime factorization.Therefore, (1024-h)^{6} is a multiple of 2^{6} exactly when (1024-h) is a multiple of 4.We now have shown that the volume of the top pyramid is a positive integer exactly when 1024-h is a multiple of both 3 and 4 .This means that the volume of the top pyramid is a positive integer exactly when 1024-h is a multiple of 12 .When h=4, we have 1024-h=1024-4=1020=12 times 85.If h=4+12, then 1024-h=12 times 84.This continues to give the possibilities h=4, h=4+12, h=4+2 times 12, h=4+3 times 12, and so on.The largest value of h is when 1024-h=12 or h=1012=4+84 times 12.Thus, the values are h=4+k times 12 where k ranges over the integers 0 through 84 .There are 85 integers h for which the volume of the upper pyramid is an integer.ANSWER: 85## Team Problems

🧠:[<think>Okay, so I have this problem here with two parts. Let me start with part (1): finding the equation of the tangent line to the graph of f(x) = x² + x at x = 1. Hmm, tangent lines... right. To find the tangent line at a point, I need two things: the slope of the tangent at that point and a point through which the tangent passes. First, the slope. The slope of the tangent to a function at a particular point is given by the derivative of the function evaluated at that point. So I need to find f'(x). Let's compute that. The derivative of x² is 2x, and the derivative of x is 1. So f'(x) = 2x + 1. Now, evaluating this at x = 1: f'(1) = 2*1 + 1 = 3. So the slope of the tangent line is 3.Next, I need a point on the tangent line. Since the tangent line touches the graph of f(x) at x = 1, I can find the corresponding y-coordinate by plugging x = 1 into f(x). Let's do that: f(1) = (1)² + 1 = 1 + 1 = 2. So the point is (1, 2).Now that I have the slope and a point, I can use the point-slope form of a line to write the equation. The point-slope formula is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point. Plugging in the values: y - 2 = 3(x - 1). Let me simplify that. Distribute the 3 on the right side: y - 2 = 3x - 3. Then, add 2 to both sides: y = 3x - 3 + 2, which simplifies to y = 3x - 1. Wait, let me check that again. If I expand 3(x - 1), it's 3x - 3, then adding 2 gives 3x - 3 + 2 = 3x -1. Yeah, that looks right. So the equation of the tangent line is y = 3x - 1. That seems straightforward. Maybe I can double-check by plugging x =1 into the tangent equation. If x=1, y=3*1 -1=2, which matches the point on f(x). And the slope is 3, which is correct. Okay, part (1) done.Now moving on to part (2): If f(x) ≥ g(x) holds for any x ∈ [-4,4], find the range of the real number m. The functions are f(x) = x² + x and g(x) = (1/3)x³ - 2x + m. So the inequality f(x) - g(x) ≥ 0 must hold for all x in [-4,4]. Let me write that out: x² + x - [(1/3)x³ - 2x + m] ≥ 0. Simplify that expression. Let's distribute the negative sign: x² + x - (1/3)x³ + 2x - m ≥ 0. Combine like terms. First, let's arrange the terms in descending order of powers of x. The highest power here is x³. So:- (1/3)x³ + x² + x + 2x - m ≥ 0.Combine the x terms: x + 2x = 3x. So:- (1/3)x³ + x² + 3x - m ≥ 0.Let me write this as:h(x) = - (1/3)x³ + x² + 3x - m ≥ 0 for all x in [-4,4].Therefore, we need h(x) ≥ 0 for all x ∈ [-4,4]. To find the range of m such that this inequality holds, we can rearrange for m. Let's rewrite h(x):h(x) = - (1/3)x³ + x² + 3x - m ≥ 0.Adding m to both sides:- (1/3)x³ + x² + 3x ≥ m.Therefore, m must be less than or equal to - (1/3)x³ + x² + 3x for all x in [-4,4]. Hence, m should be less than or equal to the minimum value of the function h(x) = - (1/3)x³ + x² + 3x on the interval [-4,4]. So, the range of m is m ≤ minimum of h(x) on [-4,4].Therefore, I need to find the minimum value of h(x) on [-4,4]. Once I find that minimum value, m must be less than or equal to that. So the task reduces to finding the minimum of h(x) over the interval [-4,4].To find the minimum of h(x) on the closed interval, we can use calculus. First, find the critical points by setting h'(x) = 0 and solve for x. Then evaluate h(x) at the critical points and at the endpoints of the interval. The smallest of those values will be the minimum.So let's compute h'(x). Given h(x) = - (1/3)x³ + x² + 3x - m. The derivative h'(x) is:h'(x) = - x² + 2x + 3.Set h'(x) = 0:- x² + 2x + 3 = 0.Multiply both sides by -1 to make it easier:x² - 2x - 3 = 0.Solve for x using quadratic formula:x = [2 ± √(4 + 12)] / 2 = [2 ± √16]/2 = [2 ± 4]/2.Thus, x = (2 + 4)/2 = 6/2 = 3, and x = (2 - 4)/2 = (-2)/2 = -1.So the critical points are at x = 3 and x = -1. Both of these points are within the interval [-4,4], so we need to consider both.Therefore, to find the minimum of h(x) on [-4,4], we evaluate h(x) at x = -4, x = -1, x = 3, and x = 4.Let's compute each:1. At x = -4:h(-4) = - (1/3)*(-4)^3 + (-4)^2 + 3*(-4) - m. Wait, no, hold on. Wait, h(x) was defined as h(x) = - (1/3)x³ + x² + 3x - m. But when we derived that h(x) ≥0 is equivalent to m ≤ - (1/3)x³ + x² + 3x, then m must be less than or equal to the minimum of that expression. Therefore, actually, h(x) is equal to that expression minus m, so the expression that must be ≥0 is - (1/3)x³ + x² + 3x - m. Therefore, to find the minimum of - (1/3)x³ + x² + 3x on the interval, then m must be ≤ that minimum. So actually, when I compute h(x) at different points, m must be ≤ the minimum of h(x) + m? Wait, maybe I need to clarify.Wait, wait. Let me go back. The original inequality is f(x) ≥ g(x). So x² + x ≥ (1/3)x³ - 2x + m. Rearranged, this is x² + x - (1/3)x³ + 2x - m ≥ 0. Which simplifies to - (1/3)x³ + x² + 3x - m ≥ 0. So, h(x) = - (1/3)x³ + x² + 3x - m. The inequality is h(x) ≥ 0 for all x in [-4,4]. Therefore, h(x) must be greater than or equal to 0 for all x in that interval. Therefore, the minimal value of h(x) on [-4,4] must be ≥ 0. Thus, the minimal value of h(x) is ≥ 0. Since h(x) = [ - (1/3)x³ + x² + 3x ] - m, so the minimal value of [ - (1/3)x³ + x² + 3x ] - m ≥ 0. Therefore, m ≤ [ - (1/3)x³ + x² + 3x ] for all x in [-4,4]. Therefore, m must be less than or equal to the minimum of [ - (1/3)x³ + x² + 3x ] over x in [-4,4]. Therefore, m ≤ min_{x ∈ [-4,4]} [ - (1/3)x³ + x² + 3x ].Therefore, if I let k(x) = - (1/3)x³ + x² + 3x, then m ≤ min k(x) over [-4,4]. Therefore, I need to find the minimum of k(x) on that interval, and m has to be less than or equal to that. So, my task is to compute k(x) at the critical points and endpoints and find the minimum value.So, k(x) = - (1/3)x³ + x² + 3x.We already found the critical points of h(x), which is the same as the critical points of k(x), since h(x) = k(x) - m, and the derivative h'(x) = k'(x). So the critical points of k(x) are x = -1 and x = 3, as found earlier.Therefore, to compute the minimum of k(x) on [-4,4], we evaluate k(x) at x = -4, x = -1, x = 3, and x =4.Let me compute each:1. x = -4:k(-4) = - (1/3)*(-4)^3 + (-4)^2 + 3*(-4).First, compute each term:- (1/3)*(-4)^3 = - (1/3)*(-64) = 64/3 ≈ 21.333...(-4)^2 = 16.3*(-4) = -12.So sum them up: 64/3 + 16 - 12.Convert 16 and 12 to thirds: 16 = 48/3, 12 = 36/3.Therefore, 64/3 + 48/3 - 36/3 = (64 + 48 - 36)/3 = (76)/3 ≈ 25.333...So k(-4) = 76/3.2. x = -1:k(-1) = - (1/3)*(-1)^3 + (-1)^2 + 3*(-1).Compute each term:- (1/3)*(-1)^3 = - (1/3)*(-1) = 1/3 ≈ 0.333...(-1)^2 = 1.3*(-1) = -3.Sum them up: 1/3 + 1 - 3 = 1/3 - 2 = -5/3 ≈ -1.666...So k(-1) = -5/3.3. x = 3:k(3) = - (1/3)*(3)^3 + (3)^2 + 3*(3).Compute each term:- (1/3)*27 = -9.9 + 9 = 18.So sum them up: -9 + 9 + 9 = 9. Wait, let me check step by step:First term: - (1/3)*(3)^3 = - (1/3)*27 = -9.Second term: (3)^2 = 9.Third term: 3*(3) = 9.Total: -9 + 9 + 9 = 9. So k(3) = 9.4. x =4:k(4) = - (1/3)*(4)^3 + (4)^2 + 3*(4).Compute each term:- (1/3)*64 = -64/3 ≈ -21.333...16 + 12 = 28.Sum them up: -64/3 + 28. Convert 28 to thirds: 84/3.So -64/3 + 84/3 = 20/3 ≈ 6.666...Thus, k(4) = 20/3.Therefore, the values of k(x) at the critical points and endpoints are:At x = -4: 76/3 ≈25.333At x = -1: -5/3 ≈-1.666At x =3: 9At x =4: 20/3≈6.666Therefore, the minimum of k(x) on [-4,4] is at x = -1, which is -5/3. Therefore, m must be less than or equal to -5/3. So the range of m is m ≤ -5/3.But wait, let me confirm again. Since the inequality is k(x) - m ≥0, which is equivalent to m ≤k(x) for all x in [-4,4]. So m has to be less than or equal to the minimal value of k(x). The minimal value of k(x) is -5/3. So m ≤ -5/3.But hold on, double-checking. If k(x) is minimal at x=-1 with k(-1) = -5/3, then setting m ≤ -5/3 ensures that for all x, k(x) - m ≥0. For example, at x=-1, k(-1) - m = (-5/3) - m ≥0. So -5/3 -m ≥0 ⇒ m ≤ -5/3. For other x, since k(x) is larger than -5/3, then if m ≤ -5/3, k(x) - m would be even larger. So yes, m must be ≤ -5/3.Therefore, the range of m is (-∞, -5/3]. So m can be any real number less than or equal to -5/3.Wait, but let me check once more. Maybe I made a mistake in computing k(-4). Let's recheck that.k(-4) = - (1/3)*(-4)^3 + (-4)^2 + 3*(-4).Compute each term:- (1/3)*(-64) = 64/3 ≈21.333(-4)^2 =163*(-4) = -12So adding them up: 64/3 +16 -12.64/3 is approximately 21.333. 16 -12 =4. So total is approximately 21.333 +4 =25.333, which is 76/3. That's correct.k(-1) = - (1/3)*(-1)^3 + (-1)^2 + 3*(-1).- (1/3)*(-1) =1/3. (-1)^2=1. 3*(-1)=-3. So 1/3 +1 -3=1/3 -2= -5/3. Correct.k(3) = - (1/3)(27) +9 +9= -9 +9+9=9. Correct.k(4)= - (1/3)(64)+16+12= -64/3 +28= -21.333+28≈6.666. Which is 20/3. Correct.So indeed, the minimal value is -5/3. Therefore, m ≤ -5/3. Hence, the range is all real numbers m such that m ≤ -5/3. So written in box form, it would be (-∞, -5/3].Wait, but just to make sure there are no other critical points or endpoints that I might have missed. The interval is [-4,4], and we found critical points at x=-1 and x=3, both within the interval. Evaluated at all critical points and endpoints. So the minimal value is indeed at x=-1. Therefore, the answer should be m ≤ -5/3.But let me check with another approach. Suppose we consider the inequality f(x) ≥ g(x) for all x in [-4,4], which is equivalent to m ≤ f(x) - [(1/3)x³ -2x]. Wait, f(x) is x² +x, so f(x) - [(1/3)x³ -2x] = x² +x - (1/3)x³ +2x = - (1/3)x³ +x² +3x. So, m ≤ - (1/3)x³ +x² +3x for all x in [-4,4]. Hence, m must be less than or equal to the minimum of that cubic function on the interval. So as we found before, the minimum is -5/3, so m ≤ -5/3. Yes, that's consistent.Therefore, the conclusion is correct. So part (2) answer is m ≤ -5/3, which in box form is boxed{(-infty, -dfrac{5}{3}]}.Just to recap, part (1) was finding the tangent line at x=1 for f(x)=x² +x, which we found to be y=3x -1. Part (2) required ensuring that f(x) is always greater than or equal to g(x) on the interval [-4,4], leading us to find the minimum of a cubic function on that interval, resulting in m needing to be less than or equal to -5/3.Final Answer(1) boxed{y = 3x - 1}(2) boxed{(-infty, -dfrac{5}{3}]}</think>(1) To find the equation of the tangent line to the graph of ( f(x) = x^2 + x ) at ( x = 1 ):- Compute the derivative ( f'(x) = 2x + 1 ).- Evaluate the derivative at ( x = 1 ): ( f'(1) = 2 cdot 1 + 1 = 3 ). This gives the slope of the tangent line.- Find the y-coordinate at ( x = 1 ): ( f(1) = 1^2 + 1 = 2 ). The point is ( (1, 2) ).- Use the point-slope form of the line equation: ( y - 2 = 3(x - 1) ). Simplifying this gives ( y = 3x - 1 ).The equation of the tangent line is (boxed{y = 3x - 1}).(2) To find the range of ( m ) such that ( f(x) geq g(x) ) for all ( x in [-4, 4] ):- Define ( h(x) = f(x) - g(x) = -frac{1}{3}x^3 + x^2 + 3x - m ).- We need ( h(x) geq 0 ) for all ( x in [-4, 4] ). This implies ( m leq -frac{1}{3}x^3 + x^2 + 3x ) for all ( x in [-4, 4] ).- Find the critical points of ( h(x) ) by solving ( h'(x) = -x^2 + 2x + 3 = 0 ). The critical points are ( x = 3 ) and ( x = -1 ).- Evaluate ( h(x) ) at the critical points and endpoints ( x = -4, -1, 3, 4 ): - ( h(-4) = frac{76}{3} ) - ( h(-1) = -frac{5}{3} ) - ( h(3) = 9 ) - ( h(4) = frac{20}{3} )- The minimum value of ( h(x) ) on ([-4, 4]) is (-frac{5}{3}) at ( x = -1 ).Thus, the range of ( m ) is (boxed{(-infty, -dfrac{5}{3}]}).,<think>Okay, let's tackle this problem step by step. First, part (1) asks for the equation of the tangent line to the graph of f(x) = x² + x at x = 1. Alright, tangent line equations generally require a point and a slope. The point is straightforward: plug x = 1 into f(x) to get the y-coordinate. Then, the slope of the tangent line at that point is the derivative of f(x) evaluated at x = 1.So let's compute f(1) first. f(1) = (1)² + 1 = 1 + 1 = 2. So the point is (1, 2). Now, the derivative of f(x) is f'(x) = 2x + 1. Plugging in x = 1, we get f'(1) = 2*1 + 1 = 3. So the slope of the tangent line is 3.Now, using the point-slope form of a line: y - y1 = m(x - x1). Plugging in the point (1, 2) and slope 3, we get y - 2 = 3(x - 1). Simplifying this, y = 3x - 3 + 2, so y = 3x - 1. Let me double-check that. If x = 1, then y = 3*1 -1 = 2, which matches the original point. And the derivative gives the slope as 3, so the equation should be correct. So part (1) seems done.Moving on to part (2): we need to find the range of m such that f(x) ≥ g(x) for all x in [-4, 4]. Given f(x) = x² + x and g(x) = (1/3)x³ - 2x + m. The inequality f(x) ≥ g(x) must hold for all x in the interval [-4, 4].So, let's set up the inequality: x² + x ≥ (1/3)x³ - 2x + m. Let's rearrange terms to bring everything to one side: x² + x - (1/3)x³ + 2x - m ≥ 0. Combining like terms: - (1/3)x³ + x² + 3x - m ≥ 0. Let's multiply both sides by -3 to make the coefficient of x³ positive, but we have to reverse the inequality sign when we multiply by a negative. So:Multiplying by -3: x³ - 3x² -9x + 3m ≤ 0. So, the inequality becomes x³ - 3x² -9x + 3m ≤ 0 for all x in [-4,4].Alternatively, maybe it's better to keep the original inequality as is and define h(x) = f(x) - g(x) = x² + x - (1/3 x³ - 2x + m) = -1/3 x³ + x² + 3x - m. Then the problem requires h(x) ≥ 0 for all x ∈ [-4,4]. Therefore, we need to ensure that h(x) ≥ 0 on the entire interval. The minimum value of h(x) on [-4,4] must be ≥ 0. Thus, we need to find the minimum of h(x) over [-4,4], set it to be ≥ 0, and solve for m.Yes, that approach makes sense. So, h(x) = -1/3 x³ + x² + 3x - m. To find the minimum of h(x) on [-4,4], we can take the derivative, find critical points, evaluate h(x) at critical points and endpoints, then ensure the minimum is ≥ 0.First, compute h'(x): derivative of h(x) is -x² + 2x + 3. Set h'(x) = 0: -x² + 2x + 3 = 0. Multiply both sides by -1: x² - 2x -3 = 0. Solving this quadratic equation: x = [2 ± sqrt(4 + 12)] / 2 = [2 ± sqrt(16)] / 2 = [2 ± 4]/2. Thus, x = (2 + 4)/2 = 6/2 = 3, and x = (2 -4)/2 = -2/2 = -1. So critical points at x = -1 and x = 3.Since we're considering the interval [-4,4], both critical points x = -1 and x = 3 are within this interval. So we need to evaluate h(x) at x = -4, -1, 3, and 4.Compute h(-4):h(-4) = -1/3*(-4)^3 + (-4)^2 + 3*(-4) - mFirst compute each term:-1/3*(-64) = 64/3 ≈ 21.333(-4)^2 = 163*(-4) = -12So h(-4) = 64/3 + 16 -12 - m = 64/3 + 4 - mConvert 4 to thirds: 12/3So h(-4) = 64/3 + 12/3 - m = 76/3 - m ≈ 25.333 - mh(-1):h(-1) = -1/3*(-1)^3 + (-1)^2 + 3*(-1) - mCompute each term:-1/3*(-1) = 1/3 ≈ 0.333(-1)^2 = 13*(-1) = -3So h(-1) = 1/3 + 1 -3 - m = (1/3 - 2) - m = (-5/3) - m ≈ -1.666 - mh(3):h(3) = -1/3*(3)^3 + (3)^2 + 3*(3) - mCompute each term:-1/3*27 = -999So h(3) = -9 + 9 + 9 - m = 9 - mh(4):h(4) = -1/3*(4)^3 + (4)^2 + 3*(4) - mCompute each term:-1/3*64 ≈ -21.3331612So h(4) = -64/3 + 16 +12 - m = (-64/3 + 28) - mConvert 28 to thirds: 84/3So h(4) = (-64/3 + 84/3) - m = 20/3 - m ≈ 6.666 - mTherefore, the critical values are:At x = -4: 76/3 - mAt x = -1: -5/3 - mAt x = 3: 9 - mAt x = 4: 20/3 - mWe need to find the minimum of these. Let's compute each value:At x = -4: 76/3 - m ≈ 25.333 - mAt x = -1: -5/3 - m ≈ -1.666 - mAt x = 3: 9 - m ≈ 9 - mAt x = 4: 20/3 - m ≈ 6.666 - mLooking at these values, the smallest one is likely at x = -1, since -5/3 - m is negative unless m is very negative, but since we need h(x) ≥ 0 everywhere, and if x=-1 gives the minimum value, then we must have that -5/3 - m ≥ 0. Wait, but -5/3 - m ≥ 0 implies that -m ≥ 5/3, so m ≤ -5/3. But if m is being subtracted, perhaps the actual minimum is when h(x) is minimized, which would be at x = -1. Let me verify.Wait, perhaps I need to check the value of h(x) at x = -1. Since h(x) = -5/3 - m, for h(x) to be ≥ 0 at x = -1, we need -5/3 - m ≥ 0 => -m ≥ 5/3 => m ≤ -5/3. But wait, but in the problem statement, we need f(x) ≥ g(x) for all x ∈ [-4,4], which is equivalent to h(x) = f(x) - g(x) ≥ 0 on that interval. Therefore, the minimal value of h(x) over [-4,4] must be ≥ 0. So we need to find the minimal value of h(x) over the interval and set it to be ≥ 0.Looking at the critical points and endpoints, h(-4) = 76/3 - m ≈ 25.333 - m, which is a large positive number minus m. Similarly, h(4) = 20/3 - m ≈6.666 - m. At x =3, h(3)=9 -m. At x=-1, h(-1)= -5/3 -m ≈-1.666 -m.So among these points, the minimal value occurs at x = -1, since that term is -5/3 -m, which is the smallest because the other terms are positive numbers minus m. Therefore, to ensure h(x) ≥0 everywhere, the minimal value is at x=-1, so set -5/3 -m ≥0 => -m ≥5/3 => m ≤ -5/3.Wait, but hold on. Let me check if x=-1 is indeed the global minimum over the interval. Maybe between the endpoints and critical points, there could be other minima? For example, maybe between -4 and -1, or between -1 and 3, or between 3 and 4? However, since we have already found all critical points (x=-1 and x=3) by setting h'(x)=0, and evaluated h(x) at these critical points as well as endpoints, so these are the candidates for minima and maxima.Therefore, the minimal value is the smallest among h(-4), h(-1), h(3), h(4). Comparing these:h(-4) =76/3 - m ≈25.333 -m,h(-1)= -5/3 -m ≈-1.666 -m,h(3)=9 -m,h(4)=20/3 -m ≈6.666 -m.Assuming m is a real number, the smallest of these is h(-1)= -5/3 -m. Therefore, to have h(x) ≥0 for all x ∈ [-4,4], we need -5/3 -m ≥0 => m ≤ -5/3.Wait, but that seems counterintuitive. If m is less than or equal to -5/3, then the minimal value of h(x) would be non-negative. Let me check with m = -5/3. Then h(-1)= -5/3 - (-5/3)=0. So if m is exactly -5/3, then h(-1)=0, which satisfies h(x) ≥0. If m is smaller than -5/3, say m = -6/3 = -2, then h(-1)= -5/3 - (-2)= -5/3 +2=1/3 >0, which also satisfies. Wait, but according to the inequality, m ≤ -5/3. So if m is less than or equal to -5/3, then -m ≥5/3, so -5/3 -m = -5/3 + (-m) ≥-5/3 +5/3=0. Wait, no. If m ≤ -5/3, then -m ≥5/3. Therefore, h(-1)= -5/3 -m = -5/3 + (-m) ≥-5/3 +5/3=0. So yes, if m ≤ -5/3, then h(-1)≥0, and since all other h(x) values at critical points and endpoints would be larger than h(-1), because h(-4)=76/3 -m, which if m ≤ -5/3, 76/3 -m would be 76/3 + |m|, which is definitely positive. Similarly, h(3)=9 -m would be 9 + |m|, also positive. So indeed, setting m ≤ -5/3 would ensure h(x) ≥0 everywhere on [-4,4].Wait, but let's confirm this by testing m = -5/3 and m less than that.If m = -5/3, then h(-1)=0, and other points:h(-4)=76/3 - (-5/3)=76/3 +5/3=81/3=27>0,h(3)=9 - (-5/3)=9 +5/3=32/3≈10.666>0,h(4)=20/3 - (-5/3)=25/3≈8.333>0.So all other points are positive. Then, if m is less than -5/3, say m=-2, then h(-1)= -5/3 -(-2)= -5/3 +2=1/3>0,h(-4)=76/3 -(-2)=76/3 +6/3=82/3≈27.333>0,h(3)=9 -(-2)=11>0,h(4)=20/3 -(-2)=20/3 +6/3=26/3≈8.666>0.So indeed, if m ≤-5/3, all h(x) values at critical points and endpoints are ≥0, and since those are the minima and maxima points, h(x) is ≥0 over the entire interval.But wait, is there a possibility that between these points, the function h(x) could dip below zero even if the critical points are okay? Let's check.Suppose m = -5/3. Then h(x) has a minimum at x=-1 of 0. So at x=-1, the function touches zero. Since h'(x) is zero there, and h''(x) can tell us concavity. Let's compute h''(x): derivative of h'(x)= -x² +2x +3, so h''(x)= -2x +2. At x=-1, h''(-1)= -2*(-1)+2=2+2=4>0, which means x=-1 is a local minimum. Since the function is concave up there, so the point at x=-1 is indeed a local minimum, and since it's the only minimum in the interval, the function is above zero everywhere else. Therefore, m ≤-5/3.Wait, but h''(x)= -2x + 2. At x=3, h''(3)= -6 +2= -4<0, so x=3 is a local maximum. So the function has a local maximum at x=3. Therefore, the critical points are x=-1 (min) and x=3 (max). So yes, the minimal value occurs at x=-1, so setting m ≤-5/3 ensures h(x) ≥0 on the entire interval.Therefore, the range of m is (-∞, -5/3].But let me check once again by considering some arbitrary point in the interval. Suppose m = -5/3. Let's pick x=0. Compute h(0)= -1/3*0 +0 +0 -m= -m=5/3>0. At x=2, h(2)= -1/3*8 +4 +6 -m= -8/3 +10 -m=22/3 -m=22/3 +5/3=27/3=9>0. So indeed, even at other points, h(x) is positive.Hence, the answer for part (2) is m ≤ -5/3, so the range is (-∞, -5/3].Wait, but let me make sure there's no mistake in the algebra.Starting with f(x) ≥g(x):x² +x ≥ (1/3)x³ -2x +mBring all terms to left:x² +x - (1/3)x³ +2x -m ≥0Combine like terms:-1/3 x³ +x² +3x -m ≥0Yes, that's correct. Then h(x)= -1/3 x³ +x² +3x -m.Then derivative h’(x)= -x² +2x +3. Setting to zero, gives x² -2x -3=0, solutions x=3 and x=-1. Correct.Calculating h(x) at critical points and endpoints. Correct.So conclusion is m ≤ -5/3. So in boxed form, that would be boxed{(-infty, -dfrac{5}{3}]}.Therefore, the answers are:(1) The tangent line is y=3x -1, so boxed: boxed{y = 3x - 1}(2) The range of m is boxed{(-infty, -dfrac{5}{3}]}Final Answer(1) The equation of the tangent line is boxed{y = 3x - 1}.(2) The range of the real number ( m ) is boxed{(-infty, -dfrac{5}{3}]}.</think>To find the equation of the tangent line to the graph of ( f(x) = x^2 + x ) at ( x = 1 ):1. Calculate ( f(1) ): [ f(1) = 1^2 + 1 = 2 ] The point on the graph is ( (1, 2) ).2. Compute the derivative ( f'(x) ): [ f'(x) = 2x + 1 ] Evaluate the derivative at ( x = 1 ): [ f'(1) = 2(1) + 1 = 3 ] The slope of the tangent line is 3.3. Use the point-slope form of the line equation: [ y - 2 = 3(x - 1) ] Simplify to get the tangent line equation: [ y = 3x - 1 ]Thus, the equation of the tangent line is (boxed{y = 3x - 1}).To find the range of ( m ) such that ( f(x) geq g(x) ) for all ( x in [-4, 4] ):1. Define ( h(x) = f(x) - g(x) ): [ h(x) = x^2 + x - left( frac{1}{3}x^3 - 2x + m right) = -frac{1}{3}x^3 + x^2 + 3x - m ]2. Find the critical points by solving ( h'(x) = 0 ): [ h'(x) = -x^2 + 2x + 3 = 0 ] Solving the quadratic equation: [ x^2 - 2x - 3 = 0 implies x = 3 text{ or } x = -1 ]3. Evaluate ( h(x) ) at the critical points and endpoints ( x = -4, -1, 3, 4 ): - At ( x = -4 ): [ h(-4) = frac{76}{3} - m ] - At ( x = -1 ): [ h(-1) = -frac{5}{3} - m ] - At ( x = 3 ): [ h(3) = 9 - m ] - At ( x = 4 ): [ h(4) = frac{20}{3} - m ]4. The minimum value of ( h(x) ) on ([-4, 4]) occurs at ( x = -1 ): [ -frac{5}{3} - m geq 0 implies m leq -frac{5}{3} ]Therefore, the range of ( m ) is (boxed{(-infty, -dfrac{5}{3}]}).]

🔑:Solution:(1) Since f(x)=x^{2}+x, then f'(x)=2x+1 and f(1)=2, thus f'(1)=3.Therefore, the equation of the tangent line at x=1 is y-2=3(x-1), which simplifies to 3x-y-1=0.(2) Let h(x)=g(x)-f(x)= frac{1}{3}x^{3}-2x+m-x^{2}-x= frac{1}{3}x^{3}-3x+m-x^{2},Thus, h'(x)=x^{2}-2x-3,When -4 0; when -1 0. To ensure f(x) geqslant g(x) always holds, we need h_{max}(x) leqslant 0,From above, the maximum value of h(x) is attained at x=-1 or x=4, and h(-1)=m+ frac{5}{3}, h(4)=m- frac{20}{3},Since m+ frac{5}{3} > m- frac{20}{3},Therefore, m+ frac{5}{3} leqslant 0,which means m leqslant - frac{5}{3}. Thus, the range of m is boxed{m leqslant - frac{5}{3}}.

🧠:[<think>Alright, let me try to figure out this problem. So, we have a circle with 2021 points marked on it. Kostya is going to mark points starting from some initial point. The way he marks them is a bit specific: he starts by marking a point, then the next one to the right, then skips one and marks the next, then skips two, and so on. Each time, the number of skips increases by one. The question is asking on which move a point will be marked for the second time. In other words, when does the first collision happen where a previously marked point is marked again.Hmm, okay. Let me break this down. So, first, we have a circle with 2021 points. Let's imagine them arranged in a circle, labeled from 0 to 2020, maybe. Kostya starts at point 0, marks it, then moves 1 step to the right to mark point 1. Then, from point 1, he skips 1 point (so moves 2 steps) to mark point 3. Then, from point 3, skips 2 points (moves 3 steps) to mark point 6, and so on. Each time, the number of steps he takes increases by 1. So, on the nth move, he moves n steps from the previous point. Wait, hold on: actually, the problem says "the next point to the right", then "skipping one", then "skipping two", etc. So, maybe the first move is move 0? Let me check.Wait, the problem says: Kostya marks a point, then marks the next point to the right, then marks the point to the right skipping one, then skipping two, etc. So, the first move is marking the starting point. Then the second move is the next point. Then the third move skips one, etc. So, maybe the first move is move 1: mark the starting point. Then move 2: next point. Move 3: skip 1. Move 4: skip 2. So, the number of skips is equal to the move number minus 2? Wait, that might complicate things. Let me rephrase.Alternatively, let's think of each move as follows: starting at some point, each subsequent move is determined by how many positions you skip. So, the first mark is at position 0. Then, move 1 step to the right (no skips), so you get to position 1. Then, move 2 steps (skipping 1 position), so from 1, move 2 steps to position 3. Then, move 3 steps (skipping 2 positions) from 3, which would be position 6. So, the nth move (starting from 0) would involve moving n steps. Wait, maybe.Wait, let's think of the first move as moving 1 step (marking the next point), then moving 2 steps (skipping one), then moving 3 steps (skipping two), etc. So, the first move is 1 step, second move is 2 steps, third is 3 steps, so on the kth move, he moves k steps. But since we start at position 0, the positions would be cumulative sums modulo 2021.Wait, maybe we can model this as a sequence of positions where each position is the previous position plus the number of steps, modulo 2021. The number of steps starts at 1 and increases by 1 each time. So, the first position is 0. Then, position 1: 0 + 1 = 1. Then position 3: 1 + 2 = 3. Then position 6: 3 + 3 = 6. Then position 10: 6 + 4 = 10, and so on. So, the position after n moves is the sum from k=1 to n of k, which is n(n+1)/2. But since we're moving around a circle with 2021 points, we take this sum modulo 2021.Therefore, the position after m moves is S(m) = (m(m+1)/2) mod 2021. We need to find the smallest m such that S(m) ≡ S(k) mod 2021 for some k < m. The smallest such m is the move where a point is marked for the second time.Alternatively, since all positions start at 0, and then we have positions 1, 3, 6, etc., we need to find the smallest m where the cumulative sum S(m) is congruent to an earlier sum S(k). Therefore, we need the smallest m where m(m+1)/2 ≡ k(k+1)/2 mod 2021 for some k < m. Then, m is the answer.So, we need to solve the congruence m(m+1)/2 ≡ k(k+1)/2 mod 2021. Rearranging, this is equivalent to m(m+1) ≡ k(k+1) mod 4042, since 2021 is odd, so multiplying both sides by 2 gives the equation modulo 4042. Therefore, m(m+1) ≡ k(k+1) mod 4042. Then, m^2 + m - k^2 - k ≡ 0 mod 4042. Which factors as (m - k)(m + k + 1) ≡ 0 mod 4042.So, we need (m - k)(m + k + 1) ≡ 0 mod 4042. Since m > k, m - k is at least 1, and m + k + 1 is greater. Therefore, the product is a multiple of 4042. Since 4042 factors into prime factors as follows: 4042 ÷ 2 = 2021, and 2021 is 43 × 47. So, 4042 = 2 × 43 × 47. Therefore, the factors m - k and m + k + 1 must multiply to a multiple of 2 × 43 × 47.We need to find the minimal m such that there exists k < m where either m - k is divisible by 4042, which is impossible since m - k < m, and 4042 is quite large. Alternatively, m - k and m + k + 1 must cover the factors of 4042.But this might be a bit complicated. Maybe another approach is better.Alternatively, since we need S(m) ≡ S(k) mod 2021, which is equivalent to m(m+1)/2 ≡ k(k+1)/2 mod 2021. So, m(m+1) ≡ k(k+1) mod 4042. Let’s consider that.Let’s denote t = m - k. Then, m = k + t. Substitute into the equation:(k + t)(k + t + 1) ≡ k(k + 1) mod 4042.Expanding the left side:(k + t)(k + t + 1) = k^2 + (2t + 1)k + t(t + 1).Subtract k(k + 1):k^2 + (2t + 1)k + t(t + 1) - k^2 - k = (2t + 1 - 1)k + t(t + 1) = 2t k + t(t + 1).Thus, 2t k + t(t + 1) ≡ 0 mod 4042.Factor out t:t(2k + t + 1) ≡ 0 mod 4042.Therefore, t(2k + t + 1) ≡ 0 mod 4042.Since t = m - k ≥ 1, we have that t divides 4042 or 2k + t + 1 divides 4042, or their combination. But this seems a bit messy.Alternatively, perhaps think of the problem in terms of the sequence S(m) = m(m + 1)/2 mod 2021. We need to find the minimal m such that S(m) repeats a previous value. The minimal such m would be the period of the sequence S(m). However, since the step is increasing each time, this isn't a linear recurrence, so the period might not be straightforward.Alternatively, think of the problem as solving the equation m(m + 1)/2 ≡ k(k + 1)/2 mod 2021, which can be rewritten as (m - k)(m + k + 1) ≡ 0 mod 4042. So, either m - k ≡ 0 mod d and m + k + 1 ≡ 0 mod (4042/d) for some divisor d of 4042. Since m > k, m - k is positive.Let’s set d = m - k, so m = k + d. Then, substituting into the second equation:(k + d) + k + 1 ≡ 0 mod (4042/d).Simplify:2k + d + 1 ≡ 0 mod (4042/d).So, 2k ≡ - (d + 1) mod (4042/d).Therefore, k ≡ (- (d + 1)/2) mod (4042/(2d)).Wait, but 2k ≡ - (d + 1) mod (4042/d) implies that 2 must divide the right-hand side. So, - (d + 1) must be even, which implies that d + 1 must be even, so d must be odd.Given that 4042 = 2 × 43 × 47. So, the divisors d of 4042 must be factors of 2 × 43 × 47.But since d divides 4042, and d is a positive integer. So, possible values of d are 1, 2, 43, 47, 86, 94, 2021, 4042. Wait, 43 × 47 = 2021, so the divisors are 1, 2, 43, 47, 86 (2×43), 94 (2×47), 2021, 4042.Therefore, possible d values are these. But since m = k + d, and we need k to be a non-negative integer less than m. For each divisor d of 4042, we can attempt to solve for k. The minimal m would be the minimal k + d over all possible divisors d where k exists.But let's check for each d:First, d must be such that - (d + 1) is divisible by 2, i.e., d is odd. So, d must be odd. From the list of divisors: 1, 43, 47, 2021. These are the odd divisors. So, d ∈ {1, 43, 47, 2021}.Let’s check each d:1. d = 1:Then, modulus is 4042 / 1 = 4042.Equation: 2k ≡ - (1 + 1) = -2 mod 4042.Thus, 2k ≡ -2 mod 4042 ⇒ 2k ≡ 4040 mod 4042 ⇒ k ≡ 2020 mod 2021.Therefore, k = 2021t + 2020. Since k must be non-negative and less than m = k + 1, but if k = 2021t + 2020, then m = 2021t + 2021. The minimal k is 2020, leading to m = 2021. But m must be less than 2021? Wait, but 2021 is the number of points. However, m can be larger, but we’re looking for the first collision. So, perhaps m=2021 is possible, but maybe there's a smaller m.Wait, but maybe d=1 is not feasible because if k=2020, then m=2021. But the positions are modulo 2021, so S(2021) = 2021*2022/2 mod 2021. Let's compute that:2021*2022/2 = (2021/2)*2022. Since 2021 is odd, 2021 divides 2021, but 2022 is 2*1011. So, (2021)*(2022)/2 = 2021*1011. Then, 2021*1011 mod 2021 is 0. But S(0) is 0, so m=2021 would give S(m)=0, which was the starting point. Therefore, the collision happens at m=2021. But maybe there's a smaller m where collision occurs. So, we need to check other divisors.2. d = 43:Modulus is 4042 / 43 = 94.Equation: 2k ≡ - (43 + 1) = -44 mod 94.So, 2k ≡ -44 mod 94 ⇒ 2k ≡ 50 mod 94 ⇒ k ≡ 25 mod 47.Therefore, k = 47t + 25. Then, m = k + 43 = 47t + 25 + 43 = 47t + 68.Since k must be non-negative and m must be as small as possible, the minimal t is 0: k =25, m=68. But wait, if t=0, k=25, m=68. Let's check if this works.Compute S(68) and S(25):S(68) = 68*69/2 mod 2021 = (68*69)/2 = 34*69 = 2346. 2346 mod 2021 = 2346 - 2021 = 325.S(25) = 25*26/2 mod 2021 = 325 mod 2021 = 325. So yes, they are equal. Therefore, m=68 and k=25 is a collision. So, move 68 would mark a point that was previously marked at move 25.So, is 68 the minimal such m? Let's check with other divisors.3. d=47:Modulus is 4042 /47= 86.Equation: 2k ≡ - (47 +1 ) = -48 mod 86.Thus, 2k ≡ -48 mod 86 ⇒ 2k ≡ 38 mod 86 ⇒ k ≡ 19 mod 43.Therefore, k=43t +19. Then, m=k +47=43t +19 +47=43t +66.Minimal t=0 gives k=19, m=66. Let's check S(66) and S(19):S(66)=66*67/2= (66*67)/2=2211. 2211 mod 2021=2211-2021=190.S(19)=19*20/2=190 mod2021=190. So, indeed, S(66)=S(19). Therefore, m=66, which is smaller than 68. So, 66 is a collision.4. d=2021:Modulus is 4042 /2021=2.Equation: 2k ≡ -(2021 +1)= -2022 mod 2.But -2022 mod 2 is 0, so 2k≡0 mod 2 ⇒k≡0 mod1. So, k can be any integer. Then, m=k +2021. Minimal k=0, m=2021. Which is the same as when d=1. So, S(2021)=0=S(0). But m=2021 is larger than the previous candidates (66 and 68).Therefore, the minimal m among these is 66. But we need to check if there's a smaller m.Wait, but are there other possible divisors? For example, maybe composite divisors? Wait, earlier I considered only the odd divisors, but perhaps d could be even? Wait, but earlier when we set d = m - k, we noticed that d must be odd because 2k + d + 1 must be even (since 2k is even, and d +1 must be even, so d must be odd). Therefore, d must be odd. Therefore, the possible divisors are only the odd ones:1,43,47,2021. So, the minimal m is 66. However, wait, when d=47, we get m=66, which is less than 68 from d=43. But is 66 indeed the minimal?Wait, but let's check if there's a collision even before 66. Let's see.For example, let's compute some positions:S(0) = 0S(1) = 1S(2) = 3S(3) = 6S(4) = 10S(5) = 15S(6) = 21S(7) = 28S(8) = 36S(9) = 45S(10) = 55Continuing:S(11) = 66S(12) = 78S(13) = 91S(14) = 105S(15) = 120S(16) = 136S(17) = 153S(18) = 171S(19) = 190S(20) = 210...Wait, S(19)=190 as we saw. Then S(66)=?Wait, instead of computing all, let's check if there's any collision before m=66.Suppose we have S(m) = m(m+1)/2 mod 2021. If there is a collision before m=66, that would mean that for some k < m <66, S(m) = S(k). But given that the modulus is 2021, which is much larger than 66, it's possible but not guaranteed. However, since the problem states that there are 2021 points, which is a prime number (Wait, 2021 is 43*47, so not prime). Hmm.But 2021 is composite, so the problem could have collisions earlier. However, since the step increases each time, the sequence S(m) is a quadratic sequence, so it's not obvious whether there are collisions early on.But according to our earlier analysis with divisors, the earliest collision is at m=66. Let me verify by computing S(66):66*67/2 = (66*67)/2 = 4422/2 = 2211. 2211 mod 2021 is 2211 - 2021 = 190.Similarly, S(19) = 19*20/2 = 190. So, yes, S(19)=190 and S(66)=190. Therefore, at move 66, the point marked is 190, which was previously marked at move 19. So, the first collision happens at move 66. But let's check if there is any collision before that.For example, let's check if S(44)= S(k) for some k <44.Compute S(44) = 44*45/2 = 990. 990 mod2021=990. Is there a k where S(k)=990?S(k) =k(k+1)/2=990 ⇒k(k+1)=1980. Let's solve k² +k -1980=0. Discriminant=1 +4*1980=7921=89². So, k=(-1 ±89)/2. k=(88)/2=44. So, k=44. But that's the same as m=44. Wait, but we are looking for k < m. So, if m=44, S(44)=990, and the equation gives k=44, which is not less than m. Therefore, no collision here. So, 990 is unique up to m=44.Another example: Let's check S(43). 43*44/2=946. 946 mod2021=946. Solve k(k+1)/2=946 ⇒k(k+1)=1892. Find k. Discriminant sqrt(1 +7568)=sqrt(7569)=87. So, k=(-1 +87)/2=43. Again, k=43, same as m=43.Similarly, S(42)=42*43/2=903. 903 mod2021=903. Solve k(k+1)/2=903 ⇒k² +k -1806=0. Discriminant=1 +7224=7225=85². k=(-1 +85)/2=42. So, same as m=42. No collision.So, seems like in the earlier terms, S(m) is unique. The first collision occurs at m=66 with k=19. Therefore, the answer should be 66. Wait, but let's confirm with another example.Check S(66)=190, which was previously at S(19)=190. So, move 66 is the first time a point is marked for the second time. Therefore, the answer is 66.But wait, the problem says "On which move will a point be marked for the second time?" So, the answer is move 66. But let me make sure that there are no collisions before 66. Let's check some intermediate values.For example, let's check S(50):S(50)=50*51/2=1275. 1275 mod2021=1275. Solve k(k+1)/2=1275. k² +k -2550=0. Discriminant=1 +10200=10201=101². So, k=(-1 +101)/2=50. So, same as m=50. No collision.How about S(60):60*61/2=1830. 1830 mod2021=1830. Solve k(k+1)/2=1830 ⇒k² +k -3660=0. Discriminant=1 +14640=14641=121². k=(-1 +121)/2=60. Again, same as m=60.Similarly, S(65)=65*66/2=2145. 2145 mod2021=2145-2021=124. Check if 124 was seen before.Looking for k such that S(k)=124. So, k(k+1)/2=124 ⇒k(k+1)=248. Let's solve:k² +k -248=0. Discriminant=1 + 992=993. Not a perfect square. So, no integer solution. Therefore, 124 is new.Then, S(66)=190, which was at S(19). So, yes, 66 is the first collision.Therefore, the answer should be 66. But let me check the case when d=43 gives m=68. Let's see what S(68) is.68*69/2=2346. 2346 mod2021=2346-2021=325. S(25)=25*26/2=325. So, m=68 collides with k=25. But since 66 is smaller than 68, 66 is the first collision.Therefore, the answer is 66.But wait, I just want to confirm once more that between m=19 and m=66, there are no other collisions. For example, check S(30):30*31/2=465. 465 mod2021=465. Solve k(k+1)/2=465 ⇒k² +k -930=0. Discriminant=1 + 3720=3721=61². So, k=(-1 +61)/2=30. So, same as m=30.Another one: S(40)=40*41/2=820. 820 mod2021=820. Solve k(k+1)/2=820 ⇒k² +k -1640=0. Discriminant=1 +6560=6561=81². k=(-1 +81)/2=40. So, same as m=40.So, up until m=66, all positions are unique except for m=66 which collides with m=19. Therefore, the first time a point is marked twice is on the 66th move.Final Answerboxed{66}</think>Kostya marks points on a circle with 2021 points, starting from a point and moving right with increasing skips each time. The position after ( m ) moves is given by ( S(m) = frac{m(m+1)}{2} mod 2021 ). We need to find the smallest ( m ) such that ( S(m) equiv S(k) mod 2021 ) for some ( k < m ).1. Position Calculation: The position after ( m ) moves is ( S(m) = frac{m(m+1)}{2} mod 2021 ).2. Equation for Collision: We need ( frac{m(m+1)}{2} equiv frac{k(k+1)}{2} mod 2021 ), which simplifies to ( m(m+1) equiv k(k+1) mod 4042 ).3. Factorization: This can be factored as ( (m - k)(m + k + 1) equiv 0 mod 4042 ). The prime factors of 4042 are 2, 43, and 47.4. Divisors Analysis: Considering the divisors of 4042, we find the smallest ( m ) by solving ( t(2k + t + 1) equiv 0 mod 4042 ) for ( t = m - k ).5. Checking Divisors: For odd divisors ( d ) (1, 43, 47, 2021), we solve for ( k ) and ( m ). The smallest ( m ) found is 66 when ( d = 47 ).After verifying, the first collision occurs at ( m = 66 ) where ( S(66) equiv S(19) mod 2021 ).[boxed{66}],<think>Okay, let's try to figure out this problem. So, we have a circle with 2021 points marked on it. Kostya starts marking points in a certain pattern: first, he marks a point, then the next one to the right, then skips one and marks the next, then skips two, and so on. The question is, on which move will a point be marked for the second time?Hmm. Let me parse that again. So, the process is like this: each time, after marking a point, he moves to the right, skipping an increasing number of points each time. The first move, he just marks a point. Wait, does the first move count as moving zero skips? Let's check.The problem says: "Kostya marks a point, then marks the next point to the right, then he marks the point to the right of the last marked point, skipping one, then the point to the right of the last marked point, skipping two, and so on." So, breaking it down step by step:1. First move: marks a point. Let's call this position 0.2. Second move: moves to the next point to the right. So from position 0, he moves to position 1.3. Third move: skips one point. So from position 1, he skips position 2 and marks position 3.4. Fourth move: skips two points. From position 3, skips 4 and 5, marks 6.5. Fifth move: skips three points. From 6, skips 7,8,9, marks 10.Wait, so the number of skips increases by one each time. So, on the nth move, he skips (n-2) points? Let me confirm.Wait, move 1: he just marks a point. So, maybe move 1 is considered the first mark. Then, move 2: next point. Move 3: skip 1, so move 2 skips 0, move 3 skips 1, move 4 skips 2, etc. So the number of skips on move k is (k-2). Because starting from move 2, each subsequent move skips one more than the previous.But actually, when you skip n points, you move (n+1) positions. For example, skipping 1 point means moving 2 positions. Wait, maybe I need to clarify the movement.If you are at position x, and you skip k points, then the next position is x + k + 1. Because you skip k points, so you pass over k points and land on the (k+1)th. So, the step size is (k + 1), where k is the number of skips. But in the problem statement, it says "skipping one", "skipping two", etc.So, move 1: mark a point (position 0).Move 2: next point to the right (position 1). So, no skips, just move 1 step.Move 3: skip one, so from position 1, skip 1 (position 2), mark position 3. So, step size 2.Move 4: skip two, from position 3, skip 2 (positions 4 and 5), mark position 6. Step size 3.Move 5: skip three, step size 4. So, the step size is increasing by 1 each time.Therefore, the step size on move m is (m - 1). Because move 2 has step size 1, move 3 step size 2, move 4 step size 3, etc. So, generalizing, on move m, the step size is (m - 1).But since we're moving on a circle with 2021 points, all positions are modulo 2021. So, the position after each move can be calculated as the cumulative sum of step sizes modulo 2021.Wait, but we need to model the positions. Let's think of the positions as numbers from 0 to 2020. Let's denote the position after the k-th move as P(k).We need to find the smallest m > 1 such that P(m) ≡ P(n) mod 2021 for some n < m. Then, the answer would be m.Alternatively, since all moves are deterministic, the process will eventually cycle, and we need to find the first time a position is revisited.But how to calculate P(k)?Wait, the first move (move 1) is at position 0.Move 2: 0 + 1 = 1.Move 3: 1 + 2 = 3.Move 4: 3 + 3 = 6.Move 5: 6 + 4 = 10.Wait, so the position after move m is the sum from i=1 to (m-1) of i. Because each move after the first adds a step size of (m-1). Wait, move 2 adds 1, move 3 adds 2, etc., so up to move m, the total steps added are 1 + 2 + ... + (m - 1). But the initial position is 0 for move 1.Wait, actually, the position after move m is the sum from i=1 to (m - 1) of i. Because each move m (starting from move 2) corresponds to a step size of (m - 1). Wait, let's verify:Move 1: position 0.Move 2: 0 + 1 = 1. Sum from 1 to 1: 1.Move 3: 1 + 2 = 3. Sum from 1 to 2: 1 + 2 = 3.Move 4: 3 + 3 = 6. Sum from 1 to 3: 1 + 2 + 3 = 6.Yes, so in general, position after move m is S(m - 1) where S(n) is the sum from 1 to n. So S(n) = n(n + 1)/2. Therefore, position after move m is (m - 1)m / 2 mod 2021.Therefore, the position after the k-th move is P(k) = (k - 1)k / 2 mod 2021.We need to find the smallest k > 1 such that P(k) ≡ P(j) mod 2021 for some j < k. The smallest such k is the answer.In other words, find the smallest k where there exists j < k such that (k - 1)k / 2 ≡ (j - 1)j / 2 mod 2021.Which simplifies to (k^2 - k)/2 - (j^2 - j)/2 ≡ 0 mod 2021.Multiplying both sides by 2: (k^2 - k - j^2 + j) ≡ 0 mod 4042.Factor the left-hand side: (k^2 - j^2) - (k - j) = (k - j)(k + j) - (k - j) = (k - j)(k + j - 1).So, (k - j)(k + j - 1) ≡ 0 mod 4042.We need to find the smallest k where such a j exists. Since 2021 is the modulus, and 4042 = 2*2021. Also, 2021 factors into primes: 2021 = 43*47. So, 4042 = 2*43*47.Therefore, we need (k - j)(k + j - 1) ≡ 0 mod 2*43*47.Since k > j, and k and j are positive integers, k - j is at least 1, and k + j -1 is at least k + 1 -1 = k.So, we need to find the smallest k such that either:1. k - j is divisible by 4042, which is impossible since k - j < k, and k would have to be at least 4042 + 1, but maybe the other factor is divisible.Or,2. The product (k - j)(k + j -1) is divisible by 4042.Given that 4042 factors into 2, 43, 47.Therefore, the product (k - j)(k + j -1) must be divisible by 2, 43, and 47.Since 43 and 47 are primes, and 2 is prime. So, we need to find the minimal k such that (k - j) and (k + j -1) cover all these prime factors.Note that (k - j) and (k + j -1) are two integers with difference (k + j -1) - (k - j) = 2j -1, which is odd. Therefore, one of them is even, and the other is odd. So, the factor of 2 must come from either (k - j) or (k + j -1), but not both, since their difference is odd.So, to have the product divisible by 2, one of the factors must be even. The other factors 43 and 47 can be distributed between the two factors.So, to find the minimal k, we need to find the smallest k where (k - j) and (k + j -1) include the factors 2, 43, 47.Given that, the minimal k would be the minimal solution to the congruence P(k) ≡ P(j) mod 2021, which corresponds to a collision in positions.Alternatively, since the positions are given by triangular numbers modulo 2021, the problem reduces to finding the minimal period of the triangular numbers modulo 2021, or the minimal k where two triangular numbers T_k and T_j are congruent modulo 2021.Triangular numbers are given by T_n = n(n + 1)/2. So, we need T_{k-1} ≡ T_{j-1} mod 2021 for some j < k.Therefore, the question is equivalent to finding the smallest k > 1 such that T_{k-1} ≡ T_{j-1} mod 2021 for some j < k.This is related to the concept of the period of the triangular numbers modulo N. However, 2021 is composite, so perhaps we can use the Chinese Remainder Theorem and solve it modulo 43 and 47, then combine the results.First, let's factor 2021 = 43 * 47. So, by the Chinese Remainder Theorem, the equation T_{k-1} ≡ T_{j-1} mod 43 and mod 47.Therefore, the minimal k is the least common multiple of the periods modulo 43 and 47.Alternatively, the minimal solution k would be the minimal k such that both congruences T_{k-1} ≡ T_{j-1} mod 43 and T_{k-1} ≡ T_{j-1} mod 47 have solutions.But perhaps it's easier to compute the minimal periods modulo 43 and 47 separately, then find their least common multiple.But how do we compute the minimal period of triangular numbers modulo a prime?Let me recall that triangular numbers modulo p can be studied by solving T_n ≡ T_m mod p, which implies n(n + 1)/2 ≡ m(m + 1)/2 mod p.Multiplying both sides by 2: n(n + 1) ≡ m(m + 1) mod 2p.Therefore, n^2 + n - m^2 - m ≡ 0 mod 2p.Which factors as (n - m)(n + m + 1) ≡ 0 mod 2p.So, similar to before. So, for primes, say p, we can analyze the possible solutions.Given that, for a prime modulus p, the minimal period of triangular numbers modulo p is the minimal k such that T_{n + k} ≡ T_n mod p for all n. But in our case, we need the minimal k where T_{k - 1} ≡ T_{j - 1} mod p for some j < k, which is the minimal collision time.This might be related to the concept of the Pisano period for Fibonacci numbers, but for triangular numbers. I don't know the exact period, but perhaps we can compute it for 43 and 47.Alternatively, we can note that the problem is equivalent to solving T_{k - 1} ≡ T_{j - 1} mod 2021, which is equivalent to (k - 1)k ≡ (j - 1)j mod 4042 (multiplying both sides by 2), which is the same as (k^2 - k) - (j^2 - j) ≡ 0 mod 4042.Which factors as (k - j)(k + j - 1) ≡ 0 mod 4042.So, as before, since 4042 = 2*43*47, we can look for solutions where (k - j)(k + j - 1) is divisible by 2, 43, and 47.Given that k and j are positive integers with k > j, we need to find the minimal k such that (k - j) is divisible by some combination of these primes, and (k + j - 1) is divisible by the remaining primes.Given the factors 2, 43, 47, we need to split them between (k - j) and (k + j - 1). Since one of (k - j) or (k + j -1) must be even (as their difference is 2j -1, which is odd), so the even factor (2) is already accounted for.Then, the other factors 43 and 47 need to be distributed between (k - j) and (k + j -1). To find the minimal k, we can set one of the factors to 43 and the other to 47.So, suppose (k - j) = 43 and (k + j -1) = 47*2 = 94. Wait, but 43 and 47 are primes. Let's see:If (k - j) = 43 and (k + j -1) = 47*2 = 94. Then, solving:k - j = 43k + j -1 = 94Adding these equations:2k -1 = 137 => 2k = 138 => k = 69.Then, j = k -43 = 26.Check if (26 + 69 -1) = 94. Yes, 26 + 69 = 95, 95 -1 = 94. Correct.Alternatively, if (k - j) = 47 and (k + j -1) = 43*2 = 86:Then,k - j = 47k + j -1 = 86Adding:2k -1 = 133 => 2k = 134 => k = 67.Then, j = 67 -47 = 20.Check: 20 + 67 -1 = 86. 20 + 67 = 87, 87 -1 = 86. Correct.So, the minimal k from these two cases is 67.But wait, is there a smaller k?Alternatively, maybe (k - j) = 43*47 = 2021. Then, (k + j -1) must be 2. But k + j -1 = 2 implies k + j = 3. But k - j = 2021, so solving:k - j = 2021k + j = 3Adding: 2k = 2024 => k = 1012, which is much larger. So, not minimal.Alternatively, other factor combinations. For example, (k - j) = 2*43 = 86 and (k + j -1) = 47.Then:k - j = 86k + j -1 = 47Adding:2k -1 = 133 => 2k = 134 => k = 67, same as before. Then j = 67 -86 = negative. Not possible.Similarly, (k - j) = 2*47 = 94, (k + j -1) = 43. Then:k - j = 94k + j -1 = 43Adding: 2k -1 = 137 => k = 69, j = 69 -94 = negative. Not possible.Alternatively, (k - j) = 43*2 = 86 and (k + j -1) = 47. As above, leads to j negative.Alternatively, (k - j) = 2 and (k + j -1) = 43*47 = 2021.Then,k - j = 2k + j -1 = 2021Adding: 2k -1 = 2023 => 2k = 2024 => k = 1012. So, same as before.So, the minimal k from these possible factor splits is 67 and 69. So, the minimal k is 67.But wait, we need to check whether these actually produce collisions.For k=67, j=20. So, check if T_{66} ≡ T_{19} mod 2021.Compute T_{66} = 66*67/2 = 2211.T_{19} = 19*20/2 = 190.2211 - 190 = 2021. So, 2211 ≡ 190 mod 2021. So, yes, 2211 -190 = 2021, which is 0 mod 2021. So, indeed, T_{66} ≡ T_{19} mod 2021. Therefore, move 67 would mark a point that was previously marked on move 20.Similarly, for k=69, j=26.T_{68} = 68*69/2 = 2346T_{25} = 25*26/2 = 3252346 - 325 = 2021. So, 2346 ≡ 325 mod 2021. Therefore, T_{68} ≡ T_{25} mod 2021. So, move 69 would revisit the point from move 26.So, between k=67 and k=69, the smaller one is 67, so the answer should be 67.But let's check if there is a smaller k. Maybe by considering that (k - j) and (k + j -1) could share factors.Wait, but since 43 and 47 are primes, the only common factors between (k - j) and (k + j -1) would be 1, since their difference is 2j -1, which is odd, so GCD(k - j, k + j -1) divides GCD(k - j, 2j -1). Since k - j and j are variables, but maybe it's 1.Alternatively, perhaps there are smaller k where (k - j) is a multiple of 43 and (k + j -1) is a multiple of 47, but not necessarily 43 and 47. Wait, but 43 and 47 are both needed to cover the modulus 2021.Wait, 2021 is 43*47. So, to have the product (k - j)(k + j -1) divisible by both 43 and 47, one of the terms must be divisible by 43 and the other by 47, or one term divisible by both. But 43 and 47 are coprime, so if a term is divisible by both, it's divisible by 2021. But that would lead to a larger k.So, the minimal solutions are when one term is 43 and the other is 47*2=94, or one term is 47 and the other is 43*2=86, leading to k=67 and k=69. Thus, the minimal k is 67.Therefore, the answer should be 67. But let me verify with an example.Wait, let's compute the positions for move 67 and move 20.Compute T_{66} = (66*67)/2 = 66*33.5 = 2211. 2211 mod 2021 = 2211 - 2021 = 190.Compute T_{19} = (19*20)/2 = 190. So yes, both are 190 mod 2021. So, move 67 marks position 190, which was previously marked on move 20. Therefore, the answer is 67.But let me check if there's a smaller k. Suppose we take the modulus 43 and modulus 47 separately.For modulus 43: find the minimal k where T_{k-1} ≡ T_{j-1} mod 43.Similarly for modulus 47. Then, the minimal k would be the LCM of the periods for 43 and 47.But perhaps for modulus 43, the minimal collision occurs at k=44, since T_{43} = (43*44)/2 = 946 ≡ 946 mod 43. 43*44/2 = 43*22 = 946. 946 divided by 43 is 22, so 946 ≡ 0 mod 43. T_{0} = 0. So, T_{43} ≡ T_{0} mod 43. So, that would be a collision at k=44 and j=1. But since we start at j=1, which is the first move. So, the collision happens at k=44 for modulus 43.Similarly, for modulus 47, the minimal collision is at k=48, since T_{47} = (47*48)/2 = 1128. 1128 mod 47: 47*24 = 1128, so T_{47} ≡ 0 mod 47, same as T_{0}. So, collision at k=48.Therefore, the periods modulo 43 and 47 are 44 and 48 respectively. Then, the minimal k where collision occurs modulo both 43 and 47 is LCM(44,48). LCM(44,48) = LCM(4*11, 16*3) = 16*3*11 = 528.But that seems way too large. However, our previous calculation gave k=67. So, there's a discrepancy here.Wait, but perhaps using the Chinese Remainder Theorem, since 67 ≡ 67 mod 43: 67 -43=24, 24 mod43=24. 67 mod47=67-47=20.But 44 mod43=1, 48 mod47=1. So, perhaps the minimal solution is not LCM(44,48), but something else.Wait, the problem is that when we use the Chinese Remainder Theorem, the collisions modulo 43 and 47 can happen at different k's, and we need a k that satisfies both congruences.But in our case, we have a collision at k=67, which is 67 mod43=24, and 67 mod47=20. However, the collision at k=67 corresponds to j=20, which may or may not relate to the modulus periods.Alternatively, perhaps the approach of splitting the modulus into 43 and 47, solving the collisions modulo each prime, and then combining the results is not straightforward here because the collisions in each modulus don't necessarily correspond to the same j.Hence, the earlier approach of considering the equation (k - j)(k + j -1) ≡0 mod4042 and finding minimal k is more direct and gives k=67.But just to ensure there isn't a smaller k, let's check for k=44, which was the modulus 43 collision. If k=44, then T_{43} = 43*44/2=946. 946 mod2021=946. Was 946 previously achieved? For j=1, T_{0}=0. 946 ≠0 mod2021, so no collision. But 946 mod43=0, so in modulus 43, it collides with T_{0}, but in modulus 47, 946 mod47=946-47*20=946-940=6≡6 mod47. So, unless there was a previous T_j ≡6 mod47, which might have occurred earlier. But to check that, we would need to see if T_{k} mod47 cycles with period 48, so within the first 48 moves, all residues mod47 are covered. But since 47 is prime, the triangular numbers mod47 would have a period of 47 or something dividing 47*2.But this is getting complicated. Since our initial calculation gave k=67, which is less than the LCM(44,48)=528, and we have a concrete example where T_{66}≡T_{19} mod2021, I think 67 is indeed the minimal k.Therefore, the answer should be the 67th move.Final AnswerThe first time a point is marked for the second time will be on the boxed{67}th move.</think>Kostya marks points on a circle with 2021 points following a specific pattern: each move skips an increasing number of points. The position after the ( k )-th move is given by the triangular number ( T_{k-1} = frac{(k-1)k}{2} mod 2021 ). We need to find the smallest ( k ) such that ( T_{k-1} equiv T_{j-1} mod 2021 ) for some ( j < k ).This translates to solving the congruence ( (k - j)(k + j - 1) equiv 0 mod 4042 ) (where 4042 = 2 * 2021). Factoring 2021 as 43 * 47, we need to split the factors 2, 43, and 47 between ( (k - j) ) and ( (k + j - 1) ).Considering the smallest ( k ), we find that:- For ( (k - j) = 47 ) and ( (k + j - 1) = 86 ), solving gives ( k = 67 ) and ( j = 20 ).- For ( (k - j) = 43 ) and ( (k + j - 1) = 94 ), solving gives ( k = 69 ) and ( j = 26 ).The minimal ( k ) is 67. Verifying, ( T_{66} equiv T_{19} mod 2021 ), confirming the collision at the 67th move.Thus, the answer is (boxed{67}).]

🔑:Answer: on the 67th move. The question is about the smallest natural b for which (a+1)+(a+2)+cdots+(b-1)+b is divisible by 2021=43 cdot 47. This sum is equal to b(b+1) / 2-a(a+1) / 2= (b-a)(b+a+1) / 2 and it is not hard to see that the smallest b=(b-a) / 2+(b+a) / 2 is achieved with the factorization (66-19)(66+19+1) / 2=47 cdot 86 / 2=47 cdot 43. In this case, b=66, i.e., this happened on the 67th move.