🧠:[<think>Okay, let's try to solve this problem step by step. So, we have two sequences, {a_n} and {b_n}. The first part is to find the general formula for {b_n}, and the second part involves finding the sum S_n of the first n terms of another sequence related to c_n, which is defined using logs of the b_n terms. Let me take this one part at a time.Starting with part (1). We know that a_1 is 100, and each subsequent term is the square of the previous one. So, a_{n+1} = (a_n)^2. That sounds like a recursive sequence where each term is the square of the prior term. Then, b_n is defined as the logarithm (base 10, I assume, since it's written as lg) of a_n. So, b_n = log(a_n). The first part asks for the general formula of {b_n}.Let me write out the first few terms of {a_n} to see the pattern.a_1 = 100a_2 = a_1^2 = 100^2 = 10,000a_3 = a_2^2 = (10,000)^2 = 100,000,000a_4 = a_3^2 = (100,000,000)^2 = 10,000,000,000,000,000So, each term is the square of the previous one, which means the exponents are growing exponentially. Since a_1 is 100, which is 10^2. Then a_2 is (10^2)^2 = 10^4, a_3 is (10^4)^2 = 10^8, a_4 is 10^16, etc. So, in general, a_n is 10^(2^{n})). Wait, let's check:a_1 is 10^2 = 10^2^1 = 10^(2^1)a_2 is 10^4 = 10^(2^2)a_3 is 10^8 = 10^(2^3)a_4 is 10^16 = 10^(2^4)Yes, so in general, a_n = 10^(2^{n}).Wait, but hold on. Let me verify with n=1: a_1 is given as 100, which is 10^2, so if n=1, then a_1 = 10^(2^1) = 10^2 = 100. Correct. Similarly, a_2 would be 10^(2^2) = 10^4 = 10,000. Correct. So, the general formula for a_n is 10 raised to the power of 2^n. So, a_n = 10^{2^{n}}.Therefore, since b_n = lg(a_n) = log_{10}(a_n). So substituting the formula for a_n:b_n = log_{10}(10^{2^{n}}) = 2^{n}, because log base 10 of 10^x is x. So, b_n = 2^n. Wait, that seems straightforward. So, part (1) answer is b_n = 2^n.Let me double-check. For n=1, b_1 = log10(a_1) = log10(100) = 2, which is 2^1. For n=2, b_2 = log10(10,000) = 4 = 2^2. For n=3, log10(100,000,000) = 8 = 2^3. Yep, that works. So part (1) is solved: b_n = 2^n.Moving on to part (2). We have c_n defined as the sum from k = n to 2n of log_2(b_k). So, c_n = log_2(b_n) + log_2(b_{n+1}) + ... + log_2(b_{2n}).Since we already know that b_k = 2^k, then log_2(b_k) = log_2(2^k) = k. Because log base 2 of 2^k is just k. Therefore, each term in the sum for c_n is just the exponent k. So, c_n = sum_{k = n}^{2n} k.That simplifies the problem. So, c_n is the sum of integers from n to 2n inclusive. Let's compute that sum.The sum of consecutive integers from a to b is given by (b(b+1)/2) - ((a-1)a/2). So, substituting a = n and b = 2n:Sum = [2n(2n + 1)/2] - [(n - 1)n/2] = [2n(2n +1) - n(n -1)] / 2Let me compute that:First term: 2n(2n +1)/2 = n(2n +1)Second term: (n -1)n/2So, the total sum is n(2n +1) - [(n(n -1))/2]Wait, actually, the formula is Sum_{k=a}^b k = Sum_{k=1}^b k - Sum_{k=1}^{a-1} k} = [b(b+1)/2] - [(a-1)a/2]So, substituting a =n, b=2n:Sum = [2n(2n +1)/2] - [(n -1)n /2] = [n(2n +1)] - [n(n -1)/2]Wait, but let me compute step by step:First part: 2n(2n +1)/2 = n(2n +1)Second part: (n -1)n /2Therefore, total sum = n(2n +1) - (n(n -1)/2)Let me compute this:Multiply out the terms:n(2n +1) = 2n^2 + n(n(n -1))/2 = (n^2 -n)/2Therefore, sum = (2n^2 +n) - (n^2 -n)/2To subtract these, let's get a common denominator:= (4n^2 + 2n)/2 - (n^2 -n)/2= [4n^2 +2n -n^2 +n]/2= [3n^2 +3n]/2= 3n(n +1)/2Therefore, c_n = 3n(n +1)/2So, that's the value of c_n. Therefore, the sequence {1/c_n} is 1 divided by 3n(n +1)/2, which is 2/(3n(n +1)).But the problem says "the sum S_n of the first n terms of the sequence {1/c_n}". So, we need to compute S_n = sum_{k=1}^n [1/c_k] = sum_{k=1}^n [2/(3k(k +1))]}So, S_n = (2/3) * sum_{k=1}^n [1/(k(k +1))]The sum of 1/(k(k +1)) is a telescoping series. Because 1/(k(k +1)) = 1/k - 1/(k +1). Let's verify:1/k - 1/(k +1) = (k +1 -k)/(k(k +1)) = 1/(k(k +1)), yes, correct.Therefore, sum_{k=1}^n [1/(k(k +1))] = sum_{k=1}^n [1/k -1/(k +1)] = [1 -1/2] + [1/2 -1/3] + ... + [1/n -1/(n +1)]All the intermediate terms cancel, leaving 1 -1/(n +1) = n/(n +1)Therefore, S_n = (2/3) * [n/(n +1)] = (2n)/(3(n +1))So, the sum S_n is 2n/(3(n +1))Let me recap to make sure I didn't skip steps. The key steps were recognizing that b_n is 2^n, so log_2(b_k) = k, which makes c_n the sum from n to 2n of k. Then, calculating that sum, which turned out to be 3n(n +1)/2. Then, taking 1/c_n gives 2/(3n(n +1)). Summing the first n terms of this sequence involves summing 2/(3k(k +1)) from k=1 to n, which telescopes to 2/3 times (1 -1/(n +1)) = 2n/(3(n +1)). Double-checking the sum from k=n to 2n of k. For example, if n=1, then sum from 1 to 2: 1 +2=3. According to the formula, 3*1*(1 +1)/2 = 3*2/2=3. Correct. If n=2, sum from 2 to4: 2+3+4=9. Formula: 3*2*(2 +1)/2= 3*2*3/2=9. Correct. So the formula for c_n is correct. Then, 1/c_n is 2/(3n(n +1)). For n=1, 1/c_1 = 2/(3*1*2)=1/3, and S_1 =1/3. According to the formula, 2*1/(3*(1 +1))=2/(6)=1/3. Correct. For n=2, S_2=1/c_1 +1/c_2 =1/3 + 2/(3*2*3)=1/3 +1/9=4/9. Formula: 2*2/(3*(2 +1))=4/9. Correct. So the formula seems correct.Thus, the final answer for part (1) is b_n =2^n, and for part (2), the sum S_n=2n/(3(n +1)). Final Answer(1) boxed{b_n = 2^n}(2) boxed{S_n = dfrac{2n}{3(n + 1)}}</think>Given sequences {a_{n}} and {b_{n}}, where a_{1}=100, a_{n+1}=a_n^2, and b_{n}=lg a_{n}.(1) To find the general formula for sequence {b_{n}}:- The sequence {a_n} is defined by a_1 = 100 and a_{n+1} = a_n^2. This results in a_n = 10^{2^n}.- Taking the logarithm base 10 of a_n, we get b_n = lg a_n = lg(10^{2^n}) = 2^n.Thus, the general formula for {b_n} is:[boxed{b_n = 2^n}](2) To find the sum S_n of the first n terms of the sequence left{frac{1}{c_n}right} where c_n = log_2 b_n + log_2 b_{n+1} + cdots + log_2 b_{2n}:- Since b_k = 2^k, we have log_2 b_k = k. Therefore, c_n is the sum of integers from n to 2n: [ c_n = sum_{k=n}^{2n} k = frac{3n(n+1)}{2} ]- The sequence left{frac{1}{c_n}right} is given by frac{2}{3n(n+1)}.- The sum S_n is the sum of the first n terms of this sequence: [ S_n = sum_{k=1}^{n} frac{2}{3k(k+1)} ] This is a telescoping series: [ sum_{k=1}^{n} left(frac{1}{k} - frac{1}{k+1}right) = 1 - frac{1}{n+1} ] Therefore, [ S_n = frac{2}{3} left(1 - frac{1}{n+1}right) = frac{2}{3} cdot frac{n}{n+1} = frac{2n}{3(n+1)} ]Thus, the sum S_n is:[boxed{S_n = dfrac{2n}{3(n + 1)}}],<think>Okay, let's see. I have two sequences here: {a_n} and {b_n}. The problem is divided into two parts. First, I need to find the general formula for {b_n}, and then use that to find the sum S_n of the first n terms of the sequence {1/c_n}, where c_n is defined as the sum from log₂ b_n to log₂ b_{2n}. Hmm, that seems a bit involved, but let's take it step by step.Starting with part (1): Find the general formula for {b_n}. Given that a₁ = 100, and a_{n+1} = (a_n)². Also, b_n = lg a_n. I need to express b_n in terms of n. First, let's recall that "lg" typically stands for logarithm base 10. So, b_n = log₁₀(a_n). Got it. Let's try to find a pattern for a_n first, then convert it to b_n using logarithm properties.Given a₁ = 100. Then, a₂ = (a₁)² = 100² = 10,000. a₃ = (a₂)² = (10,000)² = 100,000,000. Wait a second, this is 10^2, then (10^2)^2 = 10^4, then (10^4)^2 = 10^8, and so on. So, each term is squaring the previous term, which is equivalent to exponentiating 10 to the power of 2^{n-1} multiplied by the initial exponent.Wait, a₁ = 100 = 10². Then a₂ = (10²)² = 10^(2*2) = 10^4. a₃ = (10^4)^2 = 10^8. a₄ = 10^16, etc. So, in general, a_n = 10^{2^{n}}. Wait, let's check that.At n=1: a₁ = 10^{2^1} = 10² = 100, which is correct. At n=2: 10^{2²} = 10⁴ = 10,000, correct. n=3: 10^{2³}=10^8, correct. So yes, a_n = 10^{2^{n}}.Therefore, b_n = lg(a_n) = log₁₀(10^{2^{n}}) = 2^{n}, because log₁₀(10^x) = x. So, b_n = 2^{n}. That seems straightforward. So part (1) is done; the general formula for {b_n} is 2^n.Moving on to part (2): Define c_n as the sum from log₂ b_n to log₂ b_{2n}. Wait, the problem says c_n = log₂ b_n + log₂ b_{n+1} + ... + log₂ b_{2n}. So it's a sum of log base 2 of b_k, where k ranges from n to 2n. Then, we need to find the sum S_n of the first n terms of the sequence {1/c_n}. So S_n = 1/c₁ + 1/c₂ + ... + 1/c_n.Alright, let's first compute c_n. Since we already know that b_k = 2^k, so log₂ b_k = log₂ (2^k) = k. Therefore, log₂ b_k is simply k. Therefore, c_n is the sum from k = n to k = 2n of k. So, c_n = Σ_{k = n}^{2n} k.This is an arithmetic series. The first term is n, the last term is 2n, and the number of terms is (2n - n + 1) = n + 1. The formula for the sum of an arithmetic series is (number of terms)/2 * (first term + last term). Therefore, c_n = (n + 1)/2 * (n + 2n) = (n + 1)/2 * 3n = (3n(n + 1))/2.Wait, let me check that again. The number of terms from n to 2n inclusive is 2n - n + 1 = n + 1. The first term is n, the last term is 2n. So sum = (n + 1)*(n + 2n)/2 = (n + 1)*(3n)/2 = (3n(n + 1))/2. So c_n = 3n(n + 1)/2. Therefore, 1/c_n = 2/(3n(n + 1)). So the sequence {1/c_n} is 2/(3n(n + 1)). Therefore, we need to compute S_n = Σ_{k=1}^{n} [2/(3k(k + 1))]. Let's factor out the constant 2/3: S_n = (2/3) * Σ_{k=1}^{n} [1/(k(k + 1))]. The sum Σ_{k=1}^{n} [1/(k(k + 1))] is a telescoping series. Recall that 1/(k(k + 1)) = 1/k - 1/(k + 1). Let's verify:1/k - 1/(k + 1) = (k + 1 - k)/(k(k + 1)) = 1/(k(k + 1)), which is correct. Therefore, the sum telescopes:Σ_{k=1}^{n} [1/k - 1/(k + 1)] = (1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n + 1)) = 1 - 1/(n + 1) = n/(n + 1).Therefore, S_n = (2/3)*(n/(n + 1)) = (2n)/(3(n + 1)).So putting it all together:(1) The general formula for {b_n} is 2^n.(2) The sum S_n is 2n/(3(n + 1)).Wait, let me double-check part (2). Starting from c_n = Σ_{k = n}^{2n} log₂ b_k. Since b_k = 2^k, then log₂ b_k = k. So c_n is sum_{k = n}^{2n} k. The number of terms is 2n - n + 1 = n + 1. So sum is (n + 1)(n + 2n)/2 = (n + 1)(3n)/2. So c_n = 3n(n + 1)/2. Thus, 1/c_n = 2/(3n(n + 1)). Then S_n is sum_{k=1}^n [2/(3k(k + 1))] = (2/3) sum_{k=1}^n [1/(k(k + 1))] = (2/3)(1 - 1/(n + 1)) = (2/3)(n/(n + 1)) = 2n/(3(n + 1)). Yes, that seems correct.Therefore, the answers are:(1) b_n = 2^n(2) S_n = 2n/(3(n + 1))Final Answer(1) The general formula for ( {b_n} ) is (boxed{2^n}).(2) The sum ( S_n ) is (boxed{dfrac{2n}{3(n + 1)}}).</think>Given sequences ({a_n}) and ({b_n}), where (a_1 = 100), (a_{n+1} = a_n^2), and (b_n = lg a_n):1. Finding the general formula for ({b_n}): - The sequence ({a_n}) is defined by (a_{n+1} = a_n^2). Starting with (a_1 = 100 = 10^2), we observe that each term is the square of the previous term. This results in (a_n = 10^{2^n}). - Since (b_n = lg a_n), substituting (a_n = 10^{2^n}) gives (b_n = lg(10^{2^n}) = 2^n). - Therefore, the general formula for ({b_n}) is (boxed{2^n}).2. Finding the sum (S_n) of the first (n) terms of the sequence (left{frac{1}{c_n}right}): - Given (c_n = log_2 b_n + log_2 b_{n+1} + cdots + log_2 b_{2n}), and knowing (b_k = 2^k), we have (log_2 b_k = k). - Therefore, (c_n) is the sum of integers from (n) to (2n). This is an arithmetic series with the first term (n), last term (2n), and (n + 1) terms. The sum is (frac{(n + 1)(n + 2n)}{2} = frac{3n(n + 1)}{2}). - Thus, (c_n = frac{3n(n + 1)}{2}) and (frac{1}{c_n} = frac{2}{3n(n + 1)}). - The sum (S_n = sum_{k=1}^{n} frac{1}{c_k} = frac{2}{3} sum_{k=1}^{n} frac{1}{k(k + 1)}). This sum telescopes: (sum_{k=1}^{n} left(frac{1}{k} - frac{1}{k + 1}right) = 1 - frac{1}{n + 1} = frac{n}{n + 1}). - Therefore, (S_n = frac{2}{3} cdot frac{n}{n + 1} = frac{2n}{3(n + 1)}). - The sum (S_n) is (boxed{dfrac{2n}{3(n + 1)}}).]

🔑: Solution:# Part 1: Finding the General Formula for Sequence {b_{n}}Given that a_{1}=100 and a_{n+1}=a_n^2, we start by observing the nature of the sequence {a_{n}}:1. Since a_{1}=100 > 1, and a_{n+1}=a_n^2, it follows that a_{2}=a_1^2=100^2 > 1, and similarly, a_{3}=a_2^2 > 1, and so on.2. By induction, we can conclude that for any nin mathbb{N}^{*}, a_{n} > 1.Given b_{n}=lg a_{n}, we find the relationship between consecutive terms of {b_{n}}:- lg a_{n+1}=lg (a_n^2) = 2lg a_{n}- Therefore, b_{n+1}=2b_{n}Since b_{1}=lg a_{1} = lg 100 = 2, the sequence {b_{n}} is a geometric sequence with the first term 2 and common ratio 2. Thus, the general formula for {b_{n}} is:- b_{n}=2times 2^{n-1}=2^nEncapsulating the final answer for part (1):- boxed{b_{n}=2^n}# Part 2: Finding the Sum S_{n} of the First n Terms of the Sequence left{frac{1}{c_n}right}Given log _{2}b_{n}=n, we can express c_{n} as:- c_{n}=n+(n+1)+(n+2)+cdots+2nThis is an arithmetic series with the first term n, the last term 2n, and n+1 terms in total. The sum of this series is:- c_{n}=frac{(n+1)(n+2n)}{2}=frac{3n(n+1)}{2}The reciprocal of c_{n} is:- frac{1}{c_n}=frac{2}{3n(n+1)}=frac{2}{3}left(frac{1}{n}-frac{1}{n+1}right)The sum S_{n} of the first n terms of the sequence left{frac{1}{c_n}right} is:- S_n=frac{2}{3}left(left(1-frac{1}{2}right)+left(frac{1}{2}-frac{1}{3}right)+left(frac{1}{3}-frac{1}{4}right)+cdots+left(frac{1}{n}-frac{1}{n+1}right)right)- Simplifying, we get S_n=frac{2}{3}left(1-frac{1}{n+1}right)=frac{2n}{3(n+1)}Encapsulating the final answer for part (2):- boxed{S_n=frac{2n}{3(n+1)}}

🧠:[<think>Okay, so I need to figure out how many positive integers with at most 3 digits have this mirror property. The mirror property is when a number is a perfect square, and when you reverse its digits, you get another perfect square which is the square of the reversed digits of the original square root. Let me break this down.First, let me make sure I understand the examples given. The number 144 is a square of 12, and reversing its digits gives 441, which is the square of 21, which is the reverse of 12. Similarly, 100 is a square of 10, and reversing gives 001, which is 1, the square of 1 (reverse of 10 is 01, which is 1). But 49 is 7 squared, reversing gives 94, which is not a perfect square, so it doesn't have the mirror property.So the task is to count all such numbers with up to three digits. Let's start by considering all numbers N such that N is a perfect square, and when you reverse the digits of N, you get another perfect square M, which should be the square of the reverse of the square root of N. Let me formalize this:Let N be a perfect square, so N = k² for some integer k. Let rev(N) be the number obtained by reversing the digits of N. Then rev(N) should be a perfect square as well, specifically rev(N) = (rev(k))².Therefore, the conditions are:1. N is a perfect square.2. rev(N) is a perfect square.3. rev(N) = (rev(k))² where k = sqrt(N).So, we need to check for all N up to 999 (since at most 3 digits), find N such that N is a square, and rev(N) is a square, and the square root of rev(N) is the reverse of the square root of N.Wait, but maybe not exactly. The problem states: "reversing the order of its digits results in precisely the square of the number obtained by reversing the digits of its square root." So, if N = k², then rev(N) = (rev(k))². Therefore, rev(N) must equal the square of rev(k). So, both N and rev(N) must be perfect squares, with rev(N) being the square of rev(k). So, for each k such that k² is a number with up to three digits, we check if rev(k²) is equal to (rev(k))².Therefore, another way to approach this problem is to iterate over all possible k where k² is a 1, 2, or 3-digit number. Then for each k, compute rev(k) and check if (rev(k))² equals rev(k²). If yes, then count k² as a number with the mirror property.So, first, let's determine the range of k. Since N is at most 999, k can be from 1 up to sqrt(999). sqrt(900) is 30, sqrt(961) is 31, sqrt(999) is approximately 31.6. So k can be up to 31, since 31² = 961, and 32² = 1024 which is a 4-digit number. So k ranges from 1 to 31.But actually, k can also be 0? Wait, the problem says positive integers, so N must be positive. So k starts from 1. So k is 1 to 31 inclusive.Now, for each k from 1 to 31, compute k², reverse the digits of k² to get rev(k²), reverse the digits of k to get rev(k), then check if rev(k²) is equal to (rev(k))². If yes, then count that k² as having the mirror property.Additionally, the problem mentions that 100 has the mirror property. Let's verify that. 100 is 10². Reverse the digits of 100 is 001, which is 1, and reverse the digits of 10 is 01, which is 1, and 1² is 1. So yes, that works. Similarly, 1 is 1², reverse is 1, and reverse of 1 is 1, so 1²=1. So 1 also works.So, the plan is:1. Generate all k from 1 to 31.2. For each k, compute k² and rev(k).3. Reverse the digits of k² to get rev(k²).4. Check if rev(k²) equals (rev(k))².5. If yes, then include k² in the count.6. Also, check if leading zeros are handled properly. For example, when reversing 100, leading zeros become trailing zeros, which would be 001 -> 1. Similarly, reversing 10 gives 01, which is 1. So in programming terms, when reversing numbers, leading zeros are dropped. So when reversing, numbers like 100 become 001, but 001 is considered as 1. Therefore, when reversing k², we need to treat it as an integer, so leading zeros are ignored.But in terms of reversing digits, the problem says "reversing the order of its digits". For example, reversing 100 gives 001, which is 1. So, in effect, when we reverse a number, we drop leading zeros. So when we reverse k², we need to convert it to a number by reversing all digits, including leading zeros if any, but leading zeros become trailing zeros, which are not part of the number. So, for example, reversing 100 gives 1, not 001. Therefore, when reversing k², we should reverse the digits as a string, then convert back to an integer, which automatically drops leading zeros.Therefore, in code terms, for a number N, rev(N) is the integer obtained by reversing the digits of N and converting back to integer (thereby dropping leading zeros). Similarly, rev(k) is the integer obtained by reversing the digits of k and dropping leading zeros. Therefore, when we reverse k, which is a number from 1 to 31, we need to reverse its digits as a string and then convert back to integer. For example, k=10, rev(k)=01 -> 1. k=12, rev(k)=21. k=1, rev(k)=1. k=30, rev(k)=03 -> 3. So rev(k) can sometimes be a smaller number.Therefore, the steps again:For each k from 1 to 31:1. Compute k².2. Reverse the digits of k² to get rev(k²) as an integer.3. Reverse the digits of k to get rev(k) as an integer.4. Compute (rev(k))².5. Check if rev(k²) equals (rev(k))².6. If yes, then k² has the mirror property.Now, let's perform this check for each k from 1 to 31.First, let's list k from 1 to 31, and compute k², rev(k²), rev(k), (rev(k))², and check equality.Let's start with k=1:k=1:k²=1rev(k²)=1rev(k)=1(rev(k))²=11=1, so yes. So 1 is counted.k=2:k²=4rev(k²)=4rev(k)=2(rev(k))²=44=4, so yes. 4 is counted.Wait, but in the problem statement, it was mentioned that 49 does not have the mirror property. Let's check k=7:k=7, k²=49rev(k²)=94rev(k)=7(rev(k))²=4994 ≠ 49, so 49 is not counted. So why does k=2 work? Let's check:k=2, k²=4. rev(k²)=4. rev(k)=2. (rev(k))²=4. So 4=4. So 4 is a mirror number. But 4 is a 1-digit number. So 4 is acceptable. The problem didn't say single-digit numbers are excluded. The problem says "positive integers with at most 3 digits", which includes 1, 2, and 3-digit numbers.But wait, 4 is a 1-digit number, so reversing its digits gives the same number, and rev(k)=2 reversed is still 2. So 4=2², rev(4)=4=2². Wait, rev(k)=2, which is same as k. So (rev(k))²=4. So rev(k²)=4 equals (rev(k))²=4. Therefore, 4 has the mirror property. So 4 is valid.Similarly, let's check k=3:k=3:k²=9rev(k²)=9rev(k)=3(rev(k))²=99=9, so 9 is counted.k=10:k=10:k²=100rev(k²)=001=1rev(k)=01=1(rev(k))²=1²=1rev(k²)=1 equals 1. So 100 is counted. As per the problem statement.k=11:k=11:k²=121rev(k²)=121rev(k)=11(rev(k))²=121121=121, so yes. So 121 is counted.k=12:k=12:k²=144rev(k²)=441rev(k)=21(rev(k))²=441441=441, so yes. 144 is counted.k=13:k=13:k²=169rev(k²)=961rev(k)=31(rev(k))²=961961=961, so yes. 169 is counted.k=20:k=20:k²=400rev(k²)=004=4rev(k)=02=2(rev(k))²=44=4, so yes. 400 is counted.k=21:k=21:k²=441rev(k²)=144rev(k)=12(rev(k))²=144144=144, so yes. 441 is counted.k=22:k=22:k²=484rev(k²)=484rev(k)=22(rev(k))²=484484=484, so yes. 484 is counted.k=30:k=30:k²=900rev(k²)=009=9rev(k)=03=3(rev(k))²=99=9, so yes. 900 is counted.k=31:k=31:k²=961rev(k²)=169rev(k)=13(rev(k))²=169169=169, so yes. 961 is counted.Now, let's check some other k's to see if they also satisfy.k=1: 1 (yes)k=2:4 (yes)k=3:9 (yes)k=4:k=4:k²=16rev(k²)=61rev(k)=4(rev(k))²=1661≠16, so 16 is not counted.k=5:k=5:k²=25rev(k²)=52rev(k)=5(rev(k))²=2552≠25, so no.k=6:k=6:k²=36rev(k²)=63rev(k)=6(rev(k))²=3663≠36, so no.k=7:k=7:k²=49rev(k²)=94rev(k)=7(rev(k))²=4994≠49, as given in the problem. So no.k=8:k=8:k²=64rev(k²)=46rev(k)=8(rev(k))²=6446≠64, so no.k=9:k=9:k²=81rev(k²)=18rev(k)=9(rev(k))²=8118≠81, so no.k=10: done, 100 (yes)k=11:121 (yes)k=12:144 (yes)k=13:169 (yes)k=14:k=14:k²=196rev(k²)=691rev(k)=41(rev(k))²=1681691≠1681, so no.k=15:k=15:k²=225rev(k²)=522rev(k)=51(rev(k))²=2601522≠2601, so no.k=16:k=16:k²=256rev(k²)=652rev(k)=61(rev(k))²=3721652≠3721, so no.k=17:k=17:k²=289rev(k²)=982rev(k)=71(rev(k))²=5041982≠5041, so no.k=18:k=18:k²=324rev(k²)=423rev(k)=81(rev(k))²=6561423≠6561, so no.k=19:k=19:k²=361rev(k²)=163rev(k)=91(rev(k))²=8281163≠8281, so no.k=20:400 (yes)k=21:441 (yes)k=22:484 (yes)k=23:k=23:k²=529rev(k²)=925rev(k)=32(rev(k))²=1024925≠1024, so no.k=24:k=24:k²=576rev(k²)=675rev(k)=42(rev(k))²=1764675≠1764, so no.k=25:k=25:k²=625rev(k²)=526rev(k)=52(rev(k))²=2704526≠2704, so no.k=26:k=26:k²=676rev(k²)=676rev(k)=62(rev(k))²=3844676≠3844, so no.Wait, rev(k²)=676, which is same as k². So rev(k)=62, (rev(k))²=3844. 676≠3844. So no.k=27:k=27:k²=729rev(k²)=927rev(k)=72(rev(k))²=5184927≠5184, so no.k=28:k=28:k²=784rev(k²)=487rev(k)=82(rev(k))²=6724487≠6724, so no.k=29:k=29:k²=841rev(k²)=148rev(k)=92(rev(k))²=8464148≠8464, so no.k=30:900 (yes)k=31:961 (yes)So from k=1 to k=31, the numbers that satisfy the mirror property are:k=1:1k=2:4k=3:9k=10:100k=11:121k=12:144k=13:169k=20:400k=21:441k=22:484k=30:900k=31:961Wait, that's 12 numbers. Let me count them again:1. 1 (k=1)2. 4 (k=2)3. 9 (k=3)4. 100 (k=10)5. 121 (k=11)6. 144 (k=12)7. 169 (k=13)8. 400 (k=20)9. 441 (k=21)10. 484 (k=22)11. 900 (k=30)12. 961 (k=31)So total 12 numbers. But let me check if there are any others.Wait, I might have missed k=22. Let me confirm:k=22:484. rev(484)=484, rev(k)=22, (rev(k))²=484. So yes, 484 is included. So yes, 12 numbers.Wait, but let me check k=0. Wait, the problem says positive integers, so N must be positive. So k=0 would give N=0, which is not positive. So we don't consider k=0.Wait, but wait, the problem says "positive integers with at most 3 digits". So numbers from 1 to 999. So yes, 0 is excluded. So k starts from 1.But let's verify if the numbers we have are correct:1, 4, 9, 100, 121, 144, 169, 400, 441, 484, 900, 961. That's 12 numbers.But according to the problem statement, 1, 100, 144 are examples. Let me check if these are in the list. Yes, they are.Wait, but wait, let me check k=1,2,3,10,11,12,13,20,21,22,30,31. So 12 numbers. Each of these k's gives a valid N.But wait, are there any other numbers? For example, let's check k=33. But 33²=1089, which is 4 digits. So beyond k=31, it's 4 digits. So no.Wait, let's check k=1 to k=31 again. Let me check k=4 to k=9 again:k=4:16 → rev(16)=61 vs (rev(4))²=16 → 61≠16, so no.k=5:25 → rev(25)=52 vs (rev(5))²=25 → 52≠25, no.k=6:36→63≠36, no.k=7:49→94≠49, no.k=8:64→46≠64, no.k=9:81→18≠81, no.So none of k=4 to k=9 (except k=1,2,3) work.k=10: works.k=11: works.k=12: works.k=13: works.k=14: doesn't work.k=15: doesn't.k=16: no.k=17: no.k=18: no.k=19: no.k=20: yes.k=21: yes.k=22: yes.k=23: no.k=24: no.k=25: no.k=26: no.k=27: no.k=28: no.k=29: no.k=30: yes.k=31: yes.So yes, total 12 numbers.But wait, let me check k=1,2,3. For k=1:1, k=2:4, k=3:9. These are single-digit squares. When reversed, they remain the same. rev(k) is 1,2,3, so (rev(k))² is 1,4,9. rev(N) is 1,4,9. So these are equal. Therefore, they satisfy the mirror property.So according to the problem statement, these numbers are included. The problem says "the number 1 also has the mirror property". So 1,4,9 are all single-digit squares that when reversed (which is themselves) are the squares of their roots reversed (which are themselves). Therefore, they do count.Therefore, the total number is 12. But wait, the problem might have some edge cases. Let me check if numbers like 100, 400, 900 are included correctly.For example, k=10:100. rev(100)=1, rev(k)=1, (rev(k))²=1. So 1=1, correct. So 100 is included.k=20:400. rev(400)=4, rev(k)=2, 2²=4. Correct.k=30:900. rev(900)=9, rev(k)=3, 3²=9. Correct.Similarly, k=11:121. rev(121)=121, rev(k)=11, 11²=121. Correct.k=12:144. rev(144)=441, rev(k)=21, 21²=441. Correct.k=13:169. rev(169)=961, rev(k)=31, 31²=961. Correct.k=21:441. rev(441)=144, rev(k)=12, 12²=144. Correct.k=22:484. rev(484)=484, rev(k)=22, 22²=484. Correct.k=31:961. rev(961)=169, rev(k)=13, 13²=169. Correct.So all these 12 numbers check out. Therefore, the answer should be 12.But wait, let me check once again for any possible missing numbers.Wait, is there a number like 0? But no, since we need positive integers. So 0 is excluded.What about numbers like 2-digit numbers, for example, 25. k=5:25, rev(25)=52, rev(k)=5, 5²=25≠52. So no. So as before.Similarly, 3-digit numbers like 625 (25²). rev(625)=526, rev(k)=52, 52²=2704≠526. So no.Therefore, the total numbers are indeed 12.Wait, but wait, the problem says "at most 3 digits". So numbers with 1, 2, or 3 digits. The 1-digit numbers 1,4,9 are included, as we have. The 3-digit numbers like 100, 121, 144, 169, 400, 441, 484, 900, 961. Wait, that's 9 three-digit numbers, plus 3 single-digit numbers. Total 12.Wait, but 121 is a 3-digit number. rev(121)=121. rev(k)=11. 11²=121. So yes. 121 is included.Similarly, 484: rev(484)=484, rev(k)=22. 22²=484. Correct.So 12 numbers in total.Therefore, the answer should be 12.Wait, but let me check the problem statement again. It says "positive integers with at most 3 digits". So numbers from 1 to 999. So in our list, 1,4,9 are single-digit; 100,121,144,169,400,441,484,900,961 are three-digit. Wait, 121 is three digits. So 3 single-digit and 9 three-digit. Total 12. Correct.But let me just make sure that when k=10, rev(k)=01=1. So 10 reversed is 01, which is 1. Then (rev(k))²=1²=1. And rev(k²)=001=1. So 1=1. Therefore, 100 is included. Similarly, k=20: rev(k)=02=2, 2²=4, rev(400)=004=4. So yes. Similarly, k=30: rev(k)=03=3, 3²=9, rev(900)=009=9. So yes.Therefore, all 12 numbers satisfy the mirror property. Therefore, the answer should be 12.Final Answerboxed{12}</think>To determine how many positive integers with at most 3 digits enjoy the mirror property, we need to check if a number ( N ) is a perfect square, and if reversing its digits results in another perfect square which is the square of the reversed digits of its square root. We check each ( k ) from 1 to 31 (since ( 31^2 = 961 ) is the largest 3-digit perfect square) and verify if reversing ( k^2 ) gives ( (text{rev}(k))^2 ).Here are the key steps and results:1. Check each ( k ) from 1 to 31: - Calculate ( k^2 ). - Reverse the digits of ( k^2 ) to get ( text{rev}(k^2) ). - Reverse the digits of ( k ) to get ( text{rev}(k) ). - Check if ( text{rev}(k^2) = (text{rev}(k))^2 ).2. Numbers that satisfy the mirror property: - ( k = 1 ): ( 1^2 = 1 ), reversed is 1, ( text{rev}(1) = 1 ), ( 1^2 = 1 ). - ( k = 2 ): ( 2^2 = 4 ), reversed is 4, ( text{rev}(2) = 2 ), ( 2^2 = 4 ). - ( k = 3 ): ( 3^2 = 9 ), reversed is 9, ( text{rev}(3) = 3 ), ( 3^2 = 9 ). - ( k = 10 ): ( 10^2 = 100 ), reversed is 001 (1), ( text{rev}(10) = 01 (1) ), ( 1^2 = 1 ). - ( k = 11 ): ( 11^2 = 121 ), reversed is 121, ( text{rev}(11) = 11 ), ( 11^2 = 121 ). - ( k = 12 ): ( 12^2 = 144 ), reversed is 441, ( text{rev}(12) = 21 ), ( 21^2 = 441 ). - ( k = 13 ): ( 13^2 = 169 ), reversed is 961, ( text{rev}(13) = 31 ), ( 31^2 = 961 ). - ( k = 20 ): ( 20^2 = 400 ), reversed is 004 (4), ( text{rev}(20) = 02 (2) ), ( 2^2 = 4 ). - ( k = 21 ): ( 21^2 = 441 ), reversed is 144, ( text{rev}(21) = 12 ), ( 12^2 = 144 ). - ( k = 22 ): ( 22^2 = 484 ), reversed is 484, ( text{rev}(22) = 22 ), ( 22^2 = 484 ). - ( k = 30 ): ( 30^2 = 900 ), reversed is 009 (9), ( text{rev}(30) = 03 (3) ), ( 3^2 = 9 ). - ( k = 31 ): ( 31^2 = 961 ), reversed is 169, ( text{rev}(31) = 13 ), ( 13^2 = 169 ).The numbers that satisfy the mirror property are: 1, 4, 9, 100, 121, 144, 169, 400, 441, 484, 900, and 961. Thus, the total number of positive integers with at most 3 digits that enjoy the mirror property is (boxed{12}).,<think>Okay, so I need to figure out how many positive integers with at most 3 digits have this mirror property. Let me start by understanding exactly what the mirror property is. From the problem statement, a number has the mirror property if two things are true:1. The number itself is a perfect square. So, if the number is N, then N = k² for some integer k.2. When you reverse the digits of N, you get another number which is also a perfect square, and specifically, it should be the square of the reversed digits of k. So, if N = k², then reverse(N) = (reverse(k))².For example, 144 is a perfect square (12²), and reversing its digits gives 441, which is 21². And 21 is the reverse of 12. Similarly, 100 is 10², and reversing gives 001, which is 1, and 1 is 1², and 1 is the reverse of 10? Wait, hold on, reversing 10 would be 01, which is just 1. Hmm, so maybe leading zeros are ignored when reversing? Because 100 reversed is 001, which is 1, and 1 is 1², but 1 is the reverse of 10 if you consider dropping the leading zero. So, the problem must consider that when reversing the digits, leading zeros are discarded. That makes sense because otherwise, numbers like 100 would have reversed numbers like 001, which is 1, but you wouldn't write it as 001. So leading zeros are ignored in the reversed number.Similarly, the number 1 is a perfect square (1²), and reversing its digits still gives 1, which is 1², and 1 reversed is 1. So that's why 1 is included.The problem says 49 does not have this property. Let's check why. 49 is 7². Reversing the digits gives 94, which is not a perfect square. And even if we reverse the square root 7, which is still 7, 7² is 49, not 94. So 94 isn't a square, so 49 doesn't have the mirror property.So, the task is to count all positive integers N with at most 3 digits (so from 1 to 999) such that:1. N is a perfect square.2. Reverse the digits of N to get M, which is a perfect square of the reversed digits of the square root of N.First, let's think about how to approach this. Maybe we can iterate over all possible square roots k, compute k², reverse the digits of k² to get M, check if M is a square, and if so, check if its square root is the reverse of k.But since numbers can have up to 3 digits, the square roots can be up to sqrt(999) ≈ 31.6. So k can be from 1 to 31, because 31² is 961, which is a 3-digit number, and 32² is 1024, which is 4 digits, so too long.Wait, but N is allowed to have at most 3 digits, so k can be from 1 to 31. Then, for each k in 1 to 31, compute k², reverse the digits, check if the reversed number is a square, and if the square root of the reversed number is the reverse of k.Alternatively, perhaps we can generate all squares up to 999, reverse each of them, check if the reversed number is a square, and then check if the square roots are reverses of each other. That seems similar.Either way, we can generate all k from 1 to 31, compute k², reverse the digits to get M, check if M is a square, say m², then check if m is reverse(k). So the steps would be:1. For each k in 1 to 31: a. Compute N = k². b. Reverse the digits of N to get M. If reversing leads to a number with fewer digits (due to leading zeros), we still take M as the number without leading zeros. For example, reversing 100 gives 001 -> 1. c. Check if M is a perfect square. Let m = sqrt(M). If M is a perfect square, then check if m is equal to reverse(k). If yes, then count N as having the mirror property.So first, let's note that we need to reverse k and check if that's equal to m. So, m should be reverse(k), and M should be m². So, M = (reverse(k))². Therefore, the requirement is that reverse(k²) = (reverse(k))².So, the key equation is reverse(k²) = (reverse(k))².Therefore, for a number k, if reversing its square gives the square of its reversed digits, then k² has the mirror property.Therefore, we need to find all k from 1 to 31 such that reverse(k²) = [reverse(k)]².Therefore, the problem reduces to checking for each k from 1 to 31 whether reverse(k²) equals [reverse(k)]².So, perhaps the approach is:- For each k in 1 to 31: - Compute k². - Reverse the digits of k to get rk. - Compute rk². - Reverse the digits of k² to get rkk. - Check if rkk equals rk². - If yes, then k² has the mirror property.Therefore, the numbers N with the mirror property are k² where k is in 1 to 31 and reverse(k²) = [reverse(k)]².So, to compute this, perhaps it's easiest to iterate k from 1 to 31, compute all these values, and check the condition.Alternatively, perhaps there are patterns or mathematical properties that can be used to find such numbers without checking each one. Let's think about that.First, single-digit numbers:k is from 1 to 9. Then k² is 1 to 81. Reversing k is the same as k (since it's a single digit). Therefore, reverse(k²) = reverse(k²), and reverse(k)² = k². Therefore, for single-digit k, reverse(k²) must equal k². So, that requires that k² is a palindrome. Therefore, the squares of single-digit numbers that are palindromic.But 1²=1 (palindrome), 2²=4 (palindrome), 3²=9 (palindrome), ..., 9²=81 (not a palindrome). Wait, so for k=1,2,3, ...,9, their squares are 1,4,9,16,25,36,49,64,81. Among these, which are palindromic? 1,4,9, which are single-digit. 16 reversed is 61, which is not 16, so 16 is not a palindrome. Similarly, 25 reversed is 52, not a palindrome. So only the single-digit squares (1,4,9) are palindromic. However, the problem states that 1 has the mirror property, but does 4 and 9 have it?Wait, the problem statement says "the number 1 also has the mirror property". Let's check 4. 4 is 2². Reverse the digits of 4 is 4, which is 2². But reverse of the square root (2) is 2, so 2 reversed is 2, so 2²=4. So, 4 should also have the mirror property. Similarly, 9 is 3², reversed is 9, which is 3², and 3 reversed is 3. So, 9 should also have the mirror property. But according to the problem statement, the number 49 does not have the mirror property, which is 7², but 49 reversed is 94, which is not a square.Wait, but in the problem statement, the first example is 144, which is 12², reversed is 441=21². Then 100 is 10², reversed is 001=1=1². So, for 100, reverse of 10 is 01=1, and 1²=1. So that works because leading zeros are dropped. So, for the single-digit squares, reversing the digits (which are single-digit) gives the same digit, which when squared gives the same number. So 1,4,9 all have the mirror property. However, in the problem statement, it says "the number 1 also has the mirror property, while the number 49 does not". So perhaps the user is implying that 1 is included, but 49 is not. But according to my previous reasoning, 4 and 9 should also be included. Let me check.Take 4: reverse of 4 is 4. reverse of the square root (2) is 2. 2 squared is 4. So yes, 4 has the mirror property. Similarly, 9: reverse of 9 is 9. Reverse of 3 is 3. 3 squared is 9. So 9 also has the mirror property. Therefore, the numbers 1,4,9,144,100 (and others) have the mirror property.But according to the problem statement, "the number 1 also has the mirror property, as does the number 1". Wait, the problem statement says: "The number 100 also has the mirror property, as does the number 1, while the number 49 does not." So maybe they are just giving examples. So the user is not excluding 4 and 9. So maybe 1,4,9,100,144 are all valid. So in the problem statement, they just mentioned 1,100,144 as examples. So perhaps 4 and 9 are also valid. Let's check that.Wait, let's actually check the problem statement again:"The number 144 is very special in that it enjoys the mirror property: not only is it true that 144=12² (i.e., it is a perfect square), but it is also true that 441=21², meaning that reversing the order of its digits results in precisely the square of the number obtained by reversing the digits of its square root. The number 100 also has the mirror property, as does the number 1, while the number 49 does not."So, 1 is included. Then, 100 is included. The problem says "as does the number 1", so perhaps 1 is included, but are 4 and 9 also included? According to the problem's definition, if we reverse the digits of N and get a square which is the square of the reversed digits of the square root.For N=4: reverse digits of 4 is 4, which is 2². The square root of N is 2, reversed is 2. 2²=4. Therefore, it satisfies the mirror property. Similarly, N=9: reverse is 9=3², reverse of 3 is 3. So 9 also satisfies. Therefore, according to the problem's definition, 1,4,9,100,144 all have the mirror property.But the problem statement only mentions 1,100,144 as examples. Maybe because they wanted to highlight numbers with multiple digits. But according to the definition, single-digit squares where reversing the digits (which doesn't change them) must equal the square of the reversed square root (which also doesn't change them). So all single-digit squares satisfy the mirror property. Therefore, 1,4,9 are all valid. So why does the problem mention 1 as an example? Maybe they just listed some examples, not all. So perhaps all single-digit squares (1,4,9) have the mirror property. Then 100,144, etc.Therefore, our task is to count all such numbers with up to 3 digits. So first, check all single-digit squares (1,4,9), then two-digit and three-digit numbers.But let's confirm. Let's take N=1. Square root is 1. Reverse N: 1, which is 1². Reverse of the square root (1) is 1. So yes.N=4: square root 2. Reverse N: 4, which is 2². Reverse of 2 is 2. So yes.N=9: square root 3. Reverse N: 9, which is 3². Reverse of 3 is 3. So yes.So all three single-digit squares are valid.Similarly, two-digit numbers:Let's check N=25. Square root is 5. Reverse N: 52. 52 is not a perfect square. So 25 does not have the mirror property.N=36: square root 6. Reverse N: 63. 63 isn't a square.N=49: square root 7. Reverse N:94. Not a square. So 49 doesn't have it.N=64: square root 8. Reverse N:46. Not a square.N=81: square root 9. Reverse N:18. Not a square.So none of the two-digit squares except maybe 100, but 100 is a three-digit number. Wait, 100 is three digits. So perhaps there are no two-digit numbers with the mirror property.Wait, but 100 is three digits. Let's check if there are any two-digit numbers. For two-digit numbers, the square roots would be from 4 to 9 (since 4²=16, 9²=81). Wait, but we saw that reversing their squares gives numbers like 61, 52, etc., which are not squares.Wait, what about N=00? But the problem says positive integers, so 00 is excluded. So maybe there are no two-digit numbers with the mirror property. Then moving on to three-digit numbers.Three-digit squares. Their square roots are from 10 to 31, since 10²=100, 31²=961.So for k from 10 to 31, compute k², reverse the digits, check if it's a square of reverse(k). Let's go through some examples.First, k=10: k²=100. Reverse digits of 100: 001=1. Reverse(k)=01=1. 1²=1. So 100 has the mirror property.k=11: 121. Reverse digits: 121, which is a palindrome. Reverse(k)=11. 11²=121. So 121 reversed is 121, which is 11². So 121 has the mirror property.Wait, so N=121, which is 11². Reverse digits gives 121, which is 11². So reverse(k)=11, which is the same as k. So this satisfies the mirror property.Similarly, k=12: k²=144. Reverse digits:441. Reverse(k)=21. 21²=441. So yes, 144 has the mirror property.k=13: 169. Reverse digits:961. Reverse(k)=31. 31²=961. So 169 reversed is 961, which is 31². So 169 has the mirror property.k=21: k²=441. Reverse digits:144. Reverse(k)=12. 12²=144. So 441 also has the mirror property. Wait, but 441 is already considered as the reverse of 144. So if we count N=144 and N=441 as separate numbers, both have the mirror property. But since we are counting positive integers with at most 3 digits, both 144 and 441 would be counted separately. Similarly, 169 and 961.Wait, but k=21 is in the range (10 to 31). So k=21: 441, reverse digits:144=12². Reverse(k)=12. 12²=144. So yes, 441 is another number with the mirror property.Similarly, k=31: k²=961. Reverse digits:169. Reverse(k)=13. 13²=169. So 961 also has the mirror property.So, let's see. For k=10, 11,12,13,21,31, their squares have the mirror property. Let's check others.k=20: 400. Reverse digits:004=4. Reverse(k)=02=2. 2²=4. So 400 reversed is 4, which is 2², and reverse(k)=2. So 2²=4. So 400 also has the mirror property.Wait, k=20: N=400. Reverse(N)=004=4. reverse(k)=02=2. 2²=4. So yes, 400 has the mirror property.Similarly, k=30: k²=900. Reverse digits:009=9. Reverse(k)=03=3. 3²=9. So 900 has the mirror property.Wait, so k=30: 900. Reverse is 009=9. reverse(k)=03=3. 3²=9. So yes.So k=10,20,30: their squares 100,400,900. Reversed digits give 1,4,9, which are squares of 1,2,3. Reverse(k) is 1,2,3. So 1²=1, 2²=4, 3²=9. So these work.Similarly, let's check k=22: k²=484. Reverse digits:484, which is a palindrome. Reverse(k)=22. 22²=484. So 484 reversed is 484, which is 22². So 484 has the mirror property.k=22: yes, that works.k=33: 33²=1089, which is four digits, so beyond our limit. So k up to 31.k=11: 121. Reverse digits:121. Reverse(k)=11. 11²=121. So 121 is a palindrome and works.k=101: Wait, k can only go up to 31. So 31 is the maximum k.So let's check all possible k from 10 to 31:k=10: 100 -> reverse(N)=1, reverse(k)=1. 1²=1. Yes.k=11: 121 -> reverse(N)=121, reverse(k)=11. 11²=121. Yes.k=12: 144 -> reverse(N)=441, reverse(k)=21. 21²=441. Yes.k=13: 169 -> reverse(N)=961, reverse(k)=31. 31²=961. Wait, reverse(k)=31? Wait, k=13, reverse(k)=31. But 31²=961, which is the reverse of N=169. So yes.k=20: 400 -> reverse(N)=004=4, reverse(k)=02=2. 2²=4. Yes.k=21: 441 -> reverse(N)=144, reverse(k)=12. 12²=144. Yes.k=22: 484 -> reverse(N)=484, reverse(k)=22. 22²=484. Yes.k=30: 900 -> reverse(N)=009=9, reverse(k)=03=3. 3²=9. Yes.k=31: 961 -> reverse(N)=169, reverse(k)=13. 13²=169. Yes.Wait, but what about k=10, reverse(k)=01=1. So reverse(k)=1. Then reverse(N)=1, which is 1². So that works.Similarly, k=20, reverse(k)=02=2. reverse(N)=4=2².k=30, reverse(k)=03=3, reverse(N)=9=3².What about k=11: reverse(k)=11. reverse(N)=121=11². So works.k=12: reverse(k)=21. reverse(N)=441=21². Works.k=13: reverse(k)=31. reverse(N)=961=31². Wait, no: k=13, reverse(k)=31. reverse(N)=169. 31²=961, which is different from 169. Wait, hold on. Wait, k=13, N=169. reverse(N)=961. reverse(k)=31. 31²=961. Therefore, reverse(N)=961=31². So yes. So 169 reversed is 961, which is 31². So that works. Similarly, 961 reversed is 169=13². So both 169 and 961 have the mirror property.Similarly, k=21, reverse(k)=12, reverse(N)=144=12². So 441 (k=21) and 144 (k=12) both work.So, in total, for three-digit numbers, the numbers are:100,121,144,169,400,441,484,900,961.But let's check k=10,11,12,13,20,21,22,30,31.k=10:100k=11:121k=12:144k=13:169k=20:400k=21:441k=22:484k=30:900k=31:961So that's 9 numbers. Then the single-digit numbers:1,4,9. So total numbers would be 9+3=12. But wait, let's check if there are any other numbers.Wait, what about k=1,2,3: their squares are 1,4,9. These are single-digit, so we considered them. Then k=10 to 31: as above. Are there any other k in 10 to 31 whose squares reversed are squares of reversed k?Let me check k=14: 14²=196. Reverse(N)=691. Is 691 a square? 26²=676, 27²=729. So no. Reverse(k)=41. 41²=1681, which is four digits. So no.k=15:225. Reverse(N)=522. Not a square. reverse(k)=51. 51²=2601. Nope.k=16:256. Reverse(N)=652. Not a square. reverse(k)=61. 61²=3721. No.k=17:289. Reverse(N)=982. Not a square. reverse(k)=71. 71²=5041. No.k=18:324. Reverse(N)=423. Not a square. reverse(k)=81. 81²=6561. No.k=19:361. Reverse(N)=163. Not a square. reverse(k)=91. 91²=8281. No.k=23:529. Reverse(N)=925. Not a square. reverse(k)=32. 32²=1024. No.k=24:576. Reverse(N)=675. Not a square. reverse(k)=42. 42²=1764. No.k=25:625. Reverse(N)=526. Not a square. reverse(k)=52. 52²=2704. No.k=26:676. Reverse(N)=676. Palindrome. reverse(k)=62. 62²=3844. No. But wait, 676 is a palindrome, so reverse(N)=676. reverse(k)=62. 62²=3844, which is not 676. So even though reverse(N) is the same as N, unless reverse(k) squared is N, it doesn't count. So 676 is a palindrome, but reverse(k)=62, 62²=3844≠676. So 676 does not have the mirror property.k=27:729. Reverse(N)=927. Not a square. reverse(k)=72. 72²=5184. No.k=28:784. Reverse(N)=487. Not a square. reverse(k)=82. 82²=6724. No.k=29:841. Reverse(N)=148. Not a square. reverse(k)=92. 92²=8464. No.k=32:1024. Exceeds three digits.Therefore, the only three-digit numbers are the ones from k=10,11,12,13,20,21,22,30,31, which are 9 numbers. Plus the single-digit numbers 1,4,9. So total 12 numbers.Wait, but let me check k=22:484. reverse(N)=484, which is a palindrome. reverse(k)=22. 22²=484. So yes, 484 has the mirror property. Similarly, k=22 is included.Similarly, check k=10:100. reverse(N)=001=1. reverse(k)=01=1. 1²=1. So yes.So that's 9 three-digit numbers and 3 single-digit numbers. So total 12.But wait, let's check if there are any numbers with two digits. For example, N=81 is a two-digit number. 81 reversed is 18, which is not a square. But N=64 reversed is 46, not a square. N=49 reversed is 94, not a square. N=36 reversed is 63, not a square. N=25 reversed is 52, not a square. N=16 reversed is 61, not a square. So there are no two-digit numbers except maybe if we consider palindromes. For example, N=25 is not a palindrome. The only two-digit square palindrome is 121, but that's three digits. 11²=121. Wait, 121 is three digits. The two-digit palindromic squares are... Let's see: 11²=121, 22²=484, which are three digits. So two-digit palindromic squares: 0²=0, 1²=1, ..., 9²=81. So the palindromic two-digit squares would be 0,1,4,9,25,36,49,64,81. Wait, 0 isn't positive. 1,4,9 are single-digit. 25,36,49,64,81 are two-digit. But none of them are palindromic. So there are no two-digit palindromic squares. Therefore, there are no two-digit numbers with the mirror property.Therefore, all numbers with the mirror property are either single-digit squares (1,4,9) or three-digit numbers (100,121,144,169,400,441,484,900,961). So total 3 + 9 = 12 numbers.Wait, but let me check if there's any other three-digit numbers. Let's go through k=10 to 31 again:k=10:100k=11:121k=12:144k=13:169k=20:400k=21:441k=22:484k=30:900k=31:961So that's 9 numbers. What about k=101? But k can only go up to 31. So no, k=31 is the max. So those 9 numbers.Plus the single-digit squares 1,4,9. So 12 total. However, let's check the three-digit numbers again:100: yes121: yes144: yes169: yes400: yes441: yes484: yes900: yes961: yesSo that's 9 numbers. 1,4,9: 3 numbers. Total 12.But wait, wait a second. What about N=000? But it's not a positive integer. So 0 is excluded. So yes, 12 numbers. But let me verify each of the three-digit numbers:1. 100: reverse is 001=1=1². reverse(k)=reverse(10)=1. 1²=1. Yes.2. 121: reverse is 121=11². reverse(k)=reverse(11)=11. 11²=121. Yes.3. 144: reverse is 441=21². reverse(k)=reverse(12)=21. 21²=441. Yes.4. 169: reverse is 961=31². reverse(k)=reverse(13)=31. 31²=961. Yes.5. 400: reverse is 004=4=2². reverse(k)=reverse(20)=02=2. 2²=4. Yes.6. 441: reverse is 144=12². reverse(k)=reverse(21)=12. 12²=144. Yes.7. 484: reverse is 484=22². reverse(k)=reverse(22)=22. 22²=484. Yes.8. 900: reverse is 009=9=3². reverse(k)=reverse(30)=03=3. 3²=9. Yes.9. 961: reverse is 169=13². reverse(k)=reverse(31)=13. 13²=169. Yes.All these check out. Then single-digit numbers:1: reverse is 1=1². reverse(k)=reverse(1)=1. 1²=1. Yes.4: reverse is 4=2². reverse(k)=reverse(2)=2. 2²=4. Yes.9: reverse is 9=3². reverse(k)=reverse(3)=3. 3²=9. Yes.So those three are valid. Therefore, total numbers are 9+3=12.Wait, but hold on. The problem statement says "positive integers with at most 3 digits". So 1,4,9 are single-digit, and 100,121,144,169,400,441,484,900,961 are three-digit. So total 12 numbers. However, wait, the problem statement says "at most 3 digits", so numbers like 01 are not considered, as they are not positive integers. Leading zeros are not allowed. So 001 is 1, which is considered as a single-digit number. So all numbers are correctly accounted for.But let me check again in case I missed any numbers. For example, does k=0 count? No, because we are considering positive integers. So k starts from 1. So no.Wait, let's check k=1 to 31:k=1:1 (single-digit)k=2:4 (single-digit)k=3:9 (single-digit)k=10:100 (three-digit)k=11:121 (three-digit)k=12:144 (three-digit)k=13:169 (three-digit)k=20:400 (three-digit)k=21:441 (three-digit)k=22:484 (three-digit)k=30:900 (three-digit)k=31:961 (three-digit)So those are 3 single-digit and 9 three-digit numbers. So 12 total. Therefore, the answer should be 12.But wait, let me check if there's any other numbers in between. For example, k=10:100. Is there another k where reverse(k²) is a square, but maybe the square root is not reverse(k)? But according to the problem statement, it's required that reversing the digits of N gives precisely the square of reversing the digits of k. So even if reverse(N) is a square, but the square root is not reverse(k), then it doesn't count. For example, take N=121. Reverse(N)=121, which is 11². reverse(k)=reverse(11)=11. So 11²=121. So it works. If N=484, reverse(N)=484=22², reverse(k)=22. So yes.If there exists a number N where reverse(N) is a square, but sqrt(reverse(N)) is not reverse(sqrt(N)), then it would not count. For example, take N=256 (16²). Reverse(N)=652, which is not a square. So no. Take N=625 (25²). Reverse(N)=526, which is not a square. So no.Another example: N=225 (15²). Reverse(N)=522, not a square. So no.Therefore, all the numbers that satisfy reverse(N)=reverse(k)² are exactly the 12 numbers listed above.Therefore, the answer is 12. So the number of positive integers with at most 3 digits that enjoy the mirror property is 12.Final Answerboxed{12}</think>To determine how many positive integers with at most 3 digits enjoy the mirror property, we need to check if a number ( N ) is a perfect square and if reversing its digits results in another perfect square, which must be the square of the reversed digits of its square root.1. Single-digit numbers: - The single-digit perfect squares are ( 1, 4, 9 ). - Reversing these digits results in the same number, which are also perfect squares. Hence, ( 1, 4, 9 ) have the mirror property.2. Three-digit numbers: - We need to check numbers ( k^2 ) where ( k ) ranges from 10 to 31 (since ( 31^2 = 961 ) is the largest three-digit perfect square). - For each ( k ), compute ( k^2 ), reverse its digits, and check if the reversed number is the square of the reversed ( k ). - Key checks: - ( k = 10 ): ( 10^2 = 100 ), reversed is 001 = 1, which is ( 1^2 ). Reverse of 10 is 1. - ( k = 11 ): ( 11^2 = 121 ), reversed is 121, which is ( 11^2 ). Reverse of 11 is 11. - ( k = 12 ): ( 12^2 = 144 ), reversed is 441, which is ( 21^2 ). Reverse of 12 is 21. - ( k = 13 ): ( 13^2 = 169 ), reversed is 961, which is ( 31^2 ). Reverse of 13 is 31. - ( k = 20 ): ( 20^2 = 400 ), reversed is 004 = 4, which is ( 2^2 ). Reverse of 20 is 2. - ( k = 21 ): ( 21^2 = 441 ), reversed is 144, which is ( 12^2 ). Reverse of 21 is 12. - ( k = 22 ): ( 22^2 = 484 ), reversed is 484, which is ( 22^2 ). Reverse of 22 is 22. - ( k = 30 ): ( 30^2 = 900 ), reversed is 009 = 9, which is ( 3^2 ). Reverse of 30 is 3. - ( k = 31 ): ( 31^2 = 961 ), reversed is 169, which is ( 13^2 ). Reverse of 31 is 13. - These result in the three-digit numbers ( 100, 121, 144, 169, 400, 441, 484, 900, 961 ).3. Two-digit numbers: - All two-digit perfect squares (16, 25, 36, 49, 64, 81) do not satisfy the mirror property as their reverses are not perfect squares.Summarizing, the valid numbers are:- Single-digit: ( 1, 4, 9 )- Three-digit: ( 100, 121, 144, 169, 400, 441, 484, 900, 961 )Thus, the total number of positive integers with at most 3 digits that enjoy the mirror property is ( 3 + 9 = 12 ).[boxed{12}]]

🔑:13. The answer is 12. The one-digit positive numbers with the reflective property are the perfect squares between 1 and 9, that is, 1, 4, and 9. There are no two-digit numbers with the reflective property: the perfect squares of two digits have a one-digit square root, which implies that they would enjoy the reflective property only if their two digits were the same; but a square with exactly two identical digits is divisible by 11, and thus by 121, and therefore does not exist. As for the three-digit numbers ABC with the reflective property, they must be the squares of a certain two-digit number XY and the number 100C + 10B + A must be the square of 10Y + X. A two-digit number greater than 31 has a square greater than 1000, so both 10X + Y and 10Y + X must be at most 31. It follows that both X and Y must be at most 3, and if one of the digits is 3, the other must be 0 or 1. The only possibilities to consider for the square root of ABC are therefore 10, 11, 12, 13, 20, 21, 22, 30, 31, which produce the following squares with the reflective property: 100, 121, 144, 169, 400, 441, 484, 900, 961. We thus have a total of 12 numbers of at most 3 digits with the reflective property.

🧠:[<think>Okay, let me try to work through this problem step by step. So, the question is about whether a = -1 is a sufficient but not necessary condition for two lines to be perpendicular. The options are A to D, and I need to figure out which one is correct.First, I remember that two lines are perpendicular if the product of their slopes is -1. So, I need to find the slopes of both lines given in the problem and set their product equal to -1. Then, solve for 'a' to see what values make them perpendicular. The original statement claims that a = -1 is a sufficient but not necessary condition, which would mean that when a = -1, the lines are definitely perpendicular, but there might be other values of 'a' that also make them perpendicular. The answer choices depend on whether this is true or not.Let me start by finding the slopes of the two lines. The first line is 4x - (a + 1)y + 9 = 0. To find the slope, I can rewrite this in slope-intercept form (y = mx + b). Let's do that:Starting with 4x - (a + 1)y + 9 = 0.Subtract 4x and 9 from both sides:-(a + 1)y = -4x - 9.Multiply both sides by -1 to make it positive:(a + 1)y = 4x + 9.Now, divide both sides by (a + 1):y = (4 / (a + 1))x + (9 / (a + 1)).So, the slope of the first line is 4 / (a + 1). Let me call this m1 = 4 / (a + 1).Now, the second line is (a² - 1)x - a y + 6 = 0. Let's do the same process.Starting with (a² - 1)x - a y + 6 = 0.Subtract (a² - 1)x and 6 from both sides:- a y = - (a² - 1)x - 6.Multiply both sides by -1:a y = (a² - 1)x + 6.Divide both sides by a:y = [(a² - 1)/a]x + 6/a.So, the slope of the second line is (a² - 1)/a. Let's call this m2 = (a² - 1)/a.Now, for the lines to be perpendicular, the product of their slopes should be -1. Therefore:m1 * m2 = -1.Substituting the expressions for m1 and m2:[4 / (a + 1)] * [(a² - 1)/a] = -1.Let me simplify this equation. First, note that (a² - 1) factors into (a - 1)(a + 1). So, substituting that:[4 / (a + 1)] * [(a - 1)(a + 1)/a] = -1.The (a + 1) terms in the numerator and denominator will cancel out:4 * (a - 1)/a = -1.So, simplifying:4(a - 1)/a = -1.Multiply both sides by a to eliminate the denominator (assuming a ≠ 0):4(a - 1) = -a.Expanding the left side:4a - 4 = -a.Now, bring all terms to one side:4a - 4 + a = 0Combine like terms:5a - 4 = 0So:5a = 4Therefore:a = 4/5.Wait, that's interesting. So, according to this calculation, the only value of 'a' that satisfies the condition is 4/5. But the problem states that a = -1 is a sufficient but not necessary condition. However, according to this calculation, a = -1 isn't even a solution. So, maybe I made a mistake here?Let me check my steps again.Starting from the slopes:First line: m1 = 4/(a + 1)Second line: m2 = (a² - 1)/aProduct: [4/(a + 1)] * [(a² - 1)/a] = -1Then, factor a² - 1 as (a - 1)(a + 1):[4/(a + 1)] * [(a - 1)(a + 1)/a] = -1Cancel (a + 1):4*(a - 1)/a = -1So, 4(a - 1) = -a4a -4 = -a4a + a = 45a = 4 => a = 4/5Yes, that seems correct. So, the only solution is a = 4/5.But the problem says a = -1 is a sufficient condition. That would mean that when a = -1, the lines are perpendicular. However, according to this, a = -1 is not a solution. Therefore, maybe there's a mistake here.Wait, but perhaps when a = -1, the original equations of the lines change. Let me check what happens when a = -1.First, substitute a = -1 into both lines.First line becomes:4x - (-1 + 1)y + 9 = 0 => 4x - 0*y + 9 = 0 => 4x + 9 = 0. That's a vertical line (since y can be any value). The slope is undefined.Second line becomes:((-1)^2 - 1)x - (-1)y + 6 = 0 => (1 - 1)x + y + 6 = 0 => 0x + y + 6 = 0 => y + 6 = 0 => y = -6. That's a horizontal line. The slope is 0.A vertical line and a horizontal line are indeed perpendicular. So, even though when a = -1, the calculation for the slopes led to a division by zero (since m1 would have a denominator of a + 1 = 0, making the slope undefined, and m2 would have a denominator of a = -1, but in that case, the second line's slope is (a² -1)/a. When a = -1, (1 -1)/(-1) = 0/-1 = 0. So, the second line is horizontal, and the first line is vertical. So, they are perpendicular.Therefore, even though a = -1 makes the slope calculation have division by zero, the lines themselves are still perpendicular. So, in the original equation, when we set a = -1, the lines become vertical and horizontal, hence perpendicular. So, a = -1 is a valid solution. However, in our algebraic solution, we arrived at a = 4/5. So, why didn't a = -1 show up in the solution?Ah, because when we derived the condition m1 * m2 = -1, we assumed that both slopes are defined. However, in the case when a = -1, the first line becomes vertical (slope undefined) and the second becomes horizontal (slope 0). These are perpendicular, but this case isn't captured by the algebraic equation since we divided by (a + 1) and a, which would be zero when a = -1. So, we need to consider this case separately.So, in summary, the values of 'a' that make the lines perpendicular are a = -1 and a = 4/5. But when we solved the equation algebraically, we only found a = 4/5, because a = -1 causes division by zero in the slope expressions, so we have to check that case separately.Therefore, the original problem states that a = -1 is a sufficient but not necessary condition. However, according to this, a = -1 is a valid solution (sufficient condition), and there's another solution a = 4/5 (so it's not necessary, since there's another value). Therefore, the answer should be A: Sufficient but not necessary condition.But wait, the problem says "the statement given is: a = -1 is a sufficient but not necessary condition...", and the options are whether this statement is correct. So, if in reality, a = -1 is a sufficient condition (because when a = -1, the lines are indeed perpendicular) and it's not necessary because another value a = 4/5 also works, then the original statement is correct, so the answer is A.But let me verify once again.First, check a = -1:First line: 4x - ( (-1) + 1 )y + 9 = 4x - 0y + 9 = 4x + 9 = 0 => x = -9/4. Vertical line.Second line: ( (-1)^2 -1 )x - (-1)y +6 = (1 -1)x + y +6 = 0x + y +6 = 0 => y = -6. Horizontal line.Vertical and horizontal lines are perpendicular. So, yes, a = -1 works.Then, check a = 4/5:First line: slope m1 = 4/( (4/5) + 1 ) = 4/(9/5) = 4*(5/9) = 20/9.Second line: slope m2 = ( ( (4/5)^2 -1 ) ) / (4/5) = (16/25 -1)/(4/5) = (-9/25)/(4/5) = (-9/25)*(5/4) = -45/100 = -9/20.Multiply m1 * m2: (20/9)*(-9/20) = -1. Correct, so they are perpendicular.Therefore, a = -1 and a = 4/5 both make the lines perpendicular. Therefore, the condition a = -1 is sufficient (since it's one way to make them perpendicular) but not necessary (since you can also use a = 4/5). Therefore, the original statement is correct, so the answer is A.Wait, but the problem statement says "the line 4x - (a + 1)y +9 =0 to be perpendicular to the line (a² -1)x - a y +6=0". So, when they say "a = -1 is a sufficient but not necessary condition", they are claiming that setting a = -1 is enough to make them perpendicular (which is true), but there are other conditions (like a = 4/5) that also make them perpendicular, so it's not necessary to have a = -1. Therefore, the answer is A.But let me check if there's a possible mistake here. The key was that when we derived the equation, we had to exclude a = -1 because of division by zero, but in reality, a = -1 is a valid case. Therefore, the solutions are a = -1 and a = 4/5. Hence, a = -1 is one of the solutions, but since there's another solution, it's not necessary. So, the statement is correct as a sufficient but not necessary condition. Therefore, the answer is A.Final Answerboxed{A}</think>To determine if (a = -1) is a sufficient but not necessary condition for the lines (4x - (a+1)y + 9 = 0) and ((a^2 - 1)x - ay + 6 = 0) to be perpendicular, we need to check the slopes of the lines and verify the condition for perpendicularity.1. Slope of the first line: [ 4x - (a + 1)y + 9 = 0 implies y = frac{4}{a + 1}x + frac{9}{a + 1} ] The slope (m_1) is (frac{4}{a + 1}).2. Slope of the second line: [ (a^2 - 1)x - ay + 6 = 0 implies y = frac{a^2 - 1}{a}x + frac{6}{a} ] The slope (m_2) is (frac{a^2 - 1}{a}).3. Perpendicularity condition: [ m_1 cdot m_2 = -1 ] Substituting the slopes: [ left(frac{4}{a + 1}right) left(frac{a^2 - 1}{a}right) = -1 ] Simplifying: [ frac{4(a - 1)(a + 1)}{a(a + 1)} = -1 implies frac{4(a - 1)}{a} = -1 ] Solving for (a): [ 4(a - 1) = -a implies 4a - 4 = -a implies 5a = 4 implies a = frac{4}{5} ]4. Special case (a = -1): - The first line becomes (4x + 9 = 0) (vertical line). - The second line becomes (y = -6) (horizontal line). - Vertical and horizontal lines are perpendicular.Thus, (a = -1) is a sufficient condition because it makes the lines perpendicular, but it is not necessary because (a = frac{4}{5}) also works. Therefore, the correct answer is (boxed{A}).,<think>Okay, so I need to figure out whether a = -1 is a sufficient but not necessary condition for the two given lines to be perpendicular. The options are A, B, C, D, with A being the one mentioned in the problem statement. Let me start by recalling that two lines are perpendicular if the product of their slopes is -1. So, first, I need to find the slopes of both lines when they are expressed in the form y = mx + b.The first line is 4x - (a + 1)y + 9 = 0. Let me rearrange this to find the slope. Subtract 4x and 9 from both sides: -(a + 1)y = -4x - 9. Then divide both sides by -(a + 1) to get y = (4/(a + 1))x + 9/(a + 1). So the slope m1 of the first line is 4/(a + 1). The second line is (a² - 1)x - a y + 6 = 0. Let's do the same here. Subtract (a² - 1)x and 6 from both sides: -a y = -(a² - 1)x - 6. Divide both sides by -a: y = [(a² - 1)/a]x + 6/a. So the slope m2 of the second line is (a² - 1)/a.Now, for the lines to be perpendicular, the product of their slopes should be -1. So, set up the equation:m1 * m2 = -1Substituting the expressions we have:(4/(a + 1)) * ((a² - 1)/a) = -1Let me simplify this equation. First, notice that a² - 1 is a difference of squares, so it factors into (a - 1)(a + 1). Therefore:(4/(a + 1)) * [(a - 1)(a + 1)/a] = -1The (a + 1) terms in the numerator and denominator will cancel out:4*(a - 1)/a = -1Now, multiply both sides by a:4*(a - 1) = -aExpand the left side:4a - 4 = -aBring all terms to one side:4a - 4 + a = 05a - 4 = 05a = 4a = 4/5Wait, so solving the equation gives a = 4/5. But the problem says a = -1 is a sufficient but not necessary condition. Hmm. So if a = -1, does that make the lines perpendicular? Let's check that.If a = -1, substitute into the slopes.First slope m1 = 4/(a + 1) = 4/(-1 + 1) = 4/0. Oh, division by zero. That means the first line is vertical. Let's check the second line's slope when a = -1.Second slope m2 = (a² - 1)/a = ((1) - 1)/(-1) = 0/(-1) = 0. So the second line is horizontal. A vertical line and a horizontal line are indeed perpendicular. So when a = -1, the first line is vertical, second is horizontal, so they are perpendicular. Therefore, a = -1 is a valid condition for perpendicularity. But according to the equation we solved earlier, a = 4/5 is another solution. So actually, there are two values of a: 4/5 and -1 that make the lines perpendicular? Wait, but when we solved the equation, we got a = 4/5. But when a = -1, the original equation might have some undefined terms. Let's check.Wait, when we set up the equation m1*m2 = -1, but if either slope is undefined (i.e., vertical line) or zero (horizontal line), that condition still holds. For example, a vertical line (undefined slope) and a horizontal line (slope 0) are perpendicular, but the product of their slopes isn't -1 because it's undefined. So in this case, when a = -1, m1 is undefined (vertical line) and m2 is 0 (horizontal line), so they are perpendicular, even though the product of their slopes isn't technically -1 (since one is undefined). Therefore, in such cases, we have to check separately.So, in the original equation, when solving m1*m2 = -1, we found a = 4/5. But when a = -1, even though the algebra leads to division by zero, the lines are still perpendicular. Therefore, a = -1 is another solution. Therefore, the values of a that make the lines perpendicular are a = -1 and a = 4/5. Therefore, the condition a = -1 is sufficient (because it does make them perpendicular) but not necessary (since a can also be 4/5). Therefore, the correct answer is A: Sufficient but not necessary condition.Wait, but the problem states that "a = -1 is a sufficient but not necessary condition", and the options are whether that statement is correct. So the answer is A. Let me verify again.Yes, so the problem says "a = -1 is a sufficient but not necessary condition" for the lines to be perpendicular. From our analysis, a = -1 does make the lines perpendicular (sufficient), but there's another value a = 4/5 that also makes them perpendicular, so it's not necessary. Therefore, the answer is A.But let me check again for a = -1. If a = -1, first line: 4x - ( -1 + 1)y + 9 = 4x - 0y + 9 = 4x + 9 = 0, which simplifies to 4x + 9 = 0, which is x = -9/4, a vertical line. The second line: ( (-1)^2 - 1)x - (-1)y + 6 = (1 - 1)x + y + 6 = 0x + y + 6 = 0, so y = -6, a horizontal line. Vertical and horizontal lines are perpendicular. So yes, a = -1 works. Then, when we solved for a, we found a = 4/5. Let's check that. Let a = 4/5.First line: 4x - (4/5 + 1)y + 9 = 4x - (9/5)y + 9 = 0. Let's compute the slope: rearranged to y = (4/(9/5))x + ... which is (4*(5/9))x = 20/9 x. So slope is 20/9. Wait, but earlier we had m1 = 4/(a +1). If a = 4/5, then m1 = 4/(4/5 +1) = 4/(9/5) = 20/9. Then m2 = (a² -1)/a = ((16/25 -1))/(4/5) = (-9/25)/(4/5) = (-9/25)*(5/4) = -45/100 = -9/20. So m1 * m2 = (20/9)*(-9/20) = -1. So yes, that works. So they are perpendicular. Therefore, a = 4/5 is another solution. Hence, a = -1 is not the only solution, so it's sufficient but not necessary. Therefore, the answer is A. I think that's correct. Let me check if there are other possible solutions. The equation we had was 4*(a - 1)/a = -1. Solving gives 4(a -1) = -a => 4a -4 = -a => 5a =4 => a=4/5. So that's the only solution from the equation. But when a = -1, we get a vertical and horizontal line, which is perpendicular, even though the algebra led to division by zero. So total solutions are a = -1 and a = 4/5. Therefore, the original statement says a = -1 is sufficient but not necessary. So the answer is A.Final Answerboxed{A}</think>To determine whether (a = -1) is a sufficient but not necessary condition for the lines (4x - (a+1)y + 9 = 0) and ((a^2 - 1)x - ay + 6 = 0) to be perpendicular, we need to check the slopes of the lines and their product.1. Slope of the first line: [ 4x - (a+1)y + 9 = 0 implies y = frac{4}{a+1}x + frac{9}{a+1} ] The slope (m_1) is (frac{4}{a+1}).2. Slope of the second line: [ (a^2 - 1)x - ay + 6 = 0 implies y = frac{a^2 - 1}{a}x + frac{6}{a} ] The slope (m_2) is (frac{a^2 - 1}{a}).3. Condition for perpendicularity: [ m_1 cdot m_2 = -1 ] Substituting the slopes: [ left(frac{4}{a+1}right) left(frac{a^2 - 1}{a}right) = -1 ] Simplifying (a^2 - 1) as ((a-1)(a+1)): [ frac{4(a-1)(a+1)}{a(a+1)} = -1 implies frac{4(a-1)}{a} = -1 ] Solving for (a): [ 4(a - 1) = -a implies 4a - 4 = -a implies 5a = 4 implies a = frac{4}{5} ]4. Checking (a = -1): - For (a = -1), the first line becomes (4x + 9 = 0) (vertical line). - The second line becomes (y = -6) (horizontal line). - Vertical and horizontal lines are perpendicular.Thus, (a = -1) is a sufficient condition (it makes the lines perpendicular) but not necessary (since (a = frac{4}{5}) also works). Therefore, the answer is (boxed{A}).]

🔑:When a=-1, the two lines become: 4x+9=0 and y+6=0, respectively. At this time, the two lines are perpendicular to each other; When a=0, the two lines become: 4x-y+9=0 and -x+6=0, respectively. At this time, the two lines are not perpendicular; When a neq -1, 0, the slopes of the two lines are: frac{4}{a+1} and frac{a^2-1}{a}, respectively. Since the two lines are perpendicular to each other, we have frac{4}{a+1} cdot frac{a^2-1}{a} = -1, solving this gives a= frac{4}{5}. In summary, a=-1 is a sufficient but not necessary condition for the line 4x-(a+1)y+9=0 to be perpendicular to the line (a^2-1)x-ay+6=0. Therefore, the answer is: boxed{text{A}}. By discussing the value of a in different cases and using the necessary and sufficient condition for two lines to be perpendicular, we can arrive at the conclusion. This problem tests the necessary and sufficient conditions for two lines to be perpendicular, the method of discussing cases, reasoning ability, and computational skills, and is considered a medium-level question.

🧠:[<think>Okay, so I need to find the number of positive integer values of t such that the line x + y = t intersects the circle x² + y² = 20 at points A and B, and the area of triangle OAB is an integer. Hmm, let's break this down step by step.First, I know that the circle x² + y² = 20 has its center at the origin (0,0) and radius √20, which simplifies to 2√5. The line x + y = t is a straight line, and depending on the value of t, it will intersect the circle at two points, one point (if it's tangent), or no points. Since the problem mentions points A and B, we can assume that the line intersects the circle at two distinct points, so t must be such that the line is not too far from the origin.The first thing I need to do is probably find the distance from the center of the circle (which is the origin) to the line x + y = t. The formula for the distance from a point (x₀, y₀) to the line ax + by + c = 0 is |ax₀ + by₀ + c| / √(a² + b²). In this case, the line is x + y = t, which can be rewritten as x + y - t = 0. So the distance from the origin (0,0) to this line is |0 + 0 - t| / √(1² + 1²) = | -t | / √2 = |t| / √2.Since the line intersects the circle at two points, this distance must be less than the radius of the circle. Therefore, |t| / √2 < 2√5. Multiplying both sides by √2 gives |t| < 2√10. Since we're dealing with positive integer values of t, t must be less than 2√10. Let me compute 2√10 numerically to see the upper bound. √10 is approximately 3.162, so 2√10 ≈ 6.324. Therefore, t can be 1, 2, 3, 4, 5, or 6. So possible positive integer values of t are 1 through 6. But wait, the problem says "positive integer values of t that satisfy the condition", so maybe some of these t values result in the area of triangle OAB being an integer, and some don't. So I need to check which of these t values (1 to 6) make the area an integer.Now, how to find the area of triangle OAB? Since O is the origin, and A and B are points on the circle, OA and OB are both radii of the circle, each of length √20. The area of triangle OAB can be found using the formula: (1/2)*OA*OB*sinθ, where θ is the angle between OA and OB. But OA and OB are both radii, so OA = OB = √20, so the area is (1/2)*(√20)*(√20)*sinθ = (1/2)*20*sinθ = 10 sinθ. So the area is 10 sinθ. Therefore, for the area to be an integer, sinθ must be a rational number since 10 sinθ is an integer. So 10 sinθ ∈ ℤ ⇒ sinθ = k/10 where k is an integer such that |k| ≤10. But sinθ must be between -1 and 1, so k can be from -10 to 10, but since θ is the angle between two intersection points of a line and a circle, θ is between 0 and π, so sinθ is non-negative. Therefore, sinθ is between 0 and 1, so k is from 0 to 10. But since the line intersects the circle at two distinct points, θ can't be 0 or π, so sinθ is between 0 and 1, not including 0. Therefore, k can be from 1 to 10, such that sinθ = k/10, but sinθ ≤1, so k can be 1 to 10. However, the area is 10 sinθ, so the area would be integers 1 through 10. But actually, 10 sinθ can be any real number between 0 and 10, not necessarily integers. Wait, but the problem says the area is an integer, so 10 sinθ must be integer, which implies sinθ must be a rational number with denominator dividing 10. So sinθ = n/10 where n is integer from 1 to 10. But sinθ must also correspond to an actual angle θ between the two points A and B on the circle.Alternatively, maybe there's another way to compute the area of triangle OAB. Since points A and B lie on the line x + y = t and on the circle x² + y² = 20, we can find coordinates of A and B and then compute the area.Let me try this approach. Let's solve the system of equations:x + y = tx² + y² = 20We can substitute y = t - x into the circle equation:x² + (t - x)² = 20Expanding that:x² + t² - 2tx + x² = 20Combine like terms:2x² - 2tx + t² - 20 = 0Divide by 2:x² - tx + (t²/2 - 10) = 0This is a quadratic in x. Let's denote this as x² - tx + C = 0, where C = (t²/2 -10). The solutions are x = [t ± √(t² - 4*1*C)] / 2Compute the discriminant:D = t² - 4*1*(t²/2 -10) = t² - 2t² + 40 = -t² +40So D = 40 - t². For real solutions, D must be positive, so 40 - t² >0 ⇒ t² <40 ⇒ |t| <√40 ≈6.324, which confirms what I found earlier. So t can be 1,2,3,4,5,6 as positive integers.Now, the coordinates of points A and B are given by the solutions to the quadratic. Let me write them:x = [t ± √(40 - t²)] / 2And y = t - x, so y = [t ∓ √(40 - t²)] / 2Therefore, points A and B are:A: ( [t + √(40 - t²)] / 2 , [t - √(40 - t²)] / 2 )B: ( [t - √(40 - t²)] / 2 , [t + √(40 - t²)] / 2 )Now, to find the area of triangle OAB, where O is the origin. The area can be calculated using the determinant formula:Area = (1/2) |x₁y₂ - x₂y₁|Where (x₁, y₁) and (x₂, y₂) are coordinates of A and B.Let's compute x₁y₂ - x₂y₁.Take point A as (x₁, y₁) and point B as (x₂, y₂):x₁ = [t + √(40 - t²)] / 2y₁ = [t - √(40 - t²)] / 2x₂ = [t - √(40 - t²)] / 2y₂ = [t + √(40 - t²)] / 2Compute x₁y₂ - x₂y₁:= [ (t + √(40 - t²))/2 ] * [ (t + √(40 - t²))/2 ] - [ (t - √(40 - t²))/2 ] * [ (t - √(40 - t²))/2 ]Wait, let me compute this step by step.First term: x₁y₂ = [ (t + √D)/2 ] * [ (t + √D)/2 ] where D =40 - t²Second term: x₂y₁ = [ (t - √D)/2 ] * [ (t - √D)/2 ]Therefore, x₁y₂ - x₂y₁ = [ (t + √D)^2 /4 ] - [ (t - √D)^2 /4 ]Factor out 1/4:= (1/4)[ (t + √D)^2 - (t - √D)^2 ]Expand both squares:= (1/4)[ (t² + 2t√D + D) - (t² - 2t√D + D) ]Simplify inside the brackets:= (1/4)[ t² + 2t√D + D - t² + 2t√D - D ]= (1/4)[ 4t√D ]= (1/4)(4t√D) = t√DTherefore, the area is (1/2)|x₁y₂ - x₂y₁| = (1/2)| t√D | = (1/2) t√D, since t is positive and D is positive.But D is 40 - t², so the area becomes (1/2) t √(40 - t²)So the area S = (t/2) * √(40 - t²)We need this area to be an integer.So, S = (t/2) * √(40 - t²) must be an integer.Therefore, (t/2) * √(40 - t²) ∈ ℤSo our task reduces to finding positive integers t (from 1 to 6) such that (t/2) * √(40 - t²) is an integer.Let me check each t from 1 to 6:t=1:S = (1/2) * √(40 -1) = (1/2)√39 ≈ 3.122... Not an integer.t=2:S = (2/2) * √(40 -4) = 1 * √36 = 6. Which is an integer. So t=2 is acceptable.t=3:S = (3/2) * √(40 -9) = (3/2)√31 ≈ (3/2)*5.567 ≈8.351... Not integer.t=4:S = (4/2)*√(40 -16)=2*√24=2*2√6=4√6≈9.798... Not integer.t=5:S=(5/2)*√(40 -25)= (5/2)*√15≈(5/2)*3.872≈9.68... Not integer.t=6:S=(6/2)*√(40 -36)=3*√4=3*2=6. Which is an integer. So t=6 is acceptable.Wait, so t=2 and t=6 give integer areas. So that would be 2 values. But the question says "the number of all positive integer values of t that satisfy the condition". But wait, maybe I missed some? Let me check again.Wait, for t=1: √39 is irrational, multiplied by 1/2, so not integer.t=2: 6, integer.t=3: (3/2)√31, not integer.t=4: 2√24=2*2√6=4√6, which is irrational.t=5: (5/2)√15, irrational.t=6: 3*2=6, integer.So only t=2 and t=6. But wait, the problem states "positive integer values of t". So t=2 and t=6. Therefore, the number of such t is 2.But wait, the answer is 2? Let me check again, maybe there's a mistake here.Wait, another approach: Let's think back to the area formula.Earlier, I found that the area is 10 sinθ, where θ is the angle between OA and OB. For this to be an integer, sinθ must be 0.1, 0.2, ..., 1.0. But 10 sinθ is integer, so sinθ must be k/10, where k is integer from 1 to 10.But how does θ relate to t? The angle θ between OA and OB can be found using the distance between A and B.Alternatively, we can use the chord length formula. The length of chord AB is 2√(r² - d²), where r is radius and d is the distance from the center to the line. Here, r=2√5, d= t/√2.So chord length AB = 2√( (2√5)^2 - (t/√2)^2 ) = 2√(20 - t²/2 )Then, the angle θ in radians can be found via chord length: AB = 2r sin(θ/2). So:2√(20 - t²/2 ) = 2*(2√5)*sin(θ/2)Divide both sides by 2:√(20 - t²/2 ) = 2√5 sin(θ/2)Square both sides:20 - t²/2 = 20 sin²(θ/2)Then, sin²(θ/2) = (20 - t²/2)/20 = 1 - t²/(40)Therefore, sin(θ/2) = √(1 - t²/40 )But θ is the angle between OA and OB, so θ is the central angle. Then, the area of triangle OAB is (1/2)*r²*sinθ = (1/2)*(20)*sinθ = 10 sinθ, as before.So we need 10 sinθ to be integer. So sinθ must be k/10, k=1,...,10.But θ = 2 arcsin(√(1 - t²/40)) from above.So sinθ = 2 sin(θ/2) cos(θ/2) = 2 √(1 - t²/40) * √(t²/40)Wait, let's see:From sin(θ/2) = √(1 - t²/40 )Then, cos(θ/2) = √(1 - sin²(θ/2)) = √( t² /40 )Therefore, sinθ = 2 sin(θ/2) cos(θ/2) = 2 * √(1 - t²/40 ) * √( t² /40 )= 2 * √( (1 - t²/40 ) * t² /40 )= 2 * √( t²/40 - t⁴/(1600) )But this seems complicated. Maybe stick to the area formula S = (t/2)√(40 - t²)So S = (t/2)√(40 - t²). We need this to be an integer.Let me denote S as integer. So (t/2)√(40 - t²) = integer.Let me square both sides to remove the square root:(t² /4)(40 - t²) = integer²So (t²(40 - t²)) /4 must be a perfect square.Let me compute this expression for each t from 1 to 6:t=1:(1*(40 -1))/4 = (39)/4 = 9.75 Not a perfect square.t=2:(4*(40 -4))/4 = (4*36)/4 = 36. Which is 6². So 36 is a perfect square.t=3:(9*(40 -9))/4 = (9*31)/4=279/4=69.75 Not a perfect square.t=4:(16*(40 -16))/4=(16*24)/4= 16*6=96. 96 is not a perfect square.Wait, but 96=16*6, which isn't a perfect square.t=5:(25*(40 -25))/4=(25*15)/4=375/4=93.75 Not a perfect square.t=6:(36*(40 -36))/4=(36*4)/4=36. Which is 6². Perfect square.So t=2 and t=6 give (t²(40 - t²))/4 equal to 36 and 36, which are perfect squares. Therefore, S²=36, so S=6 for both. So when t=2 and t=6, the area S=6, which is integer.But wait, why did squaring give us 36, but the actual area is 6? Because S = (t/2)*√(40 - t²). Squaring S gives (t²/4)*(40 - t²) = 36 for t=2 and t=6, so S²=36, so S=6. So the area is indeed 6 for these t's.But for other t's, like t=1, the squared area is 9.75, which is not a perfect square. Therefore, only t=2 and t=6 satisfy the condition. So the number of positive integer values of t is 2.But wait, hold on. The problem says "the number of all positive integer values of t". But maybe t=0? But t is positive, so t=0 is excluded. t=1 to 6. So only t=2 and t=6. So the answer is 2? But I need to check again.Wait, let's verify with t=4. The area is 2√24. Wait, √24 is 2√6, so 2*2√6=4√6≈9.798. Which is not integer. Similarly for t=3 and t=5, the areas are not integers.But wait, there might be another way. Suppose S = (t/2)*√(40 - t²) must be integer. Let's denote that expression as k, integer.So (t/2)*√(40 - t²) = kSquare both sides:(t² /4)(40 - t²) = k²Multiply both sides by 4:t²(40 - t²) = 4k²Let me rearrange:t⁴ -40t² +4k²=0But this is a quartic equation, which might not be helpful. Alternatively, note that t²(40 - t²) must be divisible by 4, so t²(40 - t²) is a multiple of 4. Since t is integer, t² is either 0 or 1 mod 4. Since t is positive integer from 1 to 6, t² mod 4:t=1:1; t=2:0; t=3:1; t=4:0; t=5:1; t=6:0.Therefore, for t even, t² is divisible by 4; for t odd, t² ≡1 mod4.So for t even, t=2,4,6:t=2:t²=4, so 4*(40 -4)=4*36=144. 144/4=36=6².t=4:t²=16, so 16*(40 -16)=16*24=384. 384/4=96. 96 is not a perfect square.t=6:t²=36, so 36*(40 -36)=36*4=144. 144/4=36=6².So for even t=2,6, we get perfect squares, but t=4 gives 96, not a square.For odd t=1,3,5:t=1:1*(40 -1)=39. 39/4=9.75 Not integer.t=3:9*(40 -9)=9*31=279. 279/4=69.75 Not integer.t=5:25*(40 -25)=25*15=375. 375/4=93.75 Not integer.Therefore, only t=2 and t=6 produce areas that are integers.But let me think again: is there a possibility that for some t, the area is integer even though (t/2)√(40 - t²) is not? For example, maybe due to simplification. Let's check t=2:Coordinates of points A and B when t=2:x + y = 2, x² + y² =20.Solving:y=2 - xx² + (2 -x)^2=202x² -4x +4=202x² -4x -16=0x² -2x -8=0Solutions:x=(2 ±√(4 +32))/2=(2 ±√36)/2=(2 ±6)/2=4 or -2Thus, points A(4, -2) and B(-2,4)Wait, plug x=4 into y=2 -x: y= -2Plug x=-2 into y=2 -x: y=4Thus, points are (4,-2) and (-2,4)Compute the area of triangle OAB:Using determinant formula:Area = (1/2)|x1y2 - x2y1|= (1/2)|4*4 - (-2)*(-2)|= (1/2)|16 -4| = (1/2)(12) =6. Correct.Similarly, for t=6:x + y=6, x² + y²=20.y=6 -xx² + (6 -x)^2=20x² +36 -12x +x²=202x² -12x +16=0x² -6x +8=0Solutions:x=(6 ±√(36 -32))/2=(6 ±2)/2=4 or 2Thus, points A(4,2) and B(2,4)Area:(1/2)|4*4 -2*2|=(1/2)|16 -4|=6. Same result.So t=2 and t=6 both give area 6, which is integer.But wait, t=0 would give x + y=0, which intersects the circle at two points, but t must be positive, so t=0 is excluded.Is there any other possible t? Let's check t= sqrt(40) ≈6.324, but we are limited to integers up to 6. So no.Wait, but perhaps there are other t values where (t/2)√(40 - t²) is integer? For example, if t and √(40 - t²) are such that their product is even, and the square root is rational. Suppose that √(40 - t²) is rational. Then 40 - t² must be a perfect square. Let's check:For t from1-6:t=1:40 -1=39 Not square.t=2:40 -4=36 Which is 6². So √36=6.t=3:40 -9=31 Not square.t=4:40 -16=24 Not square.t=5:40 -25=15 Not square.t=6:40 -36=4 Which is 2². So √4=2.Therefore, for t=2 and t=6, √(40 - t²) is integer (6 and 2 respectively). Thus, (t/2)*√(40 - t²) becomes (2/2)*6=6 and (6/2)*2=6. Both integer. For other t, √(40 - t²) is irrational, so (t/2)*√(40 - t²) is irrational multiplied by t/2, which is a rational number only if t is even. But even then, unless √(40 - t²) is rational, the product would still be irrational. Since for t=4, √(40 -16)=√24=2√6, which is irrational. So even t=4, with t even, gives an irrational number multiplied by 4/2=2, resulting in 2*2√6=4√6, still irrational.Therefore, only t=2 and t=6 satisfy the condition where √(40 - t²) is integer, leading to the entire expression being integer. Therefore, only these two t's.Thus, the number of positive integer values of t is 2.Wait, but the problem says "the number of all positive integer values of t". The answer is 2? But I vaguely recall sometimes questions like this having more answers. Let me check once again.Wait, maybe I made a mistake in assuming that √(40 - t²) must be integer. For example, suppose t is even, say t=4, and √(40 - t²)=√24=2√6. Then (t/2)*√(40 - t²)= (4/2)*2√6=4√6, which is irrational. So not integer.Alternatively, if t=1, then (1/2)*√39≈3.122, not integer.Similarly, t=3: (3/2)*√31≈8.35, not integer.t=5: (5/2)*√15≈9.68, not integer.So only t=2 and t=6 give integer areas. Therefore, the answer is 2. Hence, the number of positive integer values of t is 2.But let me check once more if there's any other approach. Suppose we consider the area formula S = 10 sinθ. So 10 sinθ must be integer. Therefore, sinθ = k/10, where k is integer from 1 to 10. Then, the angle θ corresponds to the central angle between points A and B. But how is θ related to t?We can find θ using the distance from the origin to the line x + y =t, which is d= |t|/√2. Then, in the circle, the relationship between the central angle θ and the distance d is given by d = r cos(θ/2). Wait, no. Wait, in a circle, if you have a chord at distance d from the center, then the central angle θ satisfies:d = r cos(θ/2)Therefore, cos(θ/2) = d / rHere, d = t / √2, r = 2√5.So cos(θ/2) = (t / √2) / (2√5) )= t / (2√10 )But θ is the central angle, so θ = 2 arcsin(√(1 - (d²)/(r²)) )Wait, let me check:From chord length formula, chord length AB = 2√(r² - d²)Also, chord length AB = 2r sin(θ/2)So:2r sin(θ/2) = 2√(r² - d²)Divide both sides by 2:r sin(θ/2) = √(r² - d²)Then,sin(θ/2) = √(r² - d²)/r = √(1 - (d²)/(r²))So,θ/2 = arcsin(√(1 - (d²)/(r²)) )Therefore,θ = 2 arcsin(√(1 - (d²)/(r²)) )But in our case, d= t /√2, r=2√5So,θ = 2 arcsin(√(1 - (t² / 2 ) / (20)) )= 2 arcsin(√(1 - t² /40 ) )Therefore,sinθ = 2 sin(θ/2) cos(θ/2) = 2 √(1 - t² /40 ) * √( t² /40 )= 2 * √( (1 - t²/40 ) * t² /40 )= 2 * √( t²/40 - t⁴/1600 )But this expression seems complicated. However, the area S = 10 sinθ must be integer. So 10 sinθ = integer.But sinθ = 2 * √( t²/40 - t⁴/1600 ) * √(1 - t²/40 )Wait, maybe not helpful. Let's substitute:Let’s denote t² = k, so k is from 1 to 36 (since t is 1-6). Then,sinθ = 2 * √( k/40 - k²/1600 ) = 2 * √( (40k -k²)/1600 ) = 2 * ( √(40k -k²) ) /40 ) = ( √(40k -k²) ) /20Therefore, sinθ = √(40k -k²)/20Then, 10 sinθ = √(40k -k²)/2 must be integer.So, √(40k -k²)/2 ∈ ℤThus, √(40k -k²) must be even integer.Let me denote m = √(40k -k²), so m is even integer, and m=2n, where n is integer.Therefore, 40k -k² = m²=4n²Rearranged:k² -40k +4n²=0This is a quadratic in k:k² -40k +4n²=0Solving for k:k = [40 ± √(1600 -16n²)] / 2 = [40 ± 4√(100 -n²)] /2=20 ± 2√(100 -n²)Since k = t², and t is positive integer from1-6, k can be1,4,9,16,25,36.Therefore, we have:k =20 ± 2√(100 -n²) must be one of 1,4,9,16,25,36.But since k must be positive and 20 - 2√(100 -n²) and 20 + 2√(100 -n²). Given that 20 -2√(100 -n²) must be positive, so 20 >2√(100 -n²) ⇒10>√(100 -n²) ⇒100>100 -n² ⇒n²>0, which is always true as n is integer, n≥1.So possible k values:k=20 ± 2√(100 -n²)So we need both k=20 +2√(100 -n²) and k=20 -2√(100 -n²) to be perfect squares (since k=t²). Let's denote:Let’s assume that 20 +2√(100 -n²) and 20 -2√(100 -n²) are perfect squares.Let me denote √(100 -n²)=m, so m must be integer because k is integer (since t is integer). Therefore, 100 -n² =m², so n² +m²=100.Therefore, (n,m) are integer solutions to n² +m²=100.Furthermore, since k=20 ±2m, and k must be a perfect square (k=t²). So:Case1: k=20 +2m is a perfect square.Case2: k=20 -2m is a perfect square.So we need to find integer solutions (n,m) to n² +m²=100, and then check if 20 ±2m is a perfect square.Possible integer solutions to n² +m²=100:The integer pairs (n,m) such that n² +m²=100:(0,10),(6,8),(8,6),(10,0) and their negatives. But since n and m are non-negative (as they come from sqrt), we can consider positive pairs:(0,10),(6,8),(8,6),(10,0)Therefore, possible m values for each:For (0,10): m=10For (6,8): m=8For (8,6): m=6For (10,0): m=0Now, check each m:1. m=10:Case1: k=20 +2*10=40. 40 is not a perfect square.Case2: k=20 -2*10=0. k=0, which is t=0. Disregard as t must be positive.2. m=8:Case1: k=20 +16=36. 36 is a perfect square (t=6).Case2: k=20 -16=4. 4 is a perfect square (t=2).3. m=6:Case1: k=20 +12=32. Not a perfect square.Case2: k=20 -12=8. Not a perfect square.4. m=0:Case1: k=20 +0=20. Not a perfect square.Case2: k=20 -0=20. Not a perfect square.Therefore, the only valid solutions are when m=8, leading to k=36 and k=4, which correspond to t=6 and t=2.Therefore, only t=2 and t=6 satisfy the condition, confirming our previous result. Hence, the number of positive integer values of t is 2.Final Answerboxed{2}</think>To determine the number of positive integer values of ( t ) such that the line ( x + y = t ) intersects the circle ( x^2 + y^2 = 20 ) at points ( A ) and ( B ), and the area of triangle ( triangle OAB ) is an integer, we follow these steps:1. Distance from Origin to Line: The distance from the origin to the line ( x + y = t ) is ( frac{|t|}{sqrt{2}} ). For the line to intersect the circle, this distance must be less than the radius ( 2sqrt{5} ). Thus, ( |t| < 2sqrt{10} approx 6.324 ). Therefore, possible positive integer values of ( t ) are 1, 2, 3, 4, 5, and 6.2. Area of Triangle ( triangle OAB ): The area of triangle ( triangle OAB ) can be calculated using the formula ( frac{1}{2} times text{base} times text{height} ). Here, the base is the chord length ( AB ) and the height is the distance from the origin to the line. The area is given by: [ S = frac{1}{2} times t times sqrt{40 - t^2} ] For ( S ) to be an integer, ( frac{t}{2} times sqrt{40 - t^2} ) must be an integer.3. Checking Each ( t ): - ( t = 1 ): ( frac{1}{2} times sqrt{39} ) (not an integer) - ( t = 2 ): ( frac{2}{2} times sqrt{36} = 6 ) (integer) - ( t = 3 ): ( frac{3}{2} times sqrt{31} ) (not an integer) - ( t = 4 ): ( frac{4}{2} times sqrt{24} = 2 times 2sqrt{6} ) (not an integer) - ( t = 5 ): ( frac{5}{2} times sqrt{15} ) (not an integer) - ( t = 6 ): ( frac{6}{2} times sqrt{4} = 3 times 2 = 6 ) (integer)4. Verification with Integer Solutions: We check if ( sqrt{40 - t^2} ) is an integer for each ( t ). This is true only for ( t = 2 ) and ( t = 6 ), leading to integer areas.Thus, the number of positive integer values of ( t ) that satisfy the condition is (boxed{2}).,<think>Okay, let's try to figure out this problem. So, we have a line x + y = t intersecting the circle x² + y² = 20 at points A and B. We need to find the number of positive integer values of t such that the area of triangle OAB is an integer. First, let me recall that the area of a triangle given by three points can be calculated using the determinant formula. But in this case, points A and B are on the circle centered at the origin, and O is the origin. So, triangle OAB is a triangle with vertices at the origin, A, and B. Another thought: maybe the area can be calculated using vectors or the cross product? Since OA and OB are vectors from the origin, the area of the triangle would be half the magnitude of the cross product of vectors OA and OB. The formula for that is (1/2)|OA × OB|. But OA and OB are position vectors of points A and B. The cross product in two dimensions can be represented as |x1y2 - x2y1|, so the area would be (1/2)|x1y2 - x2y1|. That makes sense. Alternatively, since both points A and B lie on the line x + y = t and the circle x² + y² = 20, perhaps we can parametrize the points of intersection and then compute the area. Let me try that approach. Let's find the points of intersection between the line and the circle. Given x + y = t, we can express y = t - x and substitute into the circle equation:x² + (t - x)² = 20Expanding that:x² + t² - 2tx + x² = 20Combine like terms:2x² - 2tx + t² - 20 = 0Divide through by 2:x² - tx + (t²/2 - 10) = 0This is a quadratic equation in x. Let me denote this as:x² - tx + (t²/2 - 10) = 0To find the solutions, we can use the quadratic formula. The discriminant D is:D = [(-t)]² - 4 * 1 * (t²/2 - 10) = t² - 2t² + 40 = -t² + 40For real intersections, the discriminant must be non-negative:-t² + 40 ≥ 0 ⇒ t² ≤ 40 ⇒ |t| ≤ √40 ≈ 6.324Since t is a positive integer, possible values of t are 1, 2, 3, 4, 5, 6.Wait, but the question is asking for the number of positive integer values t such that the area of triangle OAB is an integer. So, even though t can be from 1 to 6, maybe not all of them result in an integer area.But before that, let's check if my calculation is correct. The discriminant D = -t² + 40 must be non-negative, so t² ≤ 40. Since t is positive integer, t can be 1, 2, 3, 4, 5, 6. So there are 6 possible values. But maybe the area is an integer only for some of them. Let's compute the area for each t and see.Alternatively, perhaps there's a formula for the area of triangle OAB in terms of t. Let me think.Alternatively, since the line x + y = t intersects the circle x² + y² = 20 at points A and B, the area of triangle OAB can be calculated using the formula for the area of a triangle formed by two points on a circle and the center. Another approach: The area can be expressed as (1/2)*|OA × OB|, which is (1/2)*|x1y2 - x2y1|. But since points A and B lie on the line x + y = t, their coordinates satisfy x + y = t. Maybe there's a relation here.Alternatively, use the formula for the area of triangle OAB using the distance from the center to the line (which is the height) and the length of AB (which is the base). Wait, no. The triangle OAB is not a triangle with base AB and height from O. The triangle OAB is formed by three points: O, A, B. The area can be calculated by the formula (1/2)*|OA||OB|sin(theta), where theta is the angle between OA and OB. Since OA and OB are radii of the circle, their lengths are sqrt(20). So OA = OB = sqrt(20). Then the area is (1/2)*(sqrt(20))^2*sin(theta) = (1/2)*20*sin(theta) = 10*sin(theta). Therefore, the area is 10*sin(theta), which must be an integer. So 10*sin(theta) must be an integer. Therefore, sin(theta) must be a rational number with denominator dividing 10. But theta is the angle between OA and OB.Alternatively, maybe we can relate the angle theta to the distance between A and B. The chord length AB can be related to theta via AB = 2*sqrt(20)*sin(theta/2). Then AB = 2*sqrt(20)*sin(theta/2). Also, the area is 10*sin(theta). Hmm. Maybe we can express the area in terms of t.Alternatively, let's find the distance from the origin to the line x + y = t. The formula for the distance from a point (x0,y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a² + b²). Here, the line is x + y = t, so it's x + y - t = 0. The distance from O(0,0) to this line is |0 + 0 - t| / sqrt(1 + 1) = | - t | / sqrt(2) = t / sqrt(2). This distance is the perpendicular distance from O to the line AB. In the case of a circle, this distance is related to the chord length AB. The formula for the length of chord AB is 2*sqrt(r² - d²), where d is the distance from the center to the line. Here, r is sqrt(20), so r² = 20, and d = t / sqrt(2), so d² = t² / 2. Then the length AB is 2*sqrt(20 - t²/2).But how does this help with the area of triangle OAB?Wait, the area of triangle OAB can also be calculated as (1/2)*AB*d, where d is the distance from O to the line AB. So Area = (1/2)*AB*(t / sqrt(2)).We have AB = 2*sqrt(20 - t²/2). So substituting:Area = (1/2)*(2*sqrt(20 - t²/2))*(t / sqrt(2)) = sqrt(20 - t²/2)*(t / sqrt(2))Simplify:sqrt(20 - t²/2) * t / sqrt(2) = t / sqrt(2) * sqrt(20 - t²/2)Combine the square roots:t * sqrt( (20 - t²/2) / 2 ) = t * sqrt(10 - t²/4 )So the area is t * sqrt(10 - t²/4 ). This must be an integer.Therefore, we have:Area = t * sqrt(10 - t²/4 ) is integer.So, our condition is that t * sqrt(10 - (t²)/4 ) must be an integer. Since t is a positive integer from 1 to 6, let's check each possible t.But before that, maybe we can square both sides to get rid of the square root. Let me denote Area = K, where K is integer.So K = t * sqrt(10 - t²/4 )Then K² = t² * (10 - t²/4 ) = 10 t² - t^4 /4Multiply both sides by 4:4K² = 40 t² - t^4Then, rearranged:t^4 -40 t² +4K² =0But this might not help directly. Alternatively, perhaps express 10 - t²/4 as a square. Let's see.Let’s denote:sqrt(10 - t²/4 ) = m/n, where m and n are integers with no common factors, then:10 - t²/4 = m²/n²But since t is an integer, t²/4 must be rational. So 10 must be rational, which it is. However, sqrt(10 - t²/4 ) must be rational for the product t*sqrt(...) to be integer. Because if sqrt(...) is irrational, then unless t is a multiple of the denominator, the product would be irrational. But t is integer, so if sqrt(10 - t²/4 ) is rational, then the product could be integer. Therefore, 10 - t²/4 must be a perfect square of a rational number.Let’s set 10 - t²/4 = (a/b)², where a and b are positive integers with gcd(a,b)=1.Then:t²/4 = 10 - (a²)/(b²)Multiply both sides by 4b²:t² b² = 40 b² -4 a²Rearranged:t² b² +4 a² =40 b²Hmm. Not sure if this helps.Alternatively, since t is integer, let’s set s = t/2. Then t = 2s, where s can be a half-integer, but since t must be integer, s must be integer or half-integer. Wait, maybe not. Alternatively, let's see:Let’s write 10 - t²/4 as (40 - t²)/4. Therefore, sqrt( (40 - t²)/4 ) = sqrt(40 - t²)/2. Therefore, the area is t * sqrt(40 - t²)/2. Therefore, Area = (t/2) * sqrt(40 - t²)Therefore, the area is (t * sqrt(40 - t² )) / 2. This must be an integer.So, (t * sqrt(40 - t² )) / 2 is integer.Let’s denote this integer as K. Then:t * sqrt(40 - t² ) = 2KSquare both sides:t² (40 - t² ) = 4K²Which simplifies to:40 t² - t^4 =4K²Or:t^4 -40 t² +4K² =0Hmm. Still a quartic equation, which might not be straightforward.Alternatively, maybe 40 - t² must be a perfect square. Let’s denote 40 - t² = m², where m is integer.Then, t² + m² =40So, we have t and m positive integers such that t² + m² =40. Then possible t are integers such that t² ≤40, so t=1,2,3,4,5,6.Let’s check which of these t have m as integer.t=1: 1 + m²=40 ⇒ m²=39 ⇒ m not integer.t=2: 4 + m²=40 ⇒ m²=36 ⇒ m=6. So m=6.t=3:9 +m²=40 ⇒ m²=31⇒ no.t=4:16 +m²=40⇒m²=24⇒no.t=5:25 +m²=40⇒m²=15⇒no.t=6:36 +m²=40⇒m²=4⇒m=2.So, t=2 and t=6 have integer m.Wait, but this is under the assumption that 40 - t² is a perfect square, which would make sqrt(40 -t²) an integer. But in our area formula, we have sqrt(40 - t²)/2 multiplied by t. So if sqrt(40 - t²) is integer, then t must be even for the area to be integer? Let's check:For t=2, sqrt(40 -4)=sqrt(36)=6. So area=(2*6)/2=6. Which is integer.For t=6, sqrt(40 -36)=sqrt(4)=2. So area=(6*2)/2=6. Also integer.But wait, in the problem statement, it's possible that sqrt(40 - t²) is rational, not necessarily integer. For example, if sqrt(40 - t²) is a fraction that when multiplied by t and divided by 2 gives an integer. So maybe there are other values of t where 40 - t² is a square of a rational number.So, perhaps more generally, 40 - t² must be a square of a rational number. Let’s write 40 - t² = (p/q)^2, where p and q are coprime integers. Then:40 - t² = p²/q² ⇒ t² =40 - p²/q² ⇒ t² = (40 q² - p²)/q²Since t is integer, the right-hand side must be an integer. Therefore, q² divides (40 q² - p²), which implies that q² divides p². But since p and q are coprime, q must be 1. Therefore, 40 - t² must be a perfect square. Hence, only t=2 and t=6 as above.But this seems conflicting with our initial conclusion. Wait, so according to this, only t=2 and t=6 would give integer areas. But that's only if sqrt(40 - t²) is integer, but maybe we need to consider cases where sqrt(40 - t²) is a multiple of sqrt(k) where k divides 40 - t². Wait, but how else can the product (t/2)*sqrt(40 - t²) be integer?Let me think. Let’s suppose that sqrt(40 - t²) is rational. Then, 40 - t² must be a perfect square. Because if sqrt(n) is rational, then n must be a perfect square. Therefore, 40 - t² must be a perfect square. Therefore, only t=2 and t=6 as above.Therefore, only t=2 and t=6. But wait, let's check for t=4. Let's compute the area:For t=4, sqrt(40 -16)=sqrt(24)=2*sqrt(6). So area=(4*2*sqrt(6))/2=4*sqrt(6). Which is irrational. So area is not integer.Similarly, t=1: sqrt(40 -1)=sqrt(39). Area=(1*sqrt(39))/2. Not integer.t=3: sqrt(40 -9)=sqrt(31). Area=(3*sqrt(31))/2. Not integer.t=5: sqrt(40 -25)=sqrt(15). Area=(5*sqrt(15))/2. Not integer.Therefore, only t=2 and t=6 give integer areas. But according to the problem statement, the answer should be the number of positive integer values of t, which is 2. However, the problem says "the number of all positive integer values of t", so maybe I'm missing something.Wait, but let's check again the area for t=2 and t=6.For t=2:Points of intersection: solving x + y =2 and x² + y²=20.Substitute y=2 -x into the circle equation:x² + (2 -x)² = 20x² +4 -4x +x²=202x² -4x +4 -20=02x² -4x -16=0Divide by 2:x² -2x -8=0Solutions: x=(2 ± sqrt(4 +32))/2=(2 ± sqrt(36))/2=(2 ±6)/2= (8)/2=4 or (-4)/2=-2So x=4 ⇒ y= -2; x=-2 ⇒ y=4. So points A(4,-2) and B(-2,4).Area of triangle OAB: (1/2)|x1y2 -x2y1|=(1/2)|4*4 - (-2)*(-2)|=(1/2)|16 -4|=(1/2)*12=6. Which is integer. Good.For t=6:Similarly, x + y=6. Substitute y=6 -x into x² + y²=20:x² + (6 -x)^2=20x² +36 -12x +x²=202x² -12x +16=0Divide by 2:x² -6x +8=0Solutions: x=(6 ± sqrt(36 -32))/2=(6 ±2)/2=4 or 2. So x=4, y=2; x=2, y=4. So points A(4,2) and B(2,4).Area of triangle OAB: (1/2)|4*4 -2*2|=(1/2)|16 -4|=6. Also integer. So t=6 works.Wait, but earlier when we considered the discriminant, we had t must be ≤ sqrt(40)≈6.324, so t=6 is allowed. So, only t=2 and t=6. But wait, according to the problem statement, the answer is 2. But the problem says "the number of all positive integer values of t". However, the original quadratic equation for x has discriminant D= -t² +40. For the line to intersect the circle at two distinct points, D must be positive. Therefore, t² <40. So t can be 1,2,3,4,5,6 (since t is positive integer). Wait, t=6: t²=36 <40, so it's allowed.But according to our calculation, only t=2 and t=6 give integer areas. But let's verify with another approach.Alternatively, using the area formula in terms of the distance from the center to the line and the length of AB.We found that the area is (1/2)*AB*d, where d is the distance from O to the line AB, which is t/sqrt(2), and AB is 2*sqrt(20 - t²/2). Therefore, Area= (1/2)*(2*sqrt(20 -t²/2))*(t/sqrt(2))= t*sqrt(20 -t²/2)/sqrt(2)= t*sqrt((20 -t²/2)/2)=t*sqrt(10 -t²/4). Which is the same as before.So, for the area to be integer, t*sqrt(10 -t²/4) must be integer. Let's test t=2: sqrt(10 -1)=sqrt(9)=3, so 2*3=6. Integer. t=6: sqrt(10 -9)=sqrt(1)=1, so 6*1=6. Integer. For other t:t=1: sqrt(10 -0.25)=sqrt(9.75)=sqrt(39/4)=sqrt(39)/2≈3.122. 1*sqrt(39)/2≈3.122/2≈1.561. Not integer.t=3: sqrt(10 -9/4)=sqrt(10 -2.25)=sqrt(7.75)=sqrt(31/4)=sqrt(31)/2≈2.783. 3*sqrt(31)/2≈8.35. Not integer.t=4: sqrt(10 -16/4)=sqrt(10 -4)=sqrt(6)≈2.449. 4*sqrt(6)/2=2*sqrt(6)≈4.899. Not integer.t=5: sqrt(10 -25/4)=sqrt(10 -6.25)=sqrt(3.75)=sqrt(15/4)=sqrt(15)/2≈1.936. 5*sqrt(15)/2≈5*3.872/2≈9.68. Not integer.Therefore, only t=2 and t=6 give integer areas. So the answer should be 2. But wait, the problem is from a math competition or similar, and often such problems might have more solutions. Maybe I made a mistake in assuming that sqrt(40 - t²) must be integer. Let's see.Suppose that t*sqrt(40 - t²)/2 is integer. Let's denote this as K. Then, t*sqrt(40 - t²)=2K. Then, t and sqrt(40 - t²) must be such that their product is even. Let's check for t=4:sqrt(40 -16)=sqrt(24)=2*sqrt(6). Then, t*sqrt(...)=4*2*sqrt(6)=8*sqrt(6). Then 8*sqrt(6)/2=4*sqrt(6). Not integer. Similarly, for t=1,3,5, as above. So no.Alternatively, maybe there's a different approach. Since the area is 6 for both t=2 and t=6, but is there another t where the area is integer?Wait, let's check t=0, but t is positive integer, so t=0 is excluded. How about t= sqrt(40)? But that's not integer.Alternatively, perhaps t=4, but as shown, area is 4*sqrt(6)≈9.798, not integer. Similarly, t=5 gives ≈9.68, close to 10 but not integer.Alternatively, maybe the area is integer for t=2 and t=6, so only two values. Therefore, the answer is 2. But let me check again.Wait, the problem statement says "the number of all positive integer values of t". So if t=2 and t=6, that's two values. However, in Chinese math problems, sometimes there might be symmetric situations. Let me confirm once again.Wait, when t=2, the line is x+y=2, intersecting the circle x²+y²=20. The chord AB is between (4,-2) and (-2,4). The area is 6. For t=6, the line x+y=6 intersects the circle at (4,2) and (2,4). The area is also 6. Wait, so both have area 6. So that's two t's. But maybe there's another t where the area is different integer.Wait, suppose t= sqrt(8). Then, but t must be integer. How about t=4? Area was 4*sqrt(6)≈9.798, which is not integer. t=3:≈8.35, not integer. t=5:≈9.68, not integer. So no. Therefore, only t=2 and t=6. Thus, the answer is 2.But wait, the problem says "the number of all positive integer values of t". But in the initial discriminant calculation, t can be from 1 to 6. But according to this, only two of them work. So the answer is 2. However, maybe I'm missing something. Let me check once again.Wait, let's consider another formula for the area of triangle OAB. The area can also be calculated using the formula:Area = (1/2) * |(x1y2 - x2y1)|Where (x1,y1) and (x2,y2) are points A and B. Since A and B lie on the line x + y = t, we can express y1 = t - x1 and y2 = t - x2.So, substituting into the area formula:Area = (1/2)|x1(t - x2) - x2(t - x1)| = (1/2)|x1 t - x1 x2 - x2 t + x1 x2| = (1/2)|t(x1 - x2)|So Area = (1/2)|t(x1 - x2)|But since x1 and x2 are roots of the quadratic equation we had earlier: x² - tx + (t²/2 -10) =0So, the difference of roots x1 - x2 is sqrt((x1 +x2)^2 -4x1x2). Since x1 +x2 = t and x1x2 = (t²/2 -10). Therefore,(x1 -x2)^2 = (x1 +x2)^2 -4x1x2 = t² -4*(t²/2 -10)=t² -2t² +40= -t² +40Therefore, |x1 -x2|=sqrt(40 -t²)Therefore, Area=(1/2)*t*sqrt(40 -t²)Which matches our previous result. So Area=(t/2)*sqrt(40 -t²)Therefore, for this to be integer, (t/2)*sqrt(40 -t²) must be integer. So, sqrt(40 -t²) must be rational because t is integer and the product is integer. Therefore, 40 -t² must be a perfect square. So 40 -t² =k², where k is integer.Therefore, we have t² +k²=40. We need to find positive integers t and k such that t² +k²=40. Then t can be from 1 to 6. Let's list all possible pairs (t,k):t=1: k²=39 ⇒ not=2: k²=36 ⇒k=6t=3: k²=31⇒not=4: k²=24⇒not=5:k²=15⇒not=6:k²=4⇒k=2Therefore, only t=2 and t=6. Hence, there are two positive integer values of t.Thus, the answer is 2. Therefore, the number of all positive integer values of t is 2.Final Answerboxed{2}</think>Given the line ( l: x + y = t ) intersecting the circle ( odot O: x^2 + y^2 = 20 ) at points ( A ) and ( B ), we need to find the number of positive integer values of ( t ) such that the area of triangle ( triangle OAB ) is an integer.1. Intersection Points and Quadratic Equation: - Substitute ( y = t - x ) into the circle equation ( x^2 + y^2 = 20 ): [ x^2 + (t - x)^2 = 20 ] - Simplify to get the quadratic equation in ( x ): [ 2x^2 - 2tx + t^2 - 20 = 0 ] - The discriminant of this quadratic equation must be non-negative for real intersections: [ -t^2 + 40 geq 0 implies t^2 leq 40 implies t leq sqrt{40} approx 6.324 ] - Possible positive integer values of ( t ) are 1, 2, 3, 4, 5, 6.2. Area Calculation: - The area of triangle ( triangle OAB ) can be calculated using the formula: [ text{Area} = frac{1}{2} left| x_1 y_2 - x_2 y_1 right| ] - Using the coordinates of points ( A ) and ( B ) on the line ( x + y = t ), the area can also be expressed as: [ text{Area} = frac{1}{2} left| t sqrt{40 - t^2} right| ] - For the area to be an integer, ( t sqrt{40 - t^2} ) must be even.3. Perfect Square Check: - We need ( 40 - t^2 ) to be a perfect square for the area to be rational. Let ( 40 - t^2 = k^2 ): [ t^2 + k^2 = 40 ] - Checking possible values of ( t ): - ( t = 1 ): ( k^2 = 39 ) (not a perfect square) - ( t = 2 ): ( k^2 = 36 ) (( k = 6 )) - ( t = 3 ): ( k^2 = 31 ) (not a perfect square) - ( t = 4 ): ( k^2 = 24 ) (not a perfect square) - ( t = 5 ): ( k^2 = 15 ) (not a perfect square) - ( t = 6 ): ( k^2 = 4 ) (( k = 2 ))4. Valid Values of ( t ): - Only ( t = 2 ) and ( t = 6 ) result in integer areas (6 in both cases).Thus, the number of all positive integer values of ( t ) that satisfy the condition is (boxed{2}).]

🔑:7.2.Let angle A O B=2 alpha, O C perp A B, with the foot of the perpendicular being C. Then O C=sqrt{20} cos alpha=frac{t}{sqrt{1+1}}=frac{t}{sqrt{2}} Rightarrow cos alpha=frac{t}{2 sqrt{10}}.Also, S_{triangle O A B}=10 sin 2 alpha leqslant 10, so sin 2 alpha=frac{k}{10}(k in{1,2, cdots, 10}).Thus, 2 cos alpha sqrt{1-cos ^{2} alpha}=frac{k}{10}.Substituting equation (1) into the above equation and simplifying, we get t^{2}left(40-t^{2}right)=4 k^{2}.It is easy to see that t is an even number. Let t=2 t_{0}. Then 4 t_{0}^{2}left(10-t_{0}^{2}right)=k^{2} Rightarrow t_{0}<sqrt{10}. Upon inspection, t_{0}=1,3 satisfy the conditions. Therefore, t=2,6.

🧠:[<think>Alright, so I have this problem here: There are n people (where n is at least 2) each writing a greeting card. Then, the cards are collected, and each person takes one card given by someone else. The question is asking how many different ways can these n people take the cards. Hmm, okay. Let me try to wrap my head around this.First off, each person writes a card, and then they all collect the cards. Then, each person has to take a card, but not their own. So, each person ends up with someone else's card. This seems familiar. Wait, isn't this similar to something called derangements? Yeah, derangements are permutations where no element appears in its original position. That must be it. So, the problem is essentially asking for the number of derangements of n elements.But let me make sure I understand derangements correctly. If I have n objects, say numbered from 1 to n, and I want to permute them such that no object is in its original position, that's a derangement. For example, if n is 2, swapping the two objects is the only derangement. If n is 3, there are two derangements. Wait, actually, for n=3, the derangements are the permutations where 1 isn't first, 2 isn't second, and 3 isn't third. Let me list them:1. 2, 3, 12. 3, 1, 2So that's two derangements. But wait, actually, for n=3, I think there are two derangements? Let me check. The total permutations are 6. The derangements should exclude the permutations where any element is in its original position. But actually, derangements exclude all permutations where at least one element is in its original position. Wait, no, derangements are permutations where no element is in its original position. So, yes, for n=3, the two permutations I listed are the only ones where none of the numbers are in their original positions. So, two derangements. Okay, so the number of derangements is given by the subfactorial function, sometimes written as !n.So, if the problem is asking for derangements, then the answer is !n. But how do we compute that? I remember there's a formula for derangements that involves factorials and alternating sums. Let me recall.The formula for derangements is:!n = n! * (1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n / n!)Alternatively, it can also be written using the floor function and Euler's number e, but I think the formula with the alternating sum is more precise for exact values.Alternatively, there's a recursive formula as well. The number of derangements satisfies the recurrence relation:!n = (n - 1) * (!(n - 1) + !(n - 2))With the base cases !1 = 0 and !2 = 1.Wait, let me confirm that. For n=1, there's only one person, and they can't take any card except their own, which is not allowed. So, !1 = 0. For n=2, there's only one way to swap the two cards, so !2 = 1. Then, for n=3, using the recurrence relation:!3 = (3 - 1) * (!2 + !1) = 2 * (1 + 0) = 2, which matches the earlier result. For n=4, it would be 3 * (!3 + !2) = 3*(2 + 1) = 9. Let me check with the formula. The formula using the alternating sum:!4 = 4! * (1 - 1/1! + 1/2! - 1/3! + 1/4!) = 24*(1 - 1 + 0.5 - 0.1667 + 0.0417) = 24*(0.375) = 9. Yeah, that's correct. So the recurrence relation works. So, either way, whether using the recursive formula or the summation formula, we can compute derangements.So, going back to the problem. The answer should be the number of derangements of n elements, which is !n. But the problem is phrased in terms of people taking cards. Each person initially has their own card, then they redistribute the cards so that no one gets their own. So, yes, exactly a derangement.But let me make sure I'm not missing anything here. The problem states "each person takes one card given by someone else." So, does that mean that each person must receive exactly one card, and no one can receive their own. So, it's a permutation of the cards with no fixed points. Yes, so derangements. So, the answer is the number of derangements, which is !n.Alternatively, sometimes derangements are also called subfactorials, so maybe the answer is the subfactorial of n.But perhaps I need to express this in terms of a formula rather than just !n. Let me recall the formula again.The number of derangements is:!n = n! * (1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n / n!)Alternatively, it can be approximated as n! / e, rounded to the nearest integer for positive n. But for an exact formula, the alternating sum is the way to go.Alternatively, the inclusion-exclusion principle can be used to derive the formula. Let me think about that. If we consider the total number of permutations, which is n!, and subtract the permutations where at least one person gets their own card.But inclusion-exclusion for derangements. The formula is:!n = n! * ∑_{k=0}^n (-1)^k / k!Which is the same as the alternating sum formula.So, the problem is straightforward once recognized as a derangement problem. But let me confirm with a small n. For n=2, there are two people. Each takes a card from someone else. So, they have to swap. That's 1 way. Using the formula:!2 = 2! * (1 - 1/1! + 1/2!) = 2*(1 - 1 + 0.5) = 2*(0.5) = 1. Correct.For n=3, as we saw earlier, 2 derangements. Using the formula:!3 = 3! * (1 - 1 + 1/2 - 1/6) = 6*(0 + 0.5 - 0.1667) = 6*(0.3333) = 2. Correct.So, the formula works.Alternatively, using inclusion-exclusion, the number of derangements can be calculated by:!n = n! - C(n,1)(n-1)! + C(n,2)(n-2)! - ... + (-1)^n C(n,n)0!But simplifying this gives the same alternating sum. Let me check for n=3:C(3,1)*2! = 3*2=6C(3,2)*1! = 3*1=3C(3,3)*0! = 1*1=1So, inclusion-exclusion would be:Total permutations (6) - permutations fixing at least 1 element (6) + permutations fixing at least 2 elements (3) - permutations fixing all 3 elements (1)So, 6 - 6 + 3 - 1 = 2. Which matches. So, that works.Hence, the number of derangements is indeed given by that formula.Therefore, the answer to the problem is the number of derangements of n elements, which is:!n = n! * [1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n / n!]Alternatively, in terms of the nearest integer function:!n = floor(n! / e + 0.5)But since the problem probably expects an exact formula, the first expression with the summation is more appropriate.But in many combinatorial contexts, it's acceptable to present the answer as !n or D(n), but maybe the problem expects the explicit formula. Let me check the problem statement again."How many different ways can these n people take the cards?"So, they might expect the answer in terms of a formula, possibly expressed with factorials and summations. But in mathematical literature, the derangement number is often denoted as !n, so that might be acceptable. However, to be thorough, it's better to write the formula.Alternatively, if they want a closed-form expression, it can also be written using the subfactorial notation, but perhaps the summation formula is the most explicit.Wait, but perhaps the problem is from a competition or textbook problem where the answer is expected to be boxed as boxed{!n} or boxed{D(n)}, but since the user is Chinese and given the context, maybe they prefer the explicit formula. Alternatively, in Chinese math problems, sometimes derangements are presented using the notation or formula. Let me check the standard answer for derangements.In English math problems, the standard answer is usually given as !n or using the formula with the summation. But since the user hasn't specified, and given that the problem says "put your final answer within boxed{}", it's safest to use the standard notation for derangements, which is !n. However, in some contexts, the formula is written as:!n = leftlfloor frac{n!}{e} + frac{1}{2} rightrfloorBut this is an approximation that holds exactly for n ≥ 1. However, since n ≥ 2 here, but actually, for n=1, the formula gives floor(1/e + 0.5) ≈ floor(0.3678 + 0.5) = floor(0.8678) = 0, which is correct. For n=2, floor(2/e + 0.5) ≈ floor(0.7358 + 0.5)=floor(1.2358)=1, correct. For n=3, floor(6/e +0.5)≈floor(2.207+0.5)=floor(2.707)=2, correct. For n=4, floor(24/e +0.5)≈floor(8.829+0.5)=floor(9.329)=9, which is correct. So, the formula holds. Therefore, the number of derangements can also be expressed as the nearest integer to n!/e, which is equal to floor(n!/e + 0.5). However, in terms of exact expressions, the alternating sum is precise.But perhaps in the context of the problem, they just want the answer as the derangement number, so using the notation !n. However, in many mathematical problems, especially Olympiad problems, the answer is often left in terms of derangements with the notation !n. However, since this is a Chinese problem, and Chinese math education often introduces derangements as 错位排列 (dislocation permutations), and the formula is taught as the summation.But given that the user hasn't specified, and given that the problem is presented in English, probably expecting the answer in the standard mathematical notation. Since derangements are commonly denoted as !n, and in some contexts as D(n), but given that the problem is presented with n ≥ 2, and the answer is expected to be boxed, I think using the standard notation !n is acceptable, but to be safe, maybe present the formula.Wait, but maybe the problem is from a combinatorics course and expects the answer in summation notation. Let me think. For example, the number of derangements can be written as:!n = n! sum_{k=0}^n frac{(-1)^k}{k!}So, in LaTeX, that would be:!n = n! sum_{k=0}^n frac{(-1)^k}{k!}But since the problem is asking for the number of different ways, perhaps they just want the formula, not necessarily the notation. So, if I have to write the answer, it's better to present the formula inside the box.Alternatively, in Chinese textbooks, derangements are sometimes introduced with a formula, and they may even use the notation D(n) = n! [1 - 1/1! + 1/2! - ... + (-1)^n /n!]. So, perhaps writing it in that form.But given that the user hasn't specified, and the problem is presented neutrally, I think the most appropriate answer is the derangement number expressed with the standard formula. However, in many cases, problems like this just expect the recognition that it's a derangement and the answer is !n, but since the user wants the answer boxed, and given that the answer is a formula, not a numerical value, I need to present it properly.Alternatively, check the problem statement again. It says "how many different ways can these n people take the cards". So, each person takes a card, and none receives their own. So, yes, derangements. Therefore, the answer is the number of derangements of n objects, which is known to be:!n = n! left( 1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + cdots + (-1)^n frac{1}{n!} right )Alternatively, it can be written in summation notation as:!n = sum_{k=0}^n (-1)^k frac{n!}{k!}But both expressions are equivalent. So, in LaTeX, that would be:boxed{!n}But is that acceptable? In some contexts, writing boxed{!n} is okay, but in others, they might prefer the formula. Alternatively, to be precise, write the formula.Alternatively, in the problem statement, if n is a specific number, say 5, then we would compute the value. But here, it's general n. So, the answer is the derangement number for n, which is standardly denoted as !n. However, in many mathematical competitions, it's acceptable to present the answer as the formula. Wait, but in the Chinese curriculum, how is this taught?Actually, in Chinese mathematics, derangements are often referred to as "错位排列" (cuò wèi pái liè), which translates to "misaligned permutations". The formula is usually given as:D(n) = n! left[ 1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + dots + (-1)^n frac{1}{n!} right ]Alternatively, sometimes written using the summation symbol. Given that the problem is presented in Chinese, but the user has translated it into English, and given that the answer requires a boxed notation, I think the most appropriate way is to present the formula using the summation notation, as that is the explicit form, which is more informative.Therefore, the answer is n! multiplied by the sum from k=0 to n of (-1)^k /k!.Expressed in LaTeX:boxed{n! sum_{k=0}^{n} frac{(-1)^k}{k!}}Alternatively, the same formula can be written as:boxed{sum_{k=0}^{n} (-1)^k frac{n!}{k!}}But both are equivalent. However, the first expression factors out the n! which might be more standard. Let me verify with references. For example, in the derangement Wikipedia page, the formula is written as:!n = n! sum_{k=0}^n frac{(-1)^k}{k!}Yes, that's the standard formula. Therefore, the correct boxed answer is:boxed{n! sum_{k=0}^{n} frac{(-1)^k}{k!}}Alternatively, if the problem expects a more compact notation, they might prefer !n, but since the question is about "how many different ways", and unless the problem specifically defines a notation for derangements, it's safer to present the formula.Therefore, I think the best answer is the formula written explicitly as above. Therefore, the final answer is:Final AnswerThe number of different ways is boxed{n! sum_{k=0}^{n} frac{(-1)^k}{k!}}.</think>The problem asks for the number of different ways ( n ) people can each take a card written by someone else, which is equivalent to finding the number of derangements of ( n ) elements. A derangement is a permutation where no element appears in its original position.The number of derangements, denoted as ( !n ), can be calculated using the formula:[!n = n! sum_{k=0}^{n} frac{(-1)^k}{k!}]This formula is derived from the inclusion-exclusion principle and provides the exact number of derangements for ( n ) elements. The summation alternates signs and includes terms up to ( frac{(-1)^n}{n!} ), multiplied by ( n! ).Thus, the number of different ways these ( n ) people can take the cards is given by:[boxed{n! sum_{k=0}^{n} frac{(-1)^k}{k!}}],<think>Okay, so the problem is about n people exchanging greeting cards. Each person writes a card, then all the cards are collected, and then each person takes one card that was written by someone else. We need to figure out how many different ways this can happen. Hmm, let's see.First, let me try to understand the problem better. Each person starts by writing a card. Then all the cards are collected into a pile or something. After that, each person has to take one card from the pile, but they can't take their own card. So essentially, we want to know how many different ways the cards can be redistributed so that no one ends up with their own card. That sounds familiar. Isn't this similar to something called derangements?Yeah, derangements! A derangement is a permutation of elements where none of the elements appear in their original position. In this case, the original position would be each person's own card. So if we think of each card as being assigned to a person, a derangement would be a permutation where no one gets their own card. So the problem is essentially asking for the number of derangements of n elements.But just to make sure, let's double-check. Let's say n is 2. If there are two people, person A and person B. Each writes a card. Then they collect the cards, and each has to take someone else's. The only possible way is for A to take B's card and B to take A's card. So that's 1 way. If we compute the derangement number for n=2, that's D(2) = 1, right? Because the permutations of two elements are (1,2) and (2,1). The derangement is (2,1), which is 1. So that matches.Let's check n=3. Suppose we have three people: A, B, C. Each writes a card. Now, how many derangements are there? Let's list all permutations of three elements and exclude those where any element is in its original position.Original positions: A=1, B=2, C=3.Permutations:1. (1,2,3) – everyone has their own card. Not allowed.2. (1,3,2) – A has their own card. Not allowed.3. (2,1,3) – C has their own card. Not allowed.4. (2,3,1) – No one has their own card. This is a derangement.5. (3,1,2) – No one has their own card. Another derangement.6. (3,2,1) – B has their own card. Not allowed.So there are 2 derangements for n=3. Let's compute D(3) using the formula. The formula for derangements is:D(n) = n! * (1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n /n!)So for n=3:D(3) = 3! * (1 - 1 + 1/2 - 1/6) = 6 * (0 + 0.5 - 0.1667) = 6 * (0.3333) ≈ 2, which matches our manual count. So that works.Therefore, the problem is indeed about derangements. So the answer should be D(n), the number of derangements of n elements.But let me think again if there's any catch here. The problem states "each person takes one card given by someone else". So does that mean that each person must receive exactly one card, and all cards must be distributed? Yes, because initially, each person writes one card, so there are n cards. Then, after collecting, they redistribute all n cards such that each person gets exactly one, and none get their own. So this is exactly a derangement.Another way to think about it: If there are n people, each taking a card, the total number of permutations is n!. But we need to subtract the permutations where at least one person gets their own card. This is a classic inclusion-exclusion problem.So using inclusion-exclusion principle:The number of derangements is equal to the total number of permutations minus the permutations where at least one person gets their own card. The formula for derangements via inclusion-exclusion is:D(n) = n! - C(n,1)(n-1)! + C(n,2)(n-2)! - ... + (-1)^k C(n,k)(n - k)! + ... + (-1)^n C(n,n)0!Simplifying this, we get the same formula as before:D(n) = n! [1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n /n!]Alternatively, there's a recursive formula for derangements:D(n) = (n - 1) * (D(n - 1) + D(n - 2))With base cases D(1) = 0, D(2) = 1.But for the purposes of this problem, since we need to present the answer as a formula, probably the closed-form expression with the alternating sum is acceptable. Alternatively, the more compact notation using the floor function and the exponential function, but I think the standard formula is the alternating sum one.Alternatively, another way to write the derangement number is:D(n) = leftlfloor frac{n!}{e} + frac{1}{2} rightrfloorBut this is an approximation that holds for large n. However, since the problem says n >= 2, and wants an exact answer, we should present the exact formula.Therefore, the number of different ways is the derangement number D(n), which can be written as:D(n) = n! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + dots + (-1)^n frac{1}{n!}right)Alternatively, using summation notation:D(n) = sum_{k=0}^n frac{(-1)^k}{k!} cdot n!So, putting it in a box, the answer would be:boxed{!n}But wait, in LaTeX, the notation for derangements is sometimes written as !n, called subfactorial. However, depending on the context, the problem might expect the formula expressed in terms of factorials and summations. But since the problem is presented in a mathematical context, and the answer requires a box, the standard notation is either !n or the formula. However, if the problem expects a formula rather than a notation, we should present the formula.In many combinatorics problems, derangements are presented either as !n or with the formula involving the alternating sum. Since the problem doesn't specify, and given that it's a mathematical problem likely expecting recognition of derangements, either could be acceptable. However, to be safe, let's present the formula.But let me check with n=4. If n=4, derangements should be 9. Let me compute using the formula:D(4) = 4! (1 - 1/1! + 1/2! - 1/3! + 1/4!) = 24 (1 -1 + 0.5 - 0.1667 + 0.0417) = 24*(0.375) = 9. So that's correct.Alternatively, if we use the recursive formula:D(4) = (4 - 1)(D(3) + D(2)) = 3*(2 + 1) = 9. Correct.So the formula works.Alternatively, another way to express D(n) is:D(n) = leftlfloor frac{n!}{e} + frac{1}{2} rightrfloorBut this is an approximation. For example, for n=4, n! = 24, 24/e ≈ 8.829, and adding 0.5 gives ≈9.329, the floor is 9. Which matches. For n=5, 120/e ≈44.145, +0.5≈44.645, floor is 44, and D(5)=44. So this formula also gives the exact value. But this is more of an interesting fact than a definition, since the derangement number is defined via the inclusion-exclusion formula.Given that, in a mathematical context, the standard answer would be the formula with the summation or the subfactorial notation. Since the problem is presented in Chinese and then translated, but in international math competitions, derangements are usually denoted as !n or given by that formula. But if the answer requires a formula, then writing it out in summation notation is better.Alternatively, the problem might accept !n as the answer. But given that the user is probably expecting a boxed formula, I think either is okay, but to be precise, since it's a counting problem, writing the formula explicitly might be better.But let me see if there's any other way to misinterpret the problem. For example, if the problem allowed people to take more than one card, but no, the problem states each person takes one card. So it's a permutation. So each person gets exactly one card, and none gets their own. So derangement is the exact answer.Therefore, the answer is the number of derangements of n elements, which is:D(n) = n! sum_{k=0}^n frac{(-1)^k}{k!}Alternatively, this can be written as:D(n) = leftlfloor frac{n!}{e} + frac{1}{2} rightrfloorBut again, the first formula is exact, while the second is an approximation that happens to be exact for positive integers n. Wait, is that the case? Let me check for n=1. For n=1, D(1)=0, and the formula gives floor(1/e + 0.5)=floor(0.3678 +0.5)=floor(0.8678)=0, which is correct. For n=2, floor(2/e +0.5)=floor(0.7358 +0.5)=floor(1.2358)=1, which is correct. For n=3, floor(6/e +0.5)=floor(2.207 +0.5)=floor(2.707)=2, which is correct. For n=4, floor(24/e +0.5)=floor(8.829 +0.5)=floor(9.329)=9, which is correct. So in fact, this formula gives exact values for all positive integers n. So actually, the number of derangements is equal to the nearest integer to n!/e, which can be written as floor(n!/e + 0.5). Therefore, another way to express D(n) is:D(n) = leftlfloor frac{n!}{e} + frac{1}{2} rightrfloorBut in a mathematical context, it's more standard to present the formula with the alternating sum. However, depending on the problem's expectation, either could be acceptable. But in many combinatorics problems, especially in exams, derangements are denoted as !n, but also they might require the formula.But since the problem is presented as asking for "different ways," and given that derangements are a well-known concept in combinatorics, the answer is the derangement number, which can be expressed in several ways. Since the problem is in Chinese, perhaps in their educational materials, they use a specific notation. However, in international standards, both the formula and the subfactorial notation are acceptable.But since the user is asking to put the answer within a box, and since in LaTeX, the subfactorial is sometimes denoted as !n, but not universally. For example, in some contexts, !n is written as D(n) or similar. However, if we check the problem statement again, it says "how many different ways can these n people take the cards?" So the answer is the number of derangements. If we can use the notation !n, that's concise. Otherwise, we need to use the formula.But considering that the user might expect the formula, as not everyone is familiar with the subfactorial notation, it's safer to present the formula. Therefore, the answer is:n! multiplied by the sum from k=0 to n of (-1)^k /k!Which in LaTeX is:boxed{ n! left( 1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + cdots + (-1)^n frac{1}{n!} right) }Alternatively, using summation notation:boxed{ sum_{k=0}^n (-1)^k frac{n!}{k!} }But the first expression is more explicit, while the second is more compact. The summation from k=0 to n of (-1)^k * n! /k! is correct because n! /k! = n(n-1)...(k+1). But the standard derangement formula is written as n! times the sum of (-1)^k /k! from k=0 to n. So:D(n) = n! sum_{k=0}^n frac{(-1)^k}{k!}Therefore, the boxed answer should be:boxed{ sum_{k=0}^n frac{(-1)^k n!}{k!} }Alternatively, since the summation index is k, starting from 0, but in the derangement formula, the sum is from k=0 to n. However, another way is:boxed{ n! left( 1 - frac{1}{1!} + frac{1}{2!} - cdots + (-1)^n frac{1}{n!} right) }But depending on the typesetting, sometimes the terms are written with the factorial in the denominator. Both are equivalent.Alternatively, since the problem might expect the notation !n, but if we use that, we need to make sure that the notation is recognized. However, in many mathematical contexts, especially in higher-level mathematics, the subfactorial notation is known. For example, in the OEIS (Online Encyclopedia of Integer Sequences), derangements are listed as subfactorial numbers, denoted as !n.But to avoid ambiguity, since the problem didn't specify the form of the answer, it's safer to provide the explicit formula. Therefore, the answer is the number of derangements, which can be written as:boxed{!n}But let me check the problem again. It says "how many different ways can these n people take the cards?" In Chinese math problems, sometimes they expect the formula expressed in terms of factorial and summation, but other times they accept the derangement notation. Given that the problem is translated, but the user is using a platform that might expect standard mathematical notation, using !n might be acceptable. However, considering that the user might not be familiar with the subfactorial notation, but since we are to box the answer, and !n is a standard notation, perhaps it's better to use the formula.Wait, but in Chinese textbooks, derangements are often introduced with the formula, and the notation might not be as common. Therefore, to be thorough, perhaps present both. But the user instructions say "put your final answer within boxed{}". So we can only put one answer.Since the problem is in the style of a combinatorial counting problem, and derangements are a standard concept, but if the answer requires the formula, then we should use that. However, considering that in mathematical literature, both the formula and the notation are used, but given that the problem says "how many different ways", which is a numerical answer, expressed in terms of n. The most precise answer is the formula with the summation. Alternatively, if the problem expects the recognition of the concept, then the derangement number !n.But since the problem is presented in a way that expects a mathematical formula, and given that the user might not know about the subfactorial notation, it's safer to present the formula. Therefore, the answer is:boxed{n! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + cdots + (-1)^n frac{1}{n!}right)}Alternatively, using the summation notation:boxed{sum_{k=0}^n (-1)^k frac{n!}{k!}}But the first version is more explicit in showing the alternating signs and the factorial terms, which might be more understandable. However, the second is more concise.But in the interest of clarity and standard presentation, I think the first formula is better. However, in many textbooks, the derangement formula is written as:!n = n! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + cdots + (-1)^n frac{1}{n!}right)Therefore, presenting this formula in the box is appropriate. Hence, the answer is:boxed{n! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + cdots + (-1)^n frac{1}{n!}right)}Alternatively, since LaTeX allows using the cdots, which is appropriate here.But let me check with n=2. Plugging in n=2:2! (1 - 1/1! + 1/2!) = 2*(1 -1 + 0.5) = 2*(0.5) =1. Correct.For n=3:6*(1 -1 + 0.5 - 1/6) =6*(0.5 -0.1667)=6*0.3333≈2. Correct.So the formula holds. Therefore, the answer is correctly represented by this formula.Alternatively, since the problem says "different ways", which is a count, and derangements are the exact count. Therefore, the answer is the derangement number, which can be expressed as the formula above. Therefore, the boxed answer should be that formula.Thus, the final answer is:boxed{n! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + cdots + (-1)^n frac{1}{n!}right)}</think>The problem involves determining the number of ways ( n ) people can each take a greeting card written by someone else, which is equivalent to finding the number of derangements of ( n ) elements. A derangement is a permutation where no element appears in its original position. The formula for the number of derangements ( D(n) ) is given by:[D(n) = n! left( 1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + cdots + (-1)^n frac{1}{n!} right)]This formula is derived using the principle of inclusion-exclusion, subtracting the permutations where at least one person gets their own card from the total permutations and adjusting for overlaps.Thus, the number of different ways ( n ) people can take the cards is:[boxed{n! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + cdots + (-1)^n frac{1}{n!}right)}]]

🔑:Let n people be denoted as A_{1}, A_{2}, cdots, A_{n}, and their corresponding greeting cards as B_{1}, B_{2}, cdots, B_{n}. Let the number of different ways for n people to take the cards be a_{n}.It is easy to see that a_{1}=0, a_{2}=1, a_{3}=2.When n geqslant 4, the process of n people taking greeting cards can be divided into two steps.(1) A_{1} has n-1 ways to choose.(2) The person corresponding to the card chosen by A_{1} then takes a card, while also considering the ways for the remaining n-2 people to take cards. Suppose A_{1} takes card B_{2}, then consider the number of ways for A_{2} to take a card, while also considering the ways for the remaining n-2 people to take cards.This can be further divided into two categories.(i) A_{2} takes B_{1}, then A_{2}, A_{3}, cdots, A_{n} have a_{n-2} ways to take cards;(ii) A_{2} does not take B_{1}, then A_{2}, A_{3}, cdots, A_{n} have a_{n-1} ways to take cards.In summary, by the principle of multiplication for step-by-step counting and the principle of addition for classification, the number of different ways for n people to take cards isbegin{array}{l}a_{n}=(n-1)left(a_{n-1}+a_{n-2}right) Rightarrow a_{n+1}=nleft(a_{n}+a_{n-1}right) Rightarrow a_{n+1}-a_{n}=n a_{n}+a_{n-1}-(n-1) a_{n-2} Rightarrow a_{n+1}-(n+1) a_{n}=a_{n-1}-(n-1) a_{n-2} . text { Let } b_{n}=a_{n+1}-(n+1) a_{n} text {. Then } b_{n}=b_{n-2} .end{array}When n is even,begin{array}{c}a_{n+1}-(n+1) a_{n}=b_{n}=b_{n-2}=cdots =b_{2}=a_{3}-3 a_{2}=-1=(-1)^{n+1} .end{array}Then a_{n+1}=(n+1) a_{n}+(-1)^{n+1}.When n is odd,begin{array}{l}a_{n+1}-(n+1) a_{n}=b_{n}=b_{n-2}=cdots =b_{1}=a_{2}-2 a_{1}=1=(-1)^{n+1} .end{array}Then a_{n+1}=(n+1) a_{n}+(-1)^{n+1}.Combining the cases for even and odd n, we havebegin{array}{l}a_{n+1}=(n+1) a_{n}+(-1)^{n+1} Rightarrow frac{a_{n+1}}{(n+1)!}=frac{a_{n}}{n!}+frac{(-1)^{n+1}}{(n+1)!} Rightarrow frac{a_{n}}{n!}=frac{a_{2}}{2!}+sum_{k=3}^{n} frac{(-1)^{k}}{k!} Rightarrow a_{n}=(n!) sum_{k=2}^{n} frac{(-1)^{k}}{k!} .end{array}

🧠:[<think>Okay, let me think. The problem is asking for the distance between the points representing -5 and -1 on the number line. The options are A: 6, B: 5, C: 4, D: 3. Hmm, distance on a number line... I remember that distance between two points is the absolute value of the difference between their coordinates. So, if we have two numbers, say a and b, the distance between them is |a - b| or |b - a|. Either way, it should give the same result because absolute value makes it positive. Let me write that down.So, the coordinates here are -5 and -1. Let's denote them as a = -5 and b = -1. Then the distance should be |a - b|. Plugging in the numbers: |-5 - (-1)|. Wait, subtracting a negative is like adding a positive. So that becomes |-5 + 1|. Let's compute that: -5 + 1 is -4. Then the absolute value of -4 is 4. So, | -4 | = 4. Therefore, the distance is 4.Alternatively, maybe I can check by counting the units on the number line. If I start at -5 and move towards -1, how many steps does it take? Let's see: from -5 to -4 is 1 unit, -4 to -3 is another, then -3 to -2 is a third, and -2 to -1 is the fourth. So that's 4 units in total. That matches the previous result. So both methods give me 4. Looking at the options, C is 4. So the answer should be C.But wait, let me double-check to make sure I didn't make a mistake. Another way to calculate distance between two numbers is the absolute difference. So, the formula is |x - y|. Let's apply that here. Let x = -5 and y = -1. Then |(-5) - (-1)| = |-5 + 1| = |-4| = 4. Alternatively, if I reverse them: |(-1) - (-5)| = |-1 + 5| = |4| = 4. Either way, same answer. So definitely 4.Another sanity check: on the number line, negative numbers are to the left of zero. So -5 is 5 units left of zero, and -1 is 1 unit left of zero. The distance between them is 5 - 1 = 4. Wait, is that right? Because from -1 to 0 is 1, and from -5 to 0 is 5. So the total distance from -5 to 0 is 5, and from -1 to 0 is 1. Therefore, the distance between -5 and -1 is 5 - 1 = 4. Yeah, that also makes sense. So subtracting their distances from zero gives the distance between them. But is this always true? Hmm, maybe only when both numbers are on the same side of zero? Let me see. For example, if one is positive and one is negative, like 3 and -2. The distance between them would be 3 - (-2) = 5, but their distances from zero are 3 and 2, which add up to 5. So in that case, you add them. So maybe when numbers are on the same side of zero, you subtract their absolute values, and when they are on opposite sides, you add. Interesting. But in this problem, both are negative, so same side. So subtracting their distances from zero (which are 5 and 1) gives 4, which is correct. So that method also works here. So that's another way to confirm.Therefore, all methods point to the answer being 4, which is option C. I don't think I made a mistake here. The answer is definitely C.Final Answerboxed{C}</think>To find the distance between the points representing (-5) and (-1) on the number line, we use the formula for the distance between two points (a) and (b) on a number line, which is (|a - b|).1. Let (a = -5) and (b = -1).2. Calculate the absolute difference: (|-5 - (-1)|).3. Simplify the expression inside the absolute value: (|-5 + 1| = |-4|).4. The absolute value of (-4) is (4).We can also verify this by counting the units on the number line:- From (-5) to (-4) is 1 unit.- From (-4) to (-3) is another unit.- From (-3) to (-2) is another unit.- From (-2) to (-1) is another unit.This gives a total of 4 units. Both methods confirm that the distance is 4.Thus, the final answer is (boxed{C}).,<think>Okay, so I need to find the distance between the points -5 and -1 on the number line. Hmm, let me think about how to do this. I remember that distance on a number line is just the absolute difference between the two points. Is that right? Yeah, because distance can't be negative, so taking the absolute value makes sense.So, if I have two numbers, say a and b, the distance between them is |a - b| or |b - a|. Either way, it should give me the same positive number. Let me test this with some numbers I know. For example, the distance between 2 and 5 is 3. Let me check: |5 - 2| = |3| = 3, and |2 - 5| = |-3| = 3. Yep, that works. So the same logic should apply to negative numbers.Alright, so applying this to -5 and -1. Let's do both ways to make sure. First, |-5 - (-1)|. Let's compute that. Subtracting a negative is like adding, so that becomes |-5 + 1| = |-4| = 4. Alternatively, |-1 - (-5)| = |-1 + 5| = |4| = 4. Both ways, I get 4. So the distance is 4. Let me check the answer choices. Option C is 4, so that should be the answer.Wait, but just to visualize it on the number line. Starting at -5, how many units do I move to get to -1? From -5 to -4 is 1 unit, -4 to -3 is another, -3 to -2 is third, and -2 to -1 is fourth. So that's 4 units. Yeah, that matches my calculation. So I think that's correct. The answer must be C:4.Final Answerboxed{C}</think>To find the distance between the points representing (-5) and (-1) on the number line, we calculate the absolute difference between the two points. The distance formula on a number line is given by:[|a - b|]where (a) and (b) are the coordinates of the points.Applying this to (-5) and (-1):[|-5 - (-1)| = |-5 + 1| = |-4| = 4]or equivalently,[|-1 - (-5)| = |-1 + 5| = |4| = 4]Both methods yield the same result, confirming that the distance is (4).Visualizing the number line, moving from (-5) to (-1) involves four units: (-5) to (-4), (-4) to (-3), (-3) to (-2), and (-2) to (-1).Thus, the final answer is (boxed{C}).]

🔑:To find the distance between the points representing -5 and -1 on the number line, we calculate the difference between the two numbers. This is done by subtracting the smaller number from the larger number, or in this case, since both numbers are negative, by subtracting -5 from -1:[begin{align*}-1 - (-5) &= -1 + 5 &= 4end{align*}]Therefore, the distance between the points representing -5 and -1 on the number line is 4. According to the options provided:A: 6B: 5C: 4D: 3The correct answer is boxed{C}.