🧠:[<think>Okay, let me try to solve these two problems step by step. I need to take my time and make sure I understand each part before moving on. Let's start with problem (1).Problem (1): Calculate the expressionThe expression is: (0.25)^0 - (1/16) - 0.75 + 4(1 - √2)^4 + √(6 - 4√2) + ln√e + 2^{2 + log₂3}Hmm, that looks like a mix of different operations: exponents, roots, logarithms, etc. Let me break it down term by term.1. First term: (0.25)^0 Any non-zero number to the power of 0 is 1. So, 0.25^0 = 1. Easy enough.2. Second term: -(1/16) That's just -1/16. Keep that in mind.3. Third term: -0.75 That's -3/4. So, so far we have 1 - 1/16 - 3/4. Let me convert these to fractions to add them up later. 1 is 16/16, so 16/16 - 1/16 - 12/16 (since 3/4 is 12/16). That would be (16 -1 -12)/16 = 3/16. Wait, maybe I should calculate all terms first before combining. Let me hold off on combining them until the end.4. Fourth term: 4(1 - √2)^4 This looks more complicated. Let's compute (1 - √2)^4 first. Maybe expand it step by step. Let's recall that (a - b)^4 can be expanded as a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4. Let me compute each term:(1 - √2)^2 = 1^2 - 2*1*√2 + (√2)^2 = 1 - 2√2 + 2 = 3 - 2√2.Then, (1 - √2)^4 = [(1 - √2)^2]^2 = (3 - 2√2)^2. Let's compute that:(3)^2 - 2*3*(2√2) + (2√2)^2 = 9 - 12√2 + 8 = 17 - 12√2.Therefore, 4*(1 - √2)^4 = 4*(17 - 12√2) = 68 - 48√2.5. Fifth term: √(6 - 4√2) This square root might simplify if the expression inside is a perfect square. Let me check if 6 - 4√2 can be written as (a - b√2)^2. Let's assume:(a - b√2)^2 = a^2 - 2ab√2 + 2b^2.Comparing to 6 -4√2, we have:a^2 + 2b^2 = 6 -2ab = -4From the second equation: 2ab = 4 ⇒ ab = 2.So we need integers a and b such that ab = 2 and a^2 + 2b^2 = 6.Let's try a=2, b=1: a^2 + 2b^2 = 4 + 2 = 6. Perfect!Therefore, √(6 - 4√2) = √( (2 - √2)^2 ) = |2 - √2|. Since 2 > √2 (≈1.414), this simplifies to 2 - √2.6. Sixth term: ln√e The natural logarithm of the square root of e. Remember that ln(e^x) = x. So, √e = e^(1/2), so ln(e^(1/2)) = 1/2. That's straightforward.7. Seventh term: 2^{2 + log₂3} This is an exponent with base 2. Let's use properties of exponents and logarithms. Remember that 2^{a + b} = 2^a * 2^b. Also, 2^{log₂3} = 3, because a^{log_a b} = b. So:2^{2 + log₂3} = 2^2 * 2^{log₂3} = 4 * 3 = 12. Nice.Now, let's compile all the computed terms:1. 12. -1/163. -0.754. 68 - 48√25. 2 - √26. 1/27. 12Now, sum all these together:1 - 1/16 - 0.75 + (68 - 48√2) + (2 - √2) + 1/2 + 12Let me convert all the constants (non-radical terms) first:1 - 1/16 - 0.75 + 68 + 2 + 1/2 + 12Convert all to fractions with denominator 16 for easy addition:1 = 16/16 -1/16 = -1/16 -0.75 = -3/4 = -12/16 68 = 68 2 = 2 1/2 = 8/16 12 = 12Wait, but 68 and 12 are whole numbers. Maybe better to compute decimals?Alternatively, group the whole numbers and fractions separately.Whole numbers: 1 + 68 + 2 + 12 = 1 + 68 is 69, 69 + 2 is 71, 71 +12 is 83.Fractions: -1/16 - 0.75 + 1/2. Let me convert -0.75 to -3/4 and 1/2 to 8/16.So fractions:-1/16 - 12/16 + 8/16 = (-1 -12 +8)/16 = (-5)/16.Therefore, total constants: 83 - 5/16 = 82 + 11/16 ≈ 82.6875.Now, the radical terms:-48√2 - √2 = (-48 -1)√2 = -49√2.So putting it all together:83 - 5/16 -49√2.Wait, but the answer might need to be simplified further. Let me check if they want a numerical value or if there's a way to combine terms. However, given the presence of √2, unless there's some cancellation, this is as simplified as it gets. Let me verify each term again to ensure no miscalculations.Wait, but the original expression is:(0.25)^0 - (1/16) - 0.75 + 4(1 - √2)^4 + √(6 -4√2) + ln√e + 2^{2 + log₂3}So, let's recount each term:1. (0.25)^0 = 1 2. -1/16 3. -0.75 4. 4*(1 - √2)^4 = 68 -48√2 5. √(6 -4√2) = 2 - √2 6. ln√e = 1/2 7. 2^{2 + log₂3} = 12 Adding up constants:1 -1/16 -0.75 +68 +2 +1/2 +12 Compute in decimals:1 = 1 -1/16 ≈ -0.0625 -0.75 = -0.75 68 = 68 2 = 2 1/2 = 0.5 12 = 12 Adding these:1 -0.0625 = 0.9375 0.9375 -0.75 = 0.1875 0.1875 +68 = 68.1875 68.1875 +2 = 70.1875 70.1875 +0.5 = 70.6875 70.6875 +12 = 82.6875Now the radical terms:-48√2 -√2 = -49√2 ≈ -49*1.4142 ≈ -69.2958So total ≈ 82.6875 -69.2958 ≈ 13.3917But maybe the exact form is better. Let me see:The entire expression is:1 -1/16 -3/4 +68 -48√2 +2 -√2 +1/2 +12Combine constants:1 +68 +2 +12 = 83Fractions: -1/16 -3/4 +1/2Convert to sixteenths:-1/16 -12/16 +8/16 = (-1 -12 +8)/16 = (-5)/16So constants total: 83 -5/16Radicals: -48√2 -√2 = -49√2So total expression: 83 -5/16 -49√2But is there a way to simplify this more? Let me check if 49√2 can be written as something else or if there is a relation, but I don't think so. Therefore, perhaps the answer is written as (83 -5/16) -49√2, which can also be written as (1328/16 -5/16) -49√2 = 1323/16 -49√2. But unless they want a combined fraction, maybe leave it as 83 -5/16 -49√2. Alternatively, convert 83 to sixteenths:83 = 83*16/16 = 1328/161328/16 -5/16 = 1323/16So total is 1323/16 -49√2. Hmm, maybe this is the form. Alternatively, if they require a decimal, but the problem says "calculate," which could mean exact form. So probably 1323/16 -49√2. Let me check once again if all terms were calculated correctly.Wait, 4*(1 -√2)^4: we calculated (1 -√2)^2 = 3 -2√2, then squared that: (3 -2√2)^2 = 9 -12√2 + 8 = 17 -12√2. Then 4*(17 -12√2) = 68 -48√2. Correct.√(6 -4√2) = 2 -√2. Correct, since (2 -√2)^2 = 4 -4√2 +2 = 6 -4√2.ln√e = 1/2. Correct.2^{2 + log₂3} = 4*3 =12. Correct.So yes, all terms are correct. So combining constants and radicals:1323/16 -49√2. If needed, convert 1323/16 to a mixed number: 1323 ÷16 =82.6875, which is 82 +11/16. So 82 11/16 -49√2. But I think fractional form is better here.Alternatively, if they prefer decimals, it's approximately 82.6875 -49*1.4142≈82.6875 -69.2958≈13.3917. But exact form is better unless specified otherwise.So, the exact value is 1323/16 -49√2. But let me check if the problem expects simplification in another way. The problem says "Calculate," which could mean exact simplified form. So 1323/16 -49√2. Alternatively, 83 -5/16 -49√2. Either is acceptable, but 1323/16 is perhaps more precise. Let me confirm the arithmetic once again.Original terms after computing each part:1 (first term) -1/16 (second term) -3/4 (third term) 68 -48√2 (fourth term) 2 -√2 (fifth term) 1/2 (sixth term) 12 (seventh term)Adding constants:1 -1/16 -3/4 +68 +2 +1/2 +12Convert to sixteenths:1 = 16/16 -1/16 = -1/16 -3/4 = -12/16 68 = 68*16/16 = 1088/16 2 = 32/16 1/2 = 8/16 12 = 192/16Adding all fractions:16/16 -1/16 -12/16 +1088/16 +32/16 +8/16 +192/16 = (16 -1 -12 +1088 +32 +8 +192)/16 Calculate numerator:16 -1 =15; 15 -12=3; 3 +1088=1091; 1091 +32=1123; 1123 +8=1131; 1131 +192=1323So total constants:1323/16.Radical terms: -48√2 (from fourth term) and -√2 (from fifth term) total -49√2.Thus, the exact answer is 1323/16 -49√2. So that's the simplified form. Therefore, unless there's a miscalculation I missed, this should be the correct result. Let me check once again:Original terms:1, -1/16, -0.75, 68 -48√2, 2 -√2, 0.5, 12.Adding constants:1 -0.0625 -0.75 +68 +2 +0.5 +12.Calculating step by step:Start with 1.1 -0.0625 = 0.93750.9375 -0.75 = 0.18750.1875 +68 =68.187568.1875 +2 =70.187570.1875 +0.5 =70.687570.6875 +12 =82.6875Yes, 82.6875 is 1323/16. 16*82=1312, 1323-1312=11, so 82 11/16, which is 82.6875. Correct.Radical terms: -48√2 -√2 = -49√2.So the exact answer is 1323/16 -49√2. If the question expects this, that's the answer. Alternatively, if they want a decimal approximation, but since the problem didn't specify, I think the exact form is better. Let me check if 1323/16 -49√2 can be simplified further, but I don't see an obvious way. So that's probably the answer.Problem (2): Given 14^a = 6, 14^b =7, express log_{42}56 in terms of a and b.Okay, so we need to express the logarithm of 56 with base 42 using a and b, where a and b are exponents such that 14^a =6 and 14^b=7.First, let's recall that log_{42}56 can be expressed using the change of base formula. Maybe express it in terms of logarithms with base 14, since we have information about 14^a and 14^b.Alternatively, express everything in terms of prime factors. Let's see.First, let's factorize the numbers:42 = 6*7 = 2*3*7 56 = 8*7 = 2^3*7 14 = 2*7Given that 14^a =6, which is 2*3, and 14^b=7.So, we have 14^a =6=2*3. Since 14=2*7, then 14^a =2^a *7^a =2*3. Therefore:2^a *7^a =2^1 *3^1.So, equating the exponents for prime factors:For prime 2: a =1 (since 2^a =2^1 ⇒ a=1)But wait, but that would imply 7^a=3^1. But 7^a=3 ⇒ a=log_7 3. But in the problem, 14^a=6. So this seems conflicting. Wait, maybe I need to approach it differently.Wait, perhaps the variables a and b are given such that 14^a=6 and 14^b=7. So, a and b are logarithms: a=log_{14}6 and b=log_{14}7.Therefore, since 14^a=6, then a=log_{14}6, and similarly b=log_{14}7.We need to find log_{42}56. Let's try to express this in terms of log_{14}6 and log_{14}7.First, recall that log_{42}56 = (ln56)/(ln42). If we can express ln56 and ln42 in terms of ln14, ln6, ln7, etc., then maybe we can relate them to a and b.Alternatively, use the change of base formula to base 14:log_{42}56 = (log_{14}56)/(log_{14}42)So, if we can compute log_{14}56 and log_{14}42 in terms of a and b.Compute log_{14}42:42 =14*3, so log_{14}42 = log_{14}(14*3) = log_{14}14 + log_{14}3 =1 + log_{14}3.But 14^a=6=2*3, and 14=2*7. So, perhaps express 3 in terms of 14 and a.Given 14^a=6=2*3. Since 14=2*7, then 14^a=(2*7)^a=2^a*7^a=6=2*3. Therefore:2^a *7^a =2*3 ⇒ 2^{a-1} *7^a=3.So, 3=2^{a-1}*7^a. Therefore, log_{14}3 = log_{14}(2^{a-1}*7^a) = (a-1)log_{14}2 +a log_{14}7.But we need to express log_{14}2 and log_{14}7 in terms of a and b.Wait, but from 14^a=6=2*3 and 14^b=7.We know that 14^b=7. Therefore, log_{14}7 =b.Also, from 14=2*7, log_{14}14=1= log_{14}(2*7)= log_{14}2 + log_{14}7= log_{14}2 +b. Therefore, log_{14}2=1 -b.Therefore, log_{14}3= (a-1)(1 -b) +a*b.Let me compute that:(a -1)(1 -b) +a*b = (a -1 -ab +b) +ab =a -1 -ab +b +ab= a -1 +b.Therefore, log_{14}3= a +b -1.Therefore, log_{14}42=1 + log_{14}3=1 +a +b -1= a +b.So log_{14}42= a +b.Now compute log_{14}56:56=14*4=14*2^2. Therefore, log_{14}56= log_{14}14 + log_{14}2^2=1 +2 log_{14}2.We already established that log_{14}2=1 -b. Therefore:log_{14}56=1 +2*(1 -b)=1 +2 -2b=3 -2b.Therefore, log_{42}56= (log_{14}56)/(log_{14}42)= (3 -2b)/(a +b).So the expression is (3 -2b)/(a +b).Wait, let me check that again.log_{42}56= (log_{14}56)/(log_{14}42)= (3 -2b)/(a +b). Yes, that seems right.But let me verify step by step.First, log_{14}56:56=14*4. 14 is 14^1, 4 is 2^2. So log_{14}56= log_{14}14 + log_{14}4=1 + log_{14}4=1 + log_{14}2^2=1 +2 log_{14}2.Since log_{14}2=1 -b, then:log_{14}56=1 +2*(1 -b)=1 +2 -2b=3 -2b.Correct.log_{14}42= a +b. As computed before, since 42=14*3, log_{14}42=1 + log_{14}3, and log_{14}3= a +b -1. Therefore, 1 + (a +b -1)=a +b. Correct.Therefore, log_{42}56= (3 -2b)/(a +b).So the answer is (3 -2b)/(a +b). Let me check if that makes sense.Alternatively, let's try plugging in actual numbers to verify. Let's choose a value for a and b.Suppose we take a=log_{14}6 and b=log_{14}7. Let me compute a and b numerically.First, compute log_{14}6:log_{14}6= ln6 /ln14≈1.7918/2.6391≈0.679.Similarly, log_{14}7= ln7/ln14≈1.9459/2.6391≈0.737.So a≈0.679, b≈0.737.Then, log_{42}56= (3 -2b)/(a +b)= (3 -2*0.737)/(0.679 +0.737)= (3 -1.474)/(1.416)=1.526/1.416≈1.077.Now, compute log_{42}56 directly:log_{42}56= ln56/ln42≈3.9120/3.7377≈1.046.Wait, there's a discrepancy. Hmm. That's a problem. So either my calculation is wrong, or the approximation is off due to rounding errors. Let me check more accurately.Compute a=log_{14}6:log_14(6)=ln6/ln14≈(1.791759)/2.639057≈0.6795b=log_{14}7=ln7/ln14≈1.945910/2.639057≈0.7373Compute (3 -2b)/(a +b)= (3 -2*0.7373)/(0.6795 +0.7373)= (3 -1.4746)/1.4168≈1.5254/1.4168≈1.0768Compute log_{42}56:ln56≈4.025351, ln42≈3.7376694.025351/3.737669≈1.0770Wait, so actually, when computed more accurately, it's about 1.0770. The direct calculation of log_{42}56≈1.0770, which matches the expression (3 -2b)/(a +b)≈1.0768. So the formula seems correct.Earlier approximation was rough. Therefore, the answer is (3 -2b)/(a +b). So that's correct.Therefore, the answer to problem (2) is (3 -2b)/(a +b).Final Answer(1) boxed{dfrac{1323}{16} - 49sqrt{2}} (2) boxed{dfrac{3 - 2b}{a + b}}</think>(1) Calculate the expression:[(0.25)^0 - left(frac{1}{16}right) - 0.75 + 4(1- sqrt{2})^4 + sqrt{6-4sqrt{2}} + ln sqrt{e} + 2^{2+log_{2}3}]Step-by-step breakdown:1. ((0.25)^0 = 1)2. (-left(frac{1}{16}right) = -frac{1}{16})3. (-0.75 = -frac{3}{4})4. (4(1 - sqrt{2})^4): - ((1 - sqrt{2})^2 = 3 - 2sqrt{2}) - ((3 - 2sqrt{2})^2 = 17 - 12sqrt{2}) - (4(17 - 12sqrt{2}) = 68 - 48sqrt{2})5. (sqrt{6 - 4sqrt{2}} = 2 - sqrt{2}) (since ((2 - sqrt{2})^2 = 6 - 4sqrt{2}))6. (ln sqrt{e} = frac{1}{2})7. (2^{2 + log_{2}3} = 2^2 cdot 2^{log_{2}3} = 4 cdot 3 = 12)Combining all terms:- Constants: (1 - frac{1}{16} - frac{3}{4} + 68 + 2 + frac{1}{2} + 12 = 83 - frac{5}{16})- Radicals: (-48sqrt{2} - sqrt{2} = -49sqrt{2})Final result:[frac{1323}{16} - 49sqrt{2}](2) Given (14^a = 6) and (14^b = 7), express (log_{42}56) in terms of (a) and (b).Step-by-step breakdown:- Use change of base formula: (log_{42}56 = frac{log_{14}56}{log_{14}42})- Factorize numbers: - (56 = 14 cdot 4 = 14 cdot 2^2) - (42 = 14 cdot 3)- Calculate logarithms: - (log_{14}56 = 1 + 2log_{14}2) - (log_{14}42 = 1 + log_{14}3)- Express (log_{14}2) and (log_{14}3) in terms of (a) and (b): - (log_{14}2 = 1 - b) - (log_{14}3 = a + b - 1)- Substitute back: - (log_{14}56 = 1 + 2(1 - b) = 3 - 2b) - (log_{14}42 = 1 + (a + b - 1) = a + b)Final result:[frac{3 - 2b}{a + b}] Final Answers(1) (boxed{dfrac{1323}{16} - 49sqrt{2}})(2) (boxed{dfrac{3 - 2b}{a + b}}),<think>Okay, let me tackle these two problems step by step. Starting with problem (1):First, I need to calculate the expression: (0.25)^0 - left(frac{1}{16}right) - 0.75 + 4(1- sqrt{2})^4 + sqrt{6-4sqrt{2}} + ln sqrt{e} + 2^{2+log_{2}3}.Let me break this down term by term.1. (0.25)^0: Any number to the power of 0 is 1, so this is 1.2. - left(frac{1}{16}right): That's just negative 1/16.3. - 0.75: Subtract 0.75.4. 4(1 - sqrt{2})^4: Hmm, this might be a bit tricky. Let me compute (1 - sqrt{2})^4 first. Let's recall that (1 - sqrt{2})^2 = 1 - 2sqrt{2} + 2 = 3 - 2sqrt{2}. Then squaring that again: (3 - 2sqrt{2})^2. Let me compute that:First, square 3: 9. Then cross terms: 2*3*(-2√2) = -12√2. Then square of the second term: (-2√2)^2 = 8. So altogether, 9 - 12√2 + 8 = 17 - 12√2. Therefore, (1 - sqrt{2})^4 = 17 - 12sqrt{2}. Multiply by 4: 4*(17 - 12√2) = 68 - 48√2.Wait, let me check that again. Wait, (3 - 2√2)^2:Yes, (a - b)^2 = a² - 2ab + b². So 3² = 9, 2ab = 2*3*2√2 = 12√2, and b² = (2√2)^2 = 8. So it's 9 - 12√2 + 8 = 17 - 12√2. Correct. Then multiply by 4: 68 - 48√2. Okay.5. sqrt{6 - 4sqrt{2}}: This looks like a nested square root. Maybe it can be simplified into the form √a - √b. Let me check:Assume √(6 - 4√2) = √a - √b. Then squaring both sides:6 - 4√2 = a + b - 2√(ab).Therefore, we need to solve:a + b = 6-2√(ab) = -4√2 → 2√(ab) = 4√2 → √(ab) = 2√2 → ab = 8.So we have a system:a + b = 6ab = 8.Solving this quadratic: x² -6x +8 =0. The roots are x = [6 ± √(36 -32)]/2 = [6 ± 2]/2 = 4 or 2. Therefore, a=4 and b=2. So √(6 -4√2) = √4 - √2 = 2 - √2. But wait, let's check:(2 - √2)^2 = 4 -4√2 +2 = 6 -4√2. Yes, correct. So the square root is 2 - √2. But since 2 - √2 is positive (√2 ≈1.414, so 2 -1.414≈0.586), which is positive, so it's valid. Therefore, this term simplifies to 2 - √2.6. ln sqrt{e}: The natural logarithm of the square root of e. Let's recall that ln(e^x) =x. So √e = e^(1/2), so ln(e^(1/2)) =1/2.7. 2^{2+log_{2}3}: Let me break this down. 2^(2 + log₂3) = 2² * 2^(log₂3). 2² is 4, and 2^(log₂3) is just 3, because 2 and log₂ are inverse functions. So 4*3=12.Now, let me put all the computed terms back into the original expression:1 (from the first term) - 1/16 - 0.75 + (68 -48√2) + (2 - √2) + 0.5 + 12.Let me compute all the constants first, then the terms with √2:Constants:1 - 1/16 - 0.75 + 68 + 2 + 0.5 + 12.Convert all to fractions or decimals for easier computation. Let's do decimals:1 is 1.01/16 is 0.06250.75 is 0.7568 is 68.02 is 2.00.5 is 0.512 is 12.0So compute:1.0 - 0.0625 = 0.93750.9375 - 0.75 = 0.18750.1875 + 68.0 = 68.187568.1875 + 2.0 = 70.187570.1875 + 0.5 = 70.687570.6875 + 12.0 = 82.6875Now, the √2 terms:From 4(1 - √2)^4: -48√2From sqrt(6 -4√2): -√2So total √2 terms: -48√2 - √2 = -49√2Therefore, the entire expression is 82.6875 -49√2.Wait, but √2 is approximately 1.4142. So if we compute this numerically, 49*1.4142 ≈49*1.4142≈69.2958. So 82.6875 -69.2958≈13.3917. Let me check if this answer is exact or if maybe I made a miscalculation in combining the terms.Wait, but the problem might expect an exact form rather than a decimal. Let me check again.Wait, the original expression has several terms that can be combined:Let me list all terms again:1. 12. -1/163. -0.75 (which is -3/4)4. 68 -48√25. 2 - √26. 1/27. 12So combine the constants:1 -1/16 -3/4 +68 +2 +1/2 +12.Convert all to sixteenths to compute:1 = 16/16-1/16 = -1/16-3/4 = -12/1668 = 68*16/16 = 1088/162 = 32/161/2 = 8/1612 = 192/16So total:16/16 -1/16 -12/16 +1088/16 +32/16 +8/16 +192/16 =(16 -1 -12 +1088 +32 +8 +192)/16Compute numerator:16 -1 =1515 -12=33 +1088=10911091 +32=11231123 +8=11311131 +192=1323So 1323/16.Then, the √2 terms:-48√2 -√2 = -49√2Therefore, the exact value is 1323/16 -49√2.Wait, 1323 divided by 16. Let me check if that reduces. 16*82=1312, so 1323-1312=11. So 1323/16=82 +11/16=82.6875, which matches the decimal earlier. So the exact form is 1323/16 -49√2. Maybe this is the answer? But let me check if I did all steps correctly.Wait, 4*(1 - √2)^4 was computed as 68 -48√2. Let me verify that step again. Expanding (1 - √2)^4:First, (1 - √2)^2 =1 -2√2 +2=3 -2√2.Then, (3 -2√2)^2=9 -12√2 +8=17 -12√2. So (1 -√2)^4=17 -12√2. Multiply by 4: 4*(17 -12√2)=68 -48√2. Correct.Then sqrt(6 -4√2)=2 -√2, as verified earlier. So that term is 2 -√2. Then ln sqrt(e)=1/2. Then 2^{2 + log₂3}=4*3=12. So all steps look correct.So combining constants: 1 -1/16 -3/4 +68 +2 +1/2 +12=1323/16.Then sqrt terms: -48√2 -√2= -49√2.Therefore, total expression is 1323/16 -49√2. Let me check if this can be simplified further. 1323 and 16: 1323 ÷ 7= 189, 189 ÷7=27. So 1323=7*7*27. 16 is 2^4. So no common factors. So 1323/16 is in simplest form. So the exact value is 1323/16 -49√2. Alternatively, as a mixed number, 82 and 11/16 minus 49√2, but likely the improper fraction is better.But let me check if the problem expects a numerical value or an exact expression. The problem says "Calculate", so it's better to present the exact form. So 1323/16 -49√2.Alternatively, let me compute the numerical value to check:1323 ÷16=82.687549√2≈49*1.4142≈69.2958So 82.6875 -69.2958≈13.3917. So approximately 13.39. But unless the problem asks for a decimal, exact form is better.Alternatively, maybe we can write 1323/16 as 82.6875, but perhaps the answer is supposed to be simplified more? Let me check again.Wait, 1323/16. Let me divide 1323 by 16:16*80=1280, 1323-1280=43. 16*2=32, remainder 11. So 82 and 11/16. So 82 11/16. Hmm. Not sure if that's necessary.Alternatively, maybe there's a different approach to simplify some terms earlier?Wait, the term 4*(1 -√2)^4: we expanded it to 68 -48√2, and sqrt(6 -4√2) is 2 -√2, so when adding these two terms: 68 -48√2 +2 -√2=70 -49√2.Then the rest of the terms: 1 -1/16 -0.75 +0.5 +12.Wait, let's recompute:Original expression:1 -1/16 -0.75 + [4*(1 -√2)^4] + [sqrt(6 -4√2)] + [ln sqrt(e)] + [2^{2 +log₂3}]Which is:1 -1/16 -0.75 + (68 -48√2) + (2 -√2) + 0.5 +12.So combining the constants:1 -1/16 -0.75 +68 +2 +0.5 +12.Let me convert all to sixteenths:1 =16/16-1/16= -1/16-0.75= -12/1668= 68*16/16=1088/162=32/160.5=8/1612=192/16Adding them up:16 -1 -12 +1088 +32 +8 +192 all over 16.As before, this totals 1323/16.Then the √2 terms: -48√2 -√2= -49√2.So indeed, the exact answer is 1323/16 -49√2.But maybe the problem expects further simplification? Let me check the original expression again to see if I interpreted all terms correctly.Original problem:(0.25)^0 - (1/16) - 0.75 +4(1 -√2)^4 + sqrt(6 -4√2) + ln sqrt(e) +2^{2 +log₂3}Yes, that's all. So unless there is a miscalculation somewhere, the answer should be 1323/16 -49√2. Let me check once more:Compute each term:1. 12. -1/163. -3/44. 68 -48√25. 2 -√26. 1/27. 12Sum constants: 1 -1/16 -3/4 +68 +2 +1/2 +12.1 -1/16 is 15/16.15/16 -3/4: 3/4 is 12/16, so 15/16 -12/16=3/16.3/16 +68= 68 +3/16=68.1875.68.1875 +2=70.1875.70.1875 +0.5=70.6875.70.6875 +12=82.6875=1323/16.Sum of √2 terms: -48√2 -√2= -49√2.So yes, the total is 1323/16 -49√2. Therefore, unless there's a simplification I missed, this is the answer. Maybe 49√2 is 49√2, and 1323/16 is as simplified as it gets. So I think that's the answer.Now, moving to problem (2):Given 14^a =6, 14^b=7, express log_{42}56 in terms of a and b.Okay, let's recall that log_{42}56 means the logarithm of 56 with base 42. We need to express this in terms of a and b, where a and b are exponents such that 14^a=6 and 14^b=7.First, let's note that 42 and 56 can be related to 14, 6, 7. Let me factorize them:42=6*7= (14^a)*(14^b)=14^{a+b}? Wait, but 6*7=42, but 14^a=6, 14^b=7, so 14^{a}*14^{b}=14^{a+b}=6*7=42. Therefore, 14^{a+b}=42. So log_{14}42 =a +b. Similarly, 56 can be expressed in terms of 14. Let me see:56=7*8=7*2^3. But 14=2*7. So 56=7*(2)^3= (14/2)*2^3=14*2^2=14*4. Wait, 14*4=56. So 56=14*4=14*2^2. Alternatively, 56=14*4.Alternatively, 56=14*4=14*(2^2). But since 14=2*7, maybe we can express 56 in terms of 14 and powers.But perhaps it's better to express everything in terms of 14, 6, and 7. Since 14^a=6, 14^b=7, perhaps we can express log_{42}56 using change of base formula.First, recall that log_{42}56 = (ln56)/(ln42). But since we have logarithms related to base 14, maybe we can use the change of base formula to base 14.log_{42}56 = (log_{14}56)/(log_{14}42). We already know log_{14}42= a +b, as 14^{a +b}=42.Now, compute log_{14}56. Let's see:56=14*4. So log_{14}56=log_{14}(14*4)=log_{14}14 +log_{14}4=1 +log_{14}4.But 4=2^2. So we need to express log_{14}4 in terms of a and b.But how?Since 14=2*7, and we know 14^b=7, which gives us log_{14}7=b. Similarly, 14^a=6. Let's see if we can express 2 in terms of 14, 6, and 7.Since 14=2*7, so 2=14/7. Therefore, 2=14^{1 -b} because 7=14^b. Wait:14^1=14=2*7=2*14^b. Therefore, 14^1=2*14^b → 2=14^{1 -b}.Therefore, 2=14^{1 -b}. Therefore, log_{14}2=1 -b.Therefore, log_{14}4=log_{14}(2^2)=2*log_{14}2=2*(1 -b)=2 -2b.Therefore, log_{14}56=1 +2 -2b=3 -2b.Hence, log_{42}56=(3 -2b)/(a +b).So the answer is (3 -2b)/(a +b).But let me check again:First, 56=14*4.log_{14}56=log_{14}(14*4)=log_{14}14 +log_{14}4=1 +log_{14}4.As 14=2*7, 4=2^2. So log_{14}4= log_{14}(2^2)=2 log_{14}2.But 2=14/7=14^{1} *14^{-b} because 7=14^b. Therefore, 2=14^{1 -b}, so log_{14}2=1 -b.Therefore, log_{14}4=2*(1 -b)=2 -2b.Therefore, log_{14}56=1 +2 -2b=3 -2b.log_{14}42= log_{14}(14^{a +b})=a +b.Therefore, log_{42}56= (3 -2b)/(a +b).Yes, that seems correct. Let me verify with another approach.Alternative approach: Express log_{42}56 using change of base to base 14.log_{42}56 = (log_{14}56)/(log_{14}42) = [log_{14}56]/[a +b].Express log_{14}56 in terms of a and b.56=14*4. As above, log_{14}56=1 +log_{14}4.We can also express 4 as 6^(something) or 7^(something), but since 6=14^a and 7=14^b, maybe there's another way. But 4=2^2, and 2=14/7=14^{1 -b}.Thus, 4=(14^{1 -b})^2=14^{2(1 -b)}=14^{2 -2b}. Therefore, log_{14}4=2 -2b.Hence, log_{14}56=1 +2 -2b=3 -2b. Therefore, same result.Therefore, the final expression is (3 -2b)/(a +b).Hence, the answer is (3 -2b)/(a +b).Final Answer(1) boxed{dfrac{1323}{16} - 49sqrt{2}} (2) boxed{dfrac{3 - 2b}{a + b}}</think>(1) Calculate the expression: [(0.25)^0 - left(frac{1}{16}right) - 0.75 + 4(1- sqrt{2})^4 + sqrt{6-4sqrt{2}} + ln sqrt{e} + 2^{2+log_{2}3}]Breaking down each term:1. ((0.25)^0 = 1)2. (-left(frac{1}{16}right) = -frac{1}{16})3. (-0.75 = -frac{3}{4})4. (4(1 - sqrt{2})^4): - ((1 - sqrt{2})^2 = 3 - 2sqrt{2}) - ((3 - 2sqrt{2})^2 = 17 - 12sqrt{2}) - (4(17 - 12sqrt{2}) = 68 - 48sqrt{2})5. (sqrt{6 - 4sqrt{2}}): - (sqrt{6 - 4sqrt{2}} = 2 - sqrt{2})6. (ln sqrt{e} = frac{1}{2})7. (2^{2+log_{2}3} = 4 cdot 3 = 12)Combining all terms:- Constants: (1 - frac{1}{16} - frac{3}{4} + 68 + 2 + frac{1}{2} + 12 = frac{1323}{16})- Terms with (sqrt{2}): (-48sqrt{2} - sqrt{2} = -49sqrt{2})Result:[boxed{frac{1323}{16} - 49sqrt{2}}](2) Given (14^a = 6) and (14^b = 7), express (log_{42}56) in terms of (a) and (b):- (42 = 14^{a+b})- (log_{42}56 = frac{log_{14}56}{log_{14}42})- (log_{14}42 = a + b)- (log_{14}56 = log_{14}(14 cdot 4) = 1 + log_{14}4)- (4 = 2^2) and (2 = 14^{1 - b})- (log_{14}4 = 2(1 - b))- (log_{14}56 = 1 + 2(1 - b) = 3 - 2b)Result:[boxed{frac{3 - 2b}{a + b}}]]

🔑:(1) The original expression equals 1 - (2^{-4})^{-frac{3}{4}} + sqrt{2} - 1 + 2 - sqrt{2} + frac{1}{2} + 2^2 times 2^{log_{2}3} = 1 - 8 + 1 + frac{1}{2} + 12 = boxed{frac{13}{2}}. (2) Since 14^a = 6 and 14^b = 7, we have log_{14}6 = a and log_{14}7 = b. Therefore, log_{42}56 = frac{log_{14}4 + log_{14}14}{log_{14}6 + log_{14}7} = frac{1 + 2log_{14}2}{a + b} = frac{1 + 2log_{14}frac{14}{7}}{a + b} = frac{3 - 2log_{14}7}{a + b} = boxed{frac{3 - 2b}{a + b}}.

🧠:[<think>Okay, let's tackle this problem step by step. The question is about rounding the number 0.05019 according to different requirements and figuring out which one is incorrect. The options are A, B, C, and D, each with different rounding instructions. Let me go through each option one by one.Starting with option A: It says 0.1 when rounded to the nearest 0.1. Hmm, rounding to the nearest tenth. The original number is 0.05019. Let me visualize the decimal places. The tenths place is 0, the hundredths is 0, thousandths is 5, ten-thousandths is 0, and hundred-thousandths is 1, 9. Wait, hold on. Let me write it out properly:0.05019Breaking it down:- 0 is in the tenths place (0.0)- 0 in the hundredths (0.00)- 5 in the thousandths (0.005)- 1 in the ten-thousandths (0.0001)- 9 in the hundred-thousandths (0.00009)But when rounding to the nearest 0.1, which is one decimal place, so we look at the tenths place. Wait, the number is 0.05019. If we need to round to the nearest 0.1, which is the first decimal digit. But here, the tenths digit is 0. The next digit is 5 (hundredths place). Wait, no, actually, tenths, hundredths, thousandths. Let me correct that:Wait, 0.05019:- The first digit after the decimal is 0 (tenths: 0.0)- Second digit is 5 (hundredths: 0.05)- Third digit is 0 (thousandths: 0.050)- Fourth digit is 1 (ten-thousandths: 0.0501)- Fifth digit is 9 (hundred-thousandths: 0.05019)So, rounding to the nearest 0.1 (tenths place), we look at the tenths digit, which is 0, and then check the next digit (hundredths place) to decide whether to round up or keep it the same. The hundredths digit is 5. According to rounding rules, if the digit after the place we're rounding to is 5 or more, we round up. So, 0.0 becomes 0.1 because the 5 in the hundredths place causes the tenths place to round up from 0 to 1. Therefore, option A says 0.1, which is correct. So A is correct. So the incorrect answer isn't A.Moving to option B: 0.10 when rounded to the nearest hundredth. So rounding to two decimal places. The original number is 0.05019. Let's look at the hundredths place, which is the second decimal digit. The second digit is 5. The next digit (thousandths place) is 0. When rounding to the nearest hundredth, we look at the thousandths digit. Since it's 0, which is less than 5, we don't round up the hundredths place. Therefore, the number would be 0.05 when rounded to the nearest hundredth. However, the option says 0.10. Wait, that's a big jump from 0.05 to 0.10. That seems incorrect. Wait, let me double-check.Wait, maybe there's confusion here. The original number is 0.05019. Let me write it as 0.05 (hundredths place). Wait, actually, 0.05019 is the same as 0.05 (to the hundredths place) plus 0.00019. So when rounding to the nearest hundredth (two decimal places), we look at the third decimal digit (thousandths place), which is 0. Since 0 is less than 5, we round down, keeping the hundredths place as 5. Therefore, rounding to two decimal places should give 0.05, not 0.10. But the option B says 0.10. That's way off. So that seems incorrect. So B is incorrect? But wait, hold on. Wait, 0.05019 is 0.05 when rounded to the nearest hundredth. But maybe the question is a trick question? Let me make sure.Alternatively, perhaps there's a miscalculation here. Let me parse the original number again: 0.05019. So:- Tenths: 0- Hundredths: 5- Thousandths: 0- Ten-thousandths: 1- Hundred-thousandths: 9Wait, so when rounding to the nearest hundredth, which is two decimal places, we look at the thousandths place to decide. The thousandths digit here is 0. Since 0 is less than 5, we round down, so the hundredths place stays 5, and the rest becomes zero. So 0.05019 rounded to the nearest hundredth is 0.05. But 0.05 is two decimal places. However, the option B says 0.10. That's clearly wrong. So B is incorrect. Therefore, option B is the incorrect one. But let me check the other options to make sure.Option C: 0.050 when rounded to the nearest thousandth. Thousandth is three decimal places. So the original number is 0.05019. Let's look at the third decimal place (thousandths place), which is 0. The next digit (ten-thousandths place) is 1. Since 1 is less than 5, we round down, keeping the thousandths place as 0. So rounding to three decimal places would be 0.050. But wait, normally when rounding to three decimal places, if the fourth is less than 5, we just truncate. So 0.05019 rounded to the nearest thousandth is 0.050. So option C is correct. But wait, 0.050 is the same as 0.05 if we consider trailing zeros. But in terms of three decimal places, it's 0.050. So C is correct.Option D: 0.0502 when rounded to the nearest 0.0001, which is four decimal places. The original number is 0.05019. Let's check. The fourth decimal place is 1, and the next digit is 9. Rounding to four decimal places: the fourth is 1, fifth is 9. Since 9 is 5 or more, we round up the fourth digit. So 1 becomes 2. Therefore, 0.0502 when rounded to the nearest 0.0001 (four decimal places). So D is correct. So D is 0.0502, which is accurate.Therefore, the incorrect rounding is option B, which says 0.10 instead of 0.05. So the answer is B.But just to make sure, let's check again:Original number: 0.05019.A: Nearest 0.1 (tenths). Tenths place is 0, next digit is 5. Since 5 is equal to 5, round up. 0.0 becomes 0.1. Correct.B: Nearest hundredth (two decimal places). The third decimal is 0 (0.05019). Since 0 < 5, we keep the second decimal as is. So 0.05, which is 0.05. But 0.10 is ten times larger. So definitely incorrect.C: Nearest thousandth (three decimal places). Fourth decimal is 1, which is less than 5. So 0.050. Correct.D: Nearest 0.0001 (four decimal places). Fifth decimal is 9, so round up the fourth. 0.0501 + 0.00009 becomes 0.0502. Correct.So yes, B is incorrect.Final Answerboxed{B}</think>To determine which rounding is incorrect for the number (0.05019), we analyze each option step by step:- Option A: Rounded to the nearest (0.1) (tenths place). The tenths place is (0) and the next digit is (5). Since (5 geq 5), we round up the tenths place to (1). Result: (0.1). This is correct.- Option B: Rounded to the nearest hundredth (two decimal places). The hundredths place is (5) and the next digit (thousandths place) is (0). Since (0 < 5), we keep the hundredths place as (5). Result: (0.05). However, the option states (0.10), which is incorrect.- Option C: Rounded to the nearest thousandth (three decimal places). The thousandths place is (0) and the next digit (ten-thousandths place) is (1). Since (1 < 5), we keep the thousandths place as (0). Result: (0.050). This is correct.- Option D: Rounded to the nearest (0.0001) (four decimal places). The ten-thousandths place is (1) and the next digit (hundred-thousandths place) is (9). Since (9 geq 5), we round up the ten-thousandths place from (1) to (2). Result: (0.0502). This is correct.The incorrect rounding is found in option B. (boxed{B}),<think>Okay, so I need to figure out which of these rounding options for the number 0.05019 is incorrect. Let's take each option one by one and check if the rounding was done properly. Let me remember the rules for rounding. If the digit after the place we're rounding to is 5 or more, we round up; otherwise, we keep it the same. But I should also be careful about trailing zeros and whether they're required or not, especially in the case of decimal places.Starting with option A: 0.1 when rounded to the nearest 0.1. Let's see. The number is 0.05019. Rounding to the nearest tenth, which is the first decimal place. The first decimal place is 0, and the next digit is 5. Since we're rounding to the nearest 0.1, the 0 is in the tenths place. The digit after that is 5 (which is in the hundredths place). Wait, actually, 0.05019 is 0.0 (tenths place) 5 (hundredths) 0 (thousandths) 1 (ten-thousandths) 9 (hundred-thousandths). So when rounding to the nearest tenth (0.1), we look at the hundredths place to decide. If the hundredths place is 5 or more, we round up the tenths place. Here, the tenths place is 0, and the hundredths is 5. So 0.0 becomes 0.1. Hmm, so A says 0.1. That seems correct. So A is correctly rounded. So maybe A is okay.Moving to option B: 0.10 when rounded to the nearest hundredth. The original number is 0.05019. Rounding to the nearest hundredth (two decimal places). The hundredths place is 5, and the next digit is 0. So the third decimal is 0. Wait, let me check. The number is 0.05019. So the tenths place is 0, hundredths is 5, thousandths is 0, ten-thousandths is 1, and hundred-thousandths is 9. So if we round to the nearest hundredth (two decimal places), we look at the thousandths place. The thousandths digit here is 0, which is less than 5, so we keep the hundredths place as it is. So 0.05 rounded to the hundredth place. But 0.05 is already two decimal places. But the answer here is 0.10. Wait, that's different. Wait, 0.05019 rounded to the nearest hundredth. Let's think again.Wait, tenths is the first decimal (0), hundredths is the second (5), thousandths is the third (0), ten-thousandths is the fourth (1), hundred-thousandths is the fifth (9). So when rounding to the nearest hundredth, we need to look at the thousandths place. If it's 5 or more, round up the hundredths place. But here, the thousandths place is 0, so we leave the hundredths as is. So 0.05. But the answer given is 0.10. That's a big difference. Wait, that seems incorrect. But hold on, maybe I made a mistake here. Let me check again.Original number: 0.05019. Rounding to the nearest hundredth. So, second decimal place. The second decimal is 5. The next digit (thousandths place) is 0. Since 0 is less than 5, we don't round up the 5. So it remains 0.05. But the answer given is 0.10, which is ten hundredths. That's 0.10 versus 0.05. So that's a big jump. So that must be incorrect. So B is incorrect. Wait, but let's check the other options to make sure.Option C: 0.050 when rounded to the nearest thousandth. So third decimal place. Original number is 0.05019. The thousandths place is 0. The next digit (ten-thousandths) is 1. Since 1 is less than 5, we keep the thousandths place as 0. So 0.050. But the number is 0.05019. So rounding to the nearest thousandth: 0.050. But wait, 0.050 is the same as 0.05, but written to three decimal places. So if we write it to the thousandth, it's 0.050. So C is correct. The trailing zero is necessary to indicate the precision. So C is okay.Option D: 0.0502 when rounded to the nearest 0.0001. The nearest ten-thousandth. The original number is 0.05019. Ten-thousandth place is the fourth decimal. The fourth decimal is 1, followed by 9 in the hundred-thousandth place. So when rounding to the nearest 0.0001, we look at the fifth decimal. Since it's 9, which is 5 or more, we round up the fourth decimal. So 1 becomes 2. Therefore, 0.05019 rounded to the nearest 0.0001 is 0.0502. So D is correct.So reviewing all options, A is correct (0.1), B is incorrect (should be 0.05 instead of 0.10), C is correct (0.050), D is correct (0.0502). Therefore, the incorrect one is B. Wait, but let me double-check option A because sometimes leading zeros can be confusing. Rounding 0.05019 to the nearest 0.1. So 0.05019 is less than 0.05, which is halfway between 0.0 and 0.1. Wait, actually, 0.05 is halfway between 0.0 and 0.1. But 0.05019 is slightly more than 0.05. Wait, hold on. Wait, 0.05019 is equal to 0.05019. So 0.05 is 0.05, and 0.05019 is 0.05019. So if we're rounding to the nearest tenth, which is 0.1. The tenths place is 0, the hundredths is 5. So the rule is if the digit after the place we're rounding to is 5 or more, we round up. So here, tenths place is 0, the next digit is 5, so we round up the tenths place to 1. Therefore, 0.05019 rounded to the nearest tenth is 0.1. So A is correct.But wait, wait a second. Is 0.05019 actually closer to 0.0 or 0.1? Let's convert to fractions. 0.05019 is 5019/100000. 0.0 is 0/10, and 0.1 is 1/10 = 10000/100000. So 5019 is closer to 0 than to 10000. Wait, that can't be right. Wait, 5019 is less than 5000? No, 5019 is more than 5000. Wait, 0.05000 is 0.05, which is halfway between 0.0 and 0.1. So 0.05019 is 0.05 plus 0.00019, which is slightly more than halfway. So when rounding to the nearest tenth, 0.05019 is just over the halfway point, so it rounds up to 0.1. So yes, A is correct. But let me confirm with the standard rounding rules. If the digit after the place we're rounding to is 5 or more, we round up. So tenths place is 0, next digit is 5 (hundredths place). So according to the rules, regardless of the following digits, if it's 5 or more, we round up. So even though the rest is 019, since the hundredths is 5, we round up the tenths. Therefore, 0.1. So A is correct.Therefore, the incorrect one is B, which should be 0.05 instead of 0.10. So the answer is B.Final Answerboxed{B}</think>To determine which rounding is incorrect for the number (0.05019), we analyze each option step by step:- Option A: (0.1) (rounded to the nearest (0.1)) - The tenths place is (0), and the next digit (hundredths place) is (5). Since (5 geq 5), we round up the tenths place to (1). Thus, (0.05019) rounded to the nearest (0.1) is (0.1). This is correct.- Option B: (0.10) (rounded to the nearest hundredth) - The hundredths place is (5), and the next digit (thousandths place) is (0). Since (0 < 5), we keep the hundredths place as (5). Therefore, (0.05019) rounded to the nearest hundredth is (0.05), not (0.10). This is incorrect.- Option C: (0.050) (rounded to the nearest thousandth) - The thousandths place is (0), and the next digit (ten-thousandths place) is (1). Since (1 < 5), we keep the thousandths place as (0). Thus, (0.05019) rounded to the nearest thousandth is (0.050). This is correct.- Option D: (0.0502) (rounded to the nearest (0.0001)) - The ten-thousandths place is (1), and the next digit (hundred-thousandths place) is (9). Since (9 geq 5), we round up the ten-thousandths place from (1) to (2). Therefore, (0.05019) rounded to the nearest (0.0001) is (0.0502). This is correct.The incorrect rounding is found in option B.(boxed{B})]

🔑:To solve this problem, we will round 0.05019 according to the different requirements specified in each option and determine which one is incorrect.Option A: Rounded to the nearest 0.1- Rounding 0.05019 to the nearest tenth (0.1) gives us 0.1 because the digit in the hundredths place (5) is 5 or more, which rounds up.- Therefore, 0.05019 approx 0.1.Option B: Rounded to the nearest hundredth- Rounding 0.05019 to the nearest hundredth (0.01) should give us 0.05 because the digit in the thousandths place (0) is less than 5, which means we do not round up.- Therefore, the correct rounding to the nearest hundredth should be 0.05019 approx 0.05, making option B incorrect since it suggests 0.10.Option C: Rounded to the nearest thousandth- Rounding 0.05019 to the nearest thousandth (0.001) gives us 0.050 because the digit in the ten-thousandths place (1) is less than 5, which means we do not round up.- Therefore, 0.05019 approx 0.050.Option D: Rounded to the nearest 0.0001- Rounding 0.05019 to the nearest ten-thousandth (0.0001) gives us 0.0502 because the digit in the hundred-thousandths place (9) is 5 or more, which rounds up.- Therefore, 0.05019 approx 0.0502.Given these explanations, the incorrect option is:boxed{B}

🧠:[<think>Okay, so I have this geometry problem here. Let me read it again to make sure I understand. The problem says: Given that the diameter of circle O is 10, so the radius must be 5, right? Because diameter is twice the radius. Then, there's a point P that moves along line l, and the shortest distance from point P to point O is 5. They're asking about the positional relationship between line l and circle O. The options are separate, tangent, intersect, or cannot be determined.Hmm. Let's visualize this. We have a circle with center O and radius 5. Then there's a line l, and point P is moving along this line. The closest that P ever gets to O is 5 units. So the minimal distance from line l to point O is 5. Wait, the distance from a point to a line is the shortest distance, which is the perpendicular distance. So if the shortest distance from O to line l is 5, which is equal to the radius of the circle. In that case, the line l would be tangent to the circle, right? Because the tangent line to a circle is the line that touches the circle at exactly one point, and the distance from the center to the tangent line is equal to the radius. If the distance was more than the radius, the line would be separate, and if it was less, the line would intersect the circle at two points. But let me double-check. The diameter is 10, so radius is 5. The shortest distance from O to line l is 5. Therefore, the line is exactly at a distance equal to the radius. That should mean the line is tangent. So the answer should be B: Tangent. Wait, but the problem says point P is moving on line l, and the shortest distance from P to O is 5. So is that the same as the distance from the line l to O? Because the shortest distance from any point on l to O would be the perpendicular distance from O to l. So if that's 5, then yes, the line is tangent. So the positional relationship is tangent.But let me think if there's any other possibility. If line l were intersecting the circle, then the closest distance from O to l would be less than 5, and the farthest points on l from O would be more than 5. But in this problem, the shortest distance is exactly 5, which would occur only if the line is tangent. If the line were separate, the shortest distance would be greater than 5. So yeah, tangent is correct.So the answer is B.Final Answerboxed{B}</think>Given that the diameter of circle ( O ) is 10, the radius of the circle is 5. Point ( P ) is a moving point on line ( l ), and the shortest distance from point ( P ) to point ( O ) is 5. The shortest distance from a point to a line is the perpendicular distance. Since the shortest distance from ( O ) to line ( l ) is 5, which is equal to the radius of the circle, line ( l ) must be tangent to the circle. A tangent line to a circle touches the circle at exactly one point and the distance from the center to the tangent line is equal to the radius. Thus, the positional relationship between line ( l ) and circle ( O ) is tangent.[boxed{B}],<think>Okay, let's see. The problem is about the positional relationship between a line l and a circle O. The diameter of circle O is 10, so the radius must be half of that, which is 5 units. The point P is moving along line l, and the shortest distance from P to point O (the center of the circle) is 5. We need to figure out if line l is separate from the circle, tangent to it, intersecting it, or if we can't determine that.First, let me recall the different positional relationships between a line and a circle. If a line is separate from the circle, that means the line doesn't touch the circle at all. The distance from the center of the circle to the line is greater than the radius. If the line is tangent to the circle, then it touches the circle at exactly one point, and the distance from the center to the line is equal to the radius. If the line intersects the circle, then it passes through the circle, crossing it at two points, and the distance from the center to the line is less than the radius.In this problem, the radius is 5, and the shortest distance from point O to point P (which is moving along line l) is also 5. Wait, point P is on line l, so the shortest distance from O to line l is 5. Because the shortest distance from a point to a line is the perpendicular distance. So if the shortest distance from O to l is equal to the radius of the circle, then the line l must be tangent to the circle O.But let me verify that again. The problem states that the shortest distance from P to O is 5. Since P is moving along line l, the minimum distance from O to any point on line l is 5. That minimum distance occurs along the perpendicular from O to line l. Therefore, the distance from O to line l is 5. Since the radius of the circle is also 5, the line l is exactly at a distance equal to the radius. Therefore, line l is tangent to circle O.So the answer should be B: Tangent. Let me check if I'm missing anything here. The diameter is 10, radius 5. The minimal distance from O to l is 5, so the line touches the circle at exactly one point. Therefore, they are tangent. The other options: if the distance was more than 5, they would be separate; if less, intersecting. Since it's exactly 5, tangent. So yes, B is the correct answer.Final Answerboxed{B}</think>The diameter of circle ( O ) is 10, so the radius is ( frac{10}{2} = 5 ). The shortest distance from point ( O ) to line ( l ) is given as 5, which is the perpendicular distance from ( O ) to ( l ). Since this distance is equal to the radius of the circle, line ( l ) must be tangent to circle ( O ). Thus, the positional relationship between line ( l ) and circle ( O ) is tangent.[boxed{B}]]

🔑:Since the diameter of circle O is 10, the radius r of circle O is 5, Since the distance d from the center O to line l is 5, we have r=d, Therefore, the positional relationship between line l and circle O is tangent, Hence, the correct choice is boxed{text{B}}.

🧠:[<think>Okay, so I need to find the minimum value of the function ( f(x) = log_2sqrt{x} cdot log_{sqrt{2}}(2x) ). Hmm, let me start by recalling some logarithm properties to simplify this expression. First, let me rewrite the square roots as exponents. The square root of x is the same as ( x^{1/2} ), right? So, ( log_2sqrt{x} = log_2 x^{1/2} ). Using the logarithm power rule, which says ( log_b a^n = n log_b a ), that should simplify to ( frac{1}{2} log_2 x ). Similarly, the other logarithm has a base of ( sqrt{2} ). Let me handle that. The base ( sqrt{2} ) can be written as ( 2^{1/2} ), so ( log_{sqrt{2}}(2x) ). Maybe I can use the change of base formula here? The change of base formula is ( log_b a = frac{log_c a}{log_c b} ), so if I choose base 2, that might help. Applying that, ( log_{sqrt{2}}(2x) = frac{log_2 (2x)}{log_2 2^{1/2}} ). The denominator simplifies because ( log_2 2^{1/2} = frac{1}{2} log_2 2 = frac{1}{2} times 1 = frac{1}{2} ). So the denominator is ( frac{1}{2} ). Therefore, the entire expression becomes ( frac{log_2 (2x)}{1/2} = 2 log_2 (2x) ). Now, let's simplify ( log_2 (2x) ). Using logarithm properties, ( log_2 (2x) = log_2 2 + log_2 x = 1 + log_2 x ). So, putting that back into the previous expression, ( 2 log_2 (2x) = 2(1 + log_2 x) = 2 + 2 log_2 x ). So now, the original function ( f(x) = log_2sqrt{x} cdot log_{sqrt{2}}(2x) ) simplifies to ( frac{1}{2} log_2 x times (2 + 2 log_2 x) ). Let me write that out:( f(x) = frac{1}{2} log_2 x times 2(1 + log_2 x) )Multiplying the constants: ( frac{1}{2} times 2 = 1 ), so that leaves:( f(x) = log_2 x times (1 + log_2 x) )Hmm, this looks like a quadratic in terms of ( log_2 x ). Let me let ( y = log_2 x ). Then the function becomes:( f(y) = y(1 + y) = y^2 + y )Now, to find the minimum value of this quadratic function. Since the coefficient of ( y^2 ) is positive (1), the parabola opens upwards, and the minimum occurs at the vertex. The vertex of a parabola given by ( f(y) = ay^2 + by + c ) is at ( y = -frac{b}{2a} ). Here, ( a = 1 ), ( b = 1 ), so the vertex is at:( y = -frac{1}{2 times 1} = -frac{1}{2} )So the minimum value of ( f(y) ) is when ( y = -frac{1}{2} ). Plugging this back into the function:( f(-frac{1}{2}) = (-frac{1}{2})^2 + (-frac{1}{2}) = frac{1}{4} - frac{1}{2} = -frac{1}{4} )Therefore, the minimum value of the original function ( f(x) ) is ( -frac{1}{4} ). But wait, I need to check if this value of ( y = -frac{1}{2} ) corresponds to a valid ( x ). Remember that ( y = log_2 x ), so solving for ( x ):( log_2 x = -frac{1}{2} implies x = 2^{-1/2} = frac{1}{sqrt{2}} )Since ( x ) must be positive, this is valid. Therefore, the minimum value of ( f(x) ) is indeed ( -frac{1}{4} ).Wait a second, let me double-check my steps to make sure I didn't make a mistake. 1. Converted ( log_2 sqrt{x} ) to ( frac{1}{2} log_2 x ): Correct.2. Converted ( log_{sqrt{2}}(2x) ) by changing base to 2: Yes, that's ( frac{log_2 (2x)}{log_2 sqrt{2}} = frac{log_2 (2x)}{1/2} = 2 log_2 (2x) ). Then expanded ( log_2 (2x) ) to ( 1 + log_2 x ): Correct. So multiplied by 2 gives ( 2 + 2 log_2 x ). Then multiplied by ( frac{1}{2} log_2 x ): Yes, which gives ( frac{1}{2} log_2 x times 2 (1 + log_2 x) ), the 2 and 1/2 cancel, leaving ( log_2 x (1 + log_2 x) ). That's right.Then substituting ( y = log_2 x ), so quadratic in y, vertex at -b/(2a) = -1/2, plug back in to get minimum value of -1/4. Then checked x is positive: yes. So all steps seem correct. Therefore, the minimum value is -1/4. Wait, just to confirm, maybe I can compute the original function at x = 1/sqrt(2) to check:First, compute ( log_2 sqrt{x} ):( sqrt{x} = (1/sqrt{2})^{1/2} = 2^{-1/4} ), so ( log_2 2^{-1/4} = -1/4 ).Then compute ( log_{sqrt{2}}(2x) ):2x = 2*(1/sqrt{2}) = 2^{1} * 2^{-1/2} = 2^{1 - 1/2} = 2^{1/2} = sqrt(2). So ( log_{sqrt{2}}(sqrt(2)) = 1 ), since sqrt(2) is the base. Therefore, f(x) = (-1/4) * 1 = -1/4. Which matches. So yes, that's correct. Therefore, the minimum value is -1/4.Final AnswerThe minimum value of the function is boxed{-dfrac{1}{4}}.</think>To find the minimum value of the function ( f(x) = log_2sqrt{x} cdot log_{sqrt{2}}(2x) ), we start by simplifying the logarithmic expressions.First, rewrite ( log_2sqrt{x} ) using the power rule of logarithms:[log_2sqrt{x} = log_2 x^{1/2} = frac{1}{2} log_2 x]Next, rewrite ( log_{sqrt{2}}(2x) ) using the change of base formula. The base ( sqrt{2} ) can be expressed as ( 2^{1/2} ):[log_{sqrt{2}}(2x) = frac{log_2 (2x)}{log_2 2^{1/2}} = frac{log_2 (2x)}{1/2} = 2 log_2 (2x)]Simplify ( log_2 (2x) ):[log_2 (2x) = log_2 2 + log_2 x = 1 + log_2 x]Thus, we have:[log_{sqrt{2}}(2x) = 2(1 + log_2 x) = 2 + 2 log_2 x]Substituting these back into the original function:[f(x) = left( frac{1}{2} log_2 x right) left( 2 + 2 log_2 x right) = log_2 x left( 1 + log_2 x right)]Let ( y = log_2 x ). The function becomes:[f(y) = y(1 + y) = y^2 + y]This is a quadratic function in ( y ). The minimum value of a quadratic function ( ay^2 + by + c ) occurs at ( y = -frac{b}{2a} ). Here, ( a = 1 ) and ( b = 1 ), so:[y = -frac{1}{2 times 1} = -frac{1}{2}]Substituting ( y = -frac{1}{2} ) back into the quadratic function:[fleft( -frac{1}{2} right) = left( -frac{1}{2} right)^2 + left( -frac{1}{2} right) = frac{1}{4} - frac{1}{2} = -frac{1}{4}]To verify, we convert ( y = -frac{1}{2} ) back to ( x ):[log_2 x = -frac{1}{2} implies x = 2^{-1/2} = frac{1}{sqrt{2}}]Checking the original function at ( x = frac{1}{sqrt{2}} ):[log_2 sqrt{frac{1}{sqrt{2}}} = log_2 2^{-1/4} = -frac{1}{4}][log_{sqrt{2}}(2 cdot frac{1}{sqrt{2}}) = log_{sqrt{2}}(sqrt{2}) = 1][fleft( frac{1}{sqrt{2}} right) = -frac{1}{4} cdot 1 = -frac{1}{4}]Thus, the minimum value of the function is boxed{-dfrac{1}{4}}.,<think>Alright, so I need to find the minimum value of the function ( f(x) = log_2sqrt{x} cdot log_{sqrt{2}}(2x) ). Hmm, okay. Let me start by understanding the components of this function. Both parts are logarithms with different bases, so maybe converting them to the same base will help simplify things. First, let's recall some logarithm properties. The logarithm of a square root can be rewritten using the exponent rule: ( log_b sqrt{a} = frac{1}{2} log_b a ). Similarly, changing the base of a logarithm can be done using the change-of-base formula: ( log_b a = frac{log_c a}{log_c b} ). Maybe I can use these properties to rewrite the function in a more manageable form.Starting with the first term: ( log_2 sqrt{x} ). Applying the exponent rule, this becomes ( frac{1}{2} log_2 x ). So that part is straightforward.Now the second term: ( log_{sqrt{2}}(2x) ). Let me try to express this in terms of base 2 logarithms so that both terms have the same base. Using the change-of-base formula, ( log_{sqrt{2}}(2x) = frac{log_2 (2x)}{log_2 sqrt{2}} ). Calculating the denominator: ( log_2 sqrt{2} ). Again, using the exponent rule, ( sqrt{2} = 2^{1/2} ), so ( log_2 2^{1/2} = frac{1}{2} ). So the denominator is 1/2. Therefore, ( log_{sqrt{2}}(2x) = frac{log_2 (2x)}{1/2} = 2 log_2 (2x) ).So, substituting back into the original function, we have:( f(x) = left( frac{1}{2} log_2 x right) cdot left( 2 log_2 (2x) right) )Multiplying the constants: ( frac{1}{2} times 2 = 1 ). Therefore, the function simplifies to:( f(x) = log_2 x cdot log_2 (2x) )That's simpler. Let's see if we can expand the product of the two logarithms. Let me denote ( y = log_2 x ). Then, ( log_2 (2x) = log_2 2 + log_2 x = 1 + y ). Therefore, the function becomes:( f(x) = y cdot (1 + y) = y^2 + y )So now, the problem reduces to finding the minimum value of the quadratic function ( y^2 + y ). But we need to remember that ( y = log_2 x ), so the domain of the original function is ( x > 0 ), since logarithms are only defined for positive numbers. Therefore, ( y ) can be any real number because ( log_2 x ) can take any real value as ( x ) varies over the positive real numbers.However, let me confirm that. For ( log_2 x ), ( x ) must be positive. For ( log_{sqrt{2}}(2x) ), the argument ( 2x ) must also be positive, which again requires ( x > 0 ). So, yes, the domain is ( x > 0 ), and ( y ) can be any real number from ( -infty ) to ( +infty ).Therefore, the function ( f(x) ) can be considered as a quadratic function in terms of ( y ): ( f(y) = y^2 + y ). To find the minimum value of a quadratic function ( ay^2 + by + c ), the vertex occurs at ( y = -frac{b}{2a} ). In this case, ( a = 1 ), ( b = 1 ), so the vertex is at ( y = -frac{1}{2} ).Since the coefficient of ( y^2 ) is positive (1), the parabola opens upwards, so the vertex is the minimum point. Therefore, the minimum value of ( f(y) ) is:( fleft( -frac{1}{2} right) = left( -frac{1}{2} right)^2 + left( -frac{1}{2} right) = frac{1}{4} - frac{1}{2} = -frac{1}{4} ).Therefore, the minimum value of the function is ( -frac{1}{4} ). But wait, we need to check if this value is attainable given the original variable ( x ). That is, there must exist an ( x > 0 ) such that ( log_2 x = -frac{1}{2} ). Solving for ( x ):( log_2 x = -frac{1}{2} Rightarrow x = 2^{-1/2} = frac{1}{sqrt{2}} ).Since ( x = frac{1}{sqrt{2}} > 0 ), this is valid. Therefore, the minimum value of ( f(x) ) is indeed ( -frac{1}{4} ).But just to be thorough, let me verify this by substituting ( x = frac{1}{sqrt{2}} ) back into the original function.First term: ( log_2 sqrt{x} = log_2 sqrt{frac{1}{sqrt{2}}} = log_2 left( frac{1}{2^{1/4}} right) = log_2 2^{-1/4} = -frac{1}{4} ).Second term: ( log_{sqrt{2}}(2x) = log_{sqrt{2}} left( 2 cdot frac{1}{sqrt{2}} right) = log_{sqrt{2}} left( frac{2}{sqrt{2}} right) = log_{sqrt{2}} ( sqrt{2} cdot sqrt{2} / sqrt{2} ) = Hmm, maybe a different approach.Let's compute ( 2x = 2 times frac{1}{sqrt{2}} = sqrt{2} ). Therefore, ( log_{sqrt{2}} (sqrt{2}) ). Since ( (sqrt{2})^1 = sqrt{2} ), this logarithm equals 1. Therefore, the second term is 1.But then the original function would be ( (-1/4) times 1 = -1/4 ), which matches our previous result. So the computation checks out.Alternatively, let's compute the second term again using the conversion we did earlier. The second term is ( log_{sqrt{2}}(2x) ). When ( x = 1/sqrt{2} ), ( 2x = sqrt{2} ), so:( log_{sqrt{2}} sqrt{2} = frac{log_2 sqrt{2}}{log_2 sqrt{2}} = frac{1/2}{1/2} = 1 ). So that also gives 1. Then multiplying by the first term, which is -1/4, gives -1/4. Yep, that's correct.So, all steps check out. Therefore, the minimum value is indeed -1/4.But just to make sure there's no mistake in the algebra, let me recap:Original function:( f(x) = log_2 sqrt{x} cdot log_{sqrt{2}}(2x) )Convert each logarithm to base 2:First term: ( log_2 sqrt{x} = frac{1}{2} log_2 x )Second term: ( log_{sqrt{2}}(2x) = frac{log_2 (2x)}{log_2 sqrt{2}} = frac{log_2 (2x)}{1/2} = 2 log_2 (2x) )Multiply them:( frac{1}{2} log_2 x cdot 2 log_2 (2x) = log_2 x cdot log_2 (2x) )Express ( log_2 (2x) ) as ( log_2 2 + log_2 x = 1 + log_2 x ). Let ( y = log_2 x ), so:( f(x) = y (1 + y) = y^2 + y )Quadratic in y, which has its vertex at ( y = -b/(2a) = -1/2 ). Plugging back in gives:( (-1/2)^2 + (-1/2) = 1/4 - 1/2 = -1/4 )Therefore, the minimum value is -1/4 when ( y = -1/2 ), which corresponds to ( x = 2^{-1/2} = 1/sqrt{2} ). All steps make sense and checks out when substituting back into the original equation.So, I think that's thorough. I considered the properties of logarithms, converted to the same base, simplified the expression, transformed it into a quadratic, found the minimum, checked if that minimum is attainable within the domain, and verified by substitution. Seems solid.Final AnswerThe minimum value of the function is boxed{-dfrac{1}{4}}.</think>To find the minimum value of the function ( f(x) = log_2sqrt{x} cdot log_{sqrt{2}}(2x) ), we start by converting the logarithms to the same base using logarithm properties.First, rewrite ( log_2sqrt{x} ):[log_2sqrt{x} = frac{1}{2} log_2 x]Next, rewrite ( log_{sqrt{2}}(2x) ) using the change-of-base formula:[log_{sqrt{2}}(2x) = frac{log_2 (2x)}{log_2 sqrt{2}} = frac{log_2 (2x)}{1/2} = 2 log_2 (2x)]Substitute these back into the original function:[f(x) = left( frac{1}{2} log_2 x right) cdot left( 2 log_2 (2x) right) = log_2 x cdot log_2 (2x)]Let ( y = log_2 x ). Then, ( log_2 (2x) = log_2 2 + log_2 x = 1 + y ). Thus, the function becomes:[f(x) = y cdot (1 + y) = y^2 + y]This is a quadratic function in ( y ). The minimum value of ( y^2 + y ) occurs at the vertex of the parabola, which is at ( y = -frac{1}{2} ). Substituting ( y = -frac{1}{2} ) into the quadratic function:[fleft( -frac{1}{2} right) = left( -frac{1}{2} right)^2 + left( -frac{1}{2} right) = frac{1}{4} - frac{1}{2} = -frac{1}{4}]We check if this value is attainable by solving for ( x ) when ( log_2 x = -frac{1}{2} ):[x = 2^{-1/2} = frac{1}{sqrt{2}}]Substituting ( x = frac{1}{sqrt{2}} ) back into the original function confirms the calculation is correct. Therefore, the minimum value of the function is (boxed{-dfrac{1}{4}}).]

🔑:First, let's simplify the function using the properties of logarithms:begin{align*}f(x) &= log_2sqrt{x} cdot log_{sqrt{2}}(2x) &= log_2 x^{1/2} cdot frac{log_2 (2x)}{log_2 sqrt{2}} &= frac{1}{2}log_2 x cdot frac{log_2 2 + log_2 x}{frac{1}{2}log_2 2} &= frac{1}{2}log_2 x cdot frac{1 + log_2 x}{frac{1}{2}} &= log_2 x cdot (1 + log_2 x) &= log_2 x + (log_2 x)^2.end{align*}Now, let y = log_2 x to further simplify the problem. Then, f(y) = y + y^2. To find the minimum value of f(y), we can take the derivative and set it equal to zero:begin{align*}f'(y) &= 1 + 2y = 0 Rightarrow y &= -frac{1}{2}.end{align*}Since the coefficient of the squared term is positive, this critical point yields the minimum value of the function. Thus, the minimum value of f(y) isf(-frac{1}{2}) = -frac{1}{2} + (-frac{1}{2})^2 = -frac{1}{2} + frac{1}{4} = boxed{-frac{1}{4}}.

🧠:[<think>Okay, let me try to figure out how to solve this problem. So, the question is about Xiaoqiang's dad setting a 4-digit password on his phone, where each digit is between 0 and 9. The condition is that the sum of the four digits must be 20. The question is asking for the maximum number of times Xiaoqiang would need to try to unlock the phone, which basically means we need to find the total number of possible 4-digit combinations that add up to 20. Right?First, let me make sure I understand the problem correctly. We need to count all 4-digit numbers (from 0000 to 9999) where each digit is between 0 and 9, inclusive, and the sum of the digits is exactly 20. The answer will be the number of such combinations, as that's the maximum number of attempts needed in the worst-case scenario.Hmm, so this seems like a problem of finding the number of non-negative integer solutions to the equation:a + b + c + d = 20where each of a, b, c, d is between 0 and 9, inclusive. Wait, but hold on, digits can be from 0 to 9, so each variable a, b, c, d must satisfy 0 ≤ a, b, c, d ≤ 9. However, if we just use the stars and bars method to find the number of solutions without considering the upper limit, it would give us the number of solutions where each digit can be any non-negative integer, but here they are restricted to 0-9. So we need to adjust for that.But before that, let me recall the stars and bars formula. The number of non-negative integer solutions to a + b + c + d = n is C(n + 3, 3), where C is the combination function. However, this is when each variable can be 0 or more. But in our case, each digit is at most 9. So we need to subtract the solutions where any digit is greater than 9.This is a classic inclusion-exclusion problem. Let me remember the formula. The number of solutions where each variable is at most 9 is equal to the total number of solutions without restrictions minus the number of solutions where at least one variable is 10 or more, plus the number of solutions where at least two variables are 10 or more, and so on.So, let's define the equation a + b + c + d = 20, with 0 ≤ a, b, c, d ≤ 9.First, compute the total number of non-negative integer solutions without restrictions: C(20 + 3, 3) = C(23, 3). Let me compute that. C(23,3) = 23! / (3! * 20!) = (23*22*21)/(3*2*1) = (23*22*21)/6. Let me compute that: 23*22=506, 506*21=10626, divided by 6: 10626/6=1771. So 1771 solutions without restrictions.But now, we need to subtract the solutions where any digit is greater than 9. Let's denote each variable as x_i. For each variable x_i, if x_i ≥10, we can substitute y_i = x_i -10, so y_i ≥0. Then the equation becomes y_i + 10 + x_j + x_k + x_l =20, where j,k,l are the other variables. So this simplifies to y_i + x_j + x_k + x_l =10. The number of non-negative solutions for this is C(10 +3,3)=C(13,3)=286. Since there are 4 variables (a,b,c,d), each could potentially exceed 9, so we have 4*286=1144 solutions where at least one digit is 10 or more.But wait, inclusion-exclusion says we need to subtract the cases where one variable is over, then add back the cases where two variables are over, subtract the cases where three variables are over, and so on.So first step: Total solutions: 1771Subtract the cases where any one variable is ≥10: 4*286=1144. So 1771 - 1144 = 627.But now, we need to add back the cases where two variables are ≥10. Because when we subtracted the cases where one variable is ≥10, we subtracted too much. For two variables, say x_i and x_j, both ≥10. Then substituting y_i =x_i -10 and y_j =x_j -10, the equation becomes y_i + y_j + 10 +10 + x_k + x_l =20. So y_i + y_j + x_k + x_l =0. The number of solutions is C(0 +3,3)=1. Wait, but y_i and y_j are ≥0, so the equation y_i + y_j + x_k + x_l =0 implies that all variables y_i, y_j, x_k, x_l are 0. So the only solution is y_i=0, y_j=0, x_k=0, x_l=0, which corresponds to x_i=10, x_j=10, x_k=0, x_l=0. But wait, but the original equation is a + b + c + d =20. If x_i and x_j are both 10, then 10+10+0+0=20. But in this case, x_i and x_j can be any pair of variables. How many such solutions are there? For two variables being 10, the number of solutions is C(4,2)=6. Each such solution corresponds to choosing two variables to set to 10 and the remaining two to 0. However, in the inclusion-exclusion principle, when we subtract the overcounts, we have to consider that each pair was subtracted twice. But wait, actually, let's go through the inclusion-exclusion steps properly.So when we subtract the 4*286, which is subtracting all cases where any one variable is ≥10, but when two variables are both ≥10, those cases have been subtracted twice. So we need to add them back once.But the number of cases where two variables are ≥10 is C(4,2)* number of solutions. Wait, let's compute the number of solutions where two specific variables are ≥10. Let's say variables a and b are ≥10. Then set a =10 + y1, b=10 + y2, where y1,y2 ≥0. Then the equation becomes y1 + y2 + c + d = 20 -10 -10=0. The number of non-negative solutions is C(0 +3,3)=1. So for each pair of variables, there's only 1 solution where both are exactly 10 and the other two are 0. Wait, so if we set two variables to 10, then the remaining two have to be 0. So for each pair of variables, there's exactly one solution. Therefore, the number of solutions where two variables are ≥10 is C(4,2)*1=6.But wait, if the equation after substitution is y1 + y2 + c + d =0, the only solution is y1=y2=c=d=0. So the original variables a=10, b=10, c=0, d=0. Therefore, each pair of variables can only be set to 10 once, and the rest are 0. Hence, 6 solutions.Therefore, in inclusion-exclusion, we need to add back these 6 solutions.So now, total is 1771 - 1144 + 6 = 1771 - 1144 = 627; 627 +6=633.Next, we need to check if there are any cases where three variables are ≥10. Let's see. Suppose three variables are ≥10. Let's set a=10+y1, b=10+y2, c=10+y3, then the equation becomes y1 + y2 + y3 + d = 20 -10*3= -10. But since y1,y2,y3,d are all ≥0, the left side is non-negative and the right side is negative. So there are no solutions in this case. Similarly, if all four variables are ≥10, the sum would be at least 40, which is way more than 20, so no solutions. Therefore, the inclusion-exclusion stops here.Therefore, the total number of solutions is 1771 - 1144 +6=633.Wait, so according to this calculation, the number of valid 4-digit passwords where the digits sum to 20 is 633. Therefore, Xiaoqiang would need to try at most 633 times.But let me verify this because sometimes with these combinatorial problems, it's easy to make a mistake.Let me check with another approach. Maybe generating functions? The generating function for a single digit is 1 + x + x^2 + ... +x^9 = (1 - x^10)/(1 - x). So the generating function for four digits would be [(1 - x^10)/(1 - x)]^4. We need the coefficient of x^20 in this expansion.To compute the coefficient of x^20 in [(1 - x^10)^4/(1 - x)^4], we can use the binomial theorem. The expansion of (1 - x^10)^4 is 1 - 4x^10 + 6x^20 - 4x^30 + x^40. Then, multiplying by 1/(1 - x)^4, which is the generating function for the number of non-negative integer solutions, so its expansion is sum_{n=0}^infty C(n +3,3)x^n.Therefore, the coefficient of x^20 in the product is:C(20 +3,3) -4*C(20 -10 +3,3) +6*C(20 -20 +3,3) -4*C(20 -30 +3,3) +C(20 -40 +3,3).But the terms where the exponent is negative will have coefficient 0. So:First term: C(23,3)=1771Second term: -4*C(13,3)= -4*286= -1144Third term: +6*C(3,3)=6*1=6Fourth term: -4*C(-7,3)=0Fifth term: +C(-17,3)=0So total coefficient is 1771 -1144 +6=633, which matches our previous result. So this confirms that the number is indeed 633. Therefore, the answer should be 633.Wait, but let me check with an example. For instance, if the sum is smaller. Let's say sum is 10. Then, using the same method, the number of solutions would be C(13,3) -4*C(3,3)=286 -4=282. Wait, but if the sum is 10, the digits can't exceed 9, but maybe some of them could. Wait, but if the sum is 10, the maximum a single digit could be is 10, but since digits are up to 9, so actually, we need to subtract the cases where a digit is 10 or more. But the number of solutions where a digit is ≥10 would be 4*C(0 +3,3)=4*1=4. So total solutions would be C(13,3)-4=286-4=282, which is correct. For example, the number of 4-digit numbers where digits sum to 10 is 282. That seems plausible.Alternatively, another check: if all digits are allowed to be up to 20, which they aren't in our problem, but just for the sake of checking, C(23,3)=1771. But in reality, with the constraints, we get 633. So 633 is less than 1771, which makes sense.Wait, but another sanity check. Let's think of the maximum possible sum with four digits each between 0-9: 9*4=36. So 20 is somewhere in the middle. The number of combinations should be less than 10000, obviously. 633 seems reasonable.Alternatively, let's consider some edge cases. For example, sum =0: only 1 solution (0000). Sum=1: 4 solutions. Sum=2: C(5,3)=10 solutions. That's correct.Wait, maybe if we compute for sum=20 with generating functions, but the result is 633. Alternatively, perhaps I made a mistake in the inclusion-exclusion.Wait, another way to compute the number of solutions where a + b + c + d=20, 0 ≤ a,b,c,d ≤9 is equivalent to 9 - a' + 9 - b' +9 -c' +9 -d' =20, where a', b', c', d' ≥0. Wait, let's see. If we set a' =9 -a, so that a' ≥0 since a ≤9. Then, the equation becomes (9 -a') + (9 -b') + (9 -c') + (9 -d') =20. Simplifying, 36 - (a' +b' +c' +d')=20, so a' +b' +c' +d'=16. Now, this is equivalent to finding the number of non-negative integer solutions to a' +b' +c' +d'=16, where each a',b',c',d' ≤9.Wait, but if each a',b',c',d' ≤9, then 16 ≤4*9=36. Wait, but the original equation is equivalent to this. Wait, but this might not be helpful. However, note that the number of solutions where a' +b' +c' +d'=16 with each a',b',c',d' ≤9 is the same as the number of solutions where a +b +c +d=20 with each a,b,c,d ≤9. So perhaps using this symmetry.But in this case, the number of solutions is equal to the number of solutions of a' +b' +c' +d'=16 with each a' ≤9. Which is the same as the original problem. So perhaps we can apply the same inclusion-exclusion here.But this might not help. Alternatively, perhaps the number of solutions for the equation a + b + c + d =20 with each variable ≤9 is equal to the number of solutions for a +b +c +d=16 with each variable ≤9. Wait, no, because the complement is 9 -a, so the sum becomes 36 - (a +b +c +d)=20, so a +b +c +d=16. So the number of solutions where the original sum is 20 is equal to the number of solutions where the complement sum is16. So if we compute the number of solutions for sum=16 with each variable ≤9, we can get the same answer.But maybe this is useful? Let me check. For example, if we compute the number of solutions for sum=16 with variables ≤9. Let's compute that.Using inclusion-exclusion again. The total number of solutions without restrictions is C(16 +3,3)=C(19,3)=969. Then subtract the cases where any variable is ≥10. So for each variable, set x_i =10 + y_i, leading to y_i + rest variables =6. So number of solutions is C(6 +3,3)=C(9,3)=84. Multiply by4: 4*84=336. Then add back the cases where two variables are ≥10. For two variables, x_i=10 +y_i and x_j=10 + y_j, leading to y_i + y_j + rest =16 -20= -4, which is impossible. So no solutions here. Therefore, inclusion-exclusion gives 969 -336=633. So same result.Therefore, the number of solutions is indeed 633. So this cross-check using the complement method gives the same answer, which reinforces that 633 is correct.Alternatively, I can also code a small script to compute this, but since I can't do that here, I need to rely on mathematical reasoning.Wait, another sanity check: the number of 4-digit numbers where digits sum to 20. The maximum digit is 9, so let's see some possible combinations.For example, three 9s and one digit 20 -27= -7, which is impossible. So the maximum number of 9s is two. Let's see:Case 1: Two 9s. Then the remaining two digits must sum to 20 -18=2. The number of solutions here is C(4,2) * number of solutions for two digits summing to2. The number of solutions for two digits is 3 (0+2,1+1,2+0). So total for this case: 6 *3=18.Case 2: One 9. Then the remaining three digits sum to 11. The number of solutions here is C(4,1)* number of solutions for three digits summing to11 with each ≤9. The number of solutions for three digits summing to11 without restrictions is C(11 +2,2)=C(13,2)=78. Then subtract the cases where any digit is ≥10. For a digit ≥10, set it to10 + y, then the equation becomes y + other two digits =11 -10=1. The number of solutions is C(1 +2,2)=3. There are 3 variables, so 3*3=9. Therefore, the number of valid solutions is 78 -9=69. Therefore, case 2:4 *69=276.Case3: No 9s. Then all digits are ≤8, and sum to20. Let's see the number of solutions for four digits ≤8 summing to20. Using inclusion-exclusion again. Total solutions without restriction:C(20 +3,3)=1771. Subtract the cases where any digit is ≥9. For a digit ≥9, set it to9 + y, so equation becomes y + other three digits =20 -9=11. Number of solutions is C(11 +2,2)=78. Multiply by4:4*78=312. Then add back cases where two digits are ≥9. Set two digits to9 + y_i, equation becomes y1 + y2 + other two digits =20 -18=2. Number of solutions is C(2 +3,3)=10. For each pair of digits, which is C(4,2)=6. So 6*10=60. Then subtract cases where three digits are ≥9. Equation becomes y1 + y2 + y3 + digit4=20 -27= -7, impossible. Similarly, four digits ≥9 sum to at least 36, which is more than20. So inclusion-exclusion gives 1771 -312 +60=1519.But wait, this is the number of solutions for four digits ≤8 summing to20. However, in our original problem, we are considering digits ≤9, but in case3 we have all digits ≤8, so this seems different. Wait, no. Wait, case3 is when there are no 9s, so all digits are ≤8. So to compute the number of solutions for four digits ≤8 summing to20, we need to use inclusion-exclusion starting with C(23,3) and subtract the cases where any digit exceeds8. But digits exceeding8 would be ≥9, so similar to our original problem but with upper limit8 instead of9.Wait, perhaps this approach is getting too convoluted. Alternatively, since we already found that the total number of solutions is 633, and if we break it down into cases with two 9s, one 9, and no 9s, and sum them up, we should get 633.From case1:18, case2:276, case3:633 -18 -276=339? Wait, let's compute 18 +276=294. 633 -294=339. So case3 would be 339.But let's compute case3 properly. Case3: no digits are 9, all digits ≤8, sum=20. So the equation is a + b + c + d=20, 0 ≤a,b,c,d ≤8.Using inclusion-exclusion here. Total solutions without restriction: C(20 +3,3)=1771. Subtract the number of solutions where any digit ≥9. For each digit, set x=9 + y, equation becomes y + other three digits=11. Number of solutions per digit: C(11 +2,2)=78. Four digits:4*78=312. Add back solutions where two digits are ≥9. Then equation becomes y1 + y2 + other two digits=20 -18=2. Number of solutions per pair: C(2 +3,3)=10. Number of pairs:6, so 6*10=60. Subtract solutions where three digits are ≥9: equation becomes y1 + y2 + y3 + x4=20 -27= -7, impossible. Add back solutions where four digits are ≥9: sum=36 + y1+y2+y3+y4=20, which is impossible. So total solutions:1771 -312 +60=1519. But this can't be right, because 1519 is way larger than the total number of solutions we found earlier (633). So clearly, this approach is incorrect.Wait, I think there's confusion here. Case3 is part of the total solutions. The total solutions with digits ≤9 is633. Breaking it into cases where there are two 9s, one 9, or no 9s. So case1: two 9s, case2: one 9, case3: no 9s. Therefore, 633=18+276+case3. So case3=633-18-276=339.But when we tried to compute case3 directly using inclusion-exclusion, we got1519, which is impossible. What's wrong here? Ah! Because in case3, all digits are ≤8, but in the inclusion-exclusion we just did, we considered digits ≤8, but the equation is a +b +c +d=20. However, 4 digits each at most8 can only sum up to32. But 20 is possible. However, the problem arises because when we compute the total solutions without restriction for a +b +c +d=20, which is1771, which includes digits up to20, but when we restrict digits to ≤8, we need to subtract all solutions where any digit is ≥9. But in this case, since our original total solutions for sum20 is1771, but we are restricting digits to ≤8, which is a more stringent condition than the original problem (which allows digits up to9). Hence, this case3 is a subset of the original problem. So if we compute the number of solutions for case3 (no 9s, digits ≤8), it's indeed a different problem, but in the context of the original problem where digits can be up to9, case3 is part of the total solutions. However, we cannot directly compare these numbers because they are different problems.Alternatively, maybe we can use the generating function again for case3. The generating function for digits ≤8 is (1 +x +x^2 + ... +x^8)^4. We need the coefficient ofx^20. This is equivalent to (1 -x^9)^4/(1 -x)^4. Expanding this, we get [1 -4x^9 +6x^18 -4x^27 +x^36]/(1 -x)^4. The coefficient ofx^20 is C(20 +3,3) -4*C(20 -9 +3,3) +6*C(20 -18 +3,3) -4*C(20 -27 +3,3)+C(20 -36 +3,3). Calculating each term:C(23,3)=1771-4*C(14,3)= -4*364= -1456+6*C(5,3)=6*10=60The rest terms are 0. So total coefficient=1771 -1456 +60=375. Wait, but earlier we had case3=339. There's a discrepancy here. Hmm. This suggests that case3 should be375, but according to the breakdown of total solutions (633=18+276+339), but generating function gives375. Therefore, something is wrong.Wait, this shows that my initial breakdown into cases might be incorrect. Let's recast the problem.Total solutions with digits ≤9 summing to20:633 (from inclusion-exclusion). This can be broken into cases based on the number of 9s:Case0: 0 nines, all digits ≤8: let's denote as S0.Case1: exactly 1 nine: S1.Case2: exactly 2 nines: S2.Case3: exactly3 nines: S3.Case4: exactly4 nines: S4.But since 4 nines would sum to36, which is more than20, so S4=0. Similarly, 3 nines sum to27, which is more than20, so S3=0. So only cases S0, S1, S2.So total solutions= S0 + S1 + S2=633.From generating functions, S0=375, as computed above. Then S1= number of solutions with exactly1 nine. Let's compute S1.Exactly1 nine: Choose which digit is9 (4 choices), then the remaining three digits sum to11, with each ≤8. The number of solutions is4*N, where N is the number of solutions to a +b +c=11, each ≤8.Compute N: total solutions without restriction: C(11 +2,2)=78. Subtract the solutions where any variable is ≥9. For each variable, set it to9 + y, then y + other two variables=11 -9=2. The number of solutions per variable isC(2 +2,2)=6. There are3 variables, so subtract3*6=18. However, if two variables are ≥9, the equation becomes y1 + y2 + third variable=11 -18= -7, which is impossible. So no solutions to add back. Therefore, N=78 -18=60. Therefore, S1=4*60=240.Case2: exactly2 nines. Choose two digits to be9 (C(4,2)=6), remaining two digits sum to20 -18=2, each ≤8. The number of solutions is6*M, where M is the number of solutions to a +b=2, each ≤8. The solutions are (0,2),(1,1),(2,0). So M=3. Therefore, S2=6*3=18.Therefore, total solutions: S0 + S1 + S2=375 +240 +18=633. Which matches our previous result. However, in my initial breakdown, I had case1 as 276, which is different from S1=240. So where did I go wrong?Wait, earlier I computed case2 as one 9, but actually, if we have exactly one9, then the remaining three digits must sum to11 and each ≤8, which we found as60, so total solutions4*60=240. But when I initially broke it down, I computed case2 as4*69=276, which is incorrect. Wait, in my initial approach, I computed the number of solutions for three digits summing to11 with each ≤9, which is different. Wait, no, because in the problem where digits can be up to9, but in case1 (exactly one9), the remaining digits have to be ≤8, otherwise, they could be9 as well, leading to more than one9. Therefore, my initial calculation was wrong because I allowed the remaining digits to be9, which would conflict with the "exactly one9" condition. Therefore, the correct number is240 for S1, and18 for S2, and375 for S0.But in my first approach, when I broke down into two9s, one9, and no9s, I mistakenly allowed the remaining digits in the one9 case to be up to9, leading to an overcount. Therefore, the correct breakdown is375 +240 +18=633, which matches the inclusion-exclusion result.So this confirms that the total number is633.Therefore, the answer is633. Xiaoqiang would need to try at most633 times to unlock the phone.Final AnswerThe maximum number of times Xiaoqiang would need to try is boxed{633}.</think>To determine the maximum number of times Xiaoqiang would need to try to unlock the phone, we need to find the number of 4-digit passwords where the sum of the digits is 20. Each digit is between 0 and 9.We start with the equation (a + b + c + d = 20) where (0 leq a, b, c, d leq 9). We use the principle of inclusion-exclusion to count the number of valid solutions.1. Total number of solutions without restrictions: [ binom{20 + 3}{3} = binom{23}{3} = 1771 ]2. Subtract solutions where any digit is ≥ 10: - For each digit, substitute (x_i = 10 + y_i), leading to the equation (y_i + x_j + x_k + x_l = 10). - Number of solutions for each digit: (binom{10 + 3}{3} = binom{13}{3} = 286) - Total for 4 digits: (4 times 286 = 1144)3. Add back solutions where two digits are ≥ 10: - For each pair of digits, substitute (x_i = 10 + y_i) and (x_j = 10 + y_j), leading to the equation (y_i + y_j + x_k + x_l = 0). - Number of solutions for each pair: 1 - Total for 6 pairs: (6 times 1 = 6)4. No solutions exist where three or more digits are ≥ 10 because the sum would be negative.Using inclusion-exclusion, the total number of valid solutions is:[1771 - 1144 + 6 = 633]To verify, we use generating functions. The generating function for a single digit is (frac{1 - x^{10}}{1 - x}). Expanding (left(frac{1 - x^{10}}{1 - x}right)^4) and finding the coefficient of (x^{20}) confirms the result:[binom{23}{3} - 4binom{13}{3} + 6binom{3}{3} = 1771 - 1144 + 6 = 633]Thus, the maximum number of times Xiaoqiang would need to try is (boxed{633}).,<think>Okay, so the problem is about figuring out how many different 4-digit passwords there are where each digit is between 0 and 9, and the sum of the digits is exactly 20. Xiaoqiang needs to try all possible combinations in the worst case, so we need to find the total number of such combinations. Hmm, let's think about how to approach this.First, this seems like a combinatorics problem, specifically a problem of finding the number of non-negative integer solutions to an equation with constraints. The equation here is a + b + c + d = 20, where each of a, b, c, d represents a digit in the password. Since each digit can be from 0 to 9, we have the constraints that 0 ≤ a, b, c, d ≤ 9.Wait, but if we're talking about digits in a password, they can be 0, right? So leading zeros are allowed. For example, a password like 0992 would be valid. So each digit is independent and can range from 0 to 9. So the problem reduces to finding the number of non-negative integer solutions to a + b + c + d = 20 with each variable ≤ 9.But normally, without constraints, the number of solutions to a + b + c + d = 20 would be C(20 + 4 - 1, 4 - 1) = C(23, 3). But that's if each digit can be any non-negative integer. However, here each digit is limited to 0-9, so we need to subtract the solutions where at least one digit is greater than 9.This is a classic stars and bars problem with restrictions. The formula for the number of solutions is the inclusion-exclusion principle. Let me recall the formula:Number of solutions = C(n + k - 1, k - 1) - C(k, 1)*C(n - (m+1) + k - 1, k - 1) + C(k, 2)*C(n - 2*(m+1) + k - 1, k - 1) - ... Where n is the total sum (20 in this case), k is the number of variables (4 digits), and m is the maximum value each variable can take (9). But this formula is applicable when each variable has the same upper limit.Wait, so in our case, each digit can be at most 9, so the maximum m is 9. So the formula would subtract the cases where one or more digits exceed 9. Let's apply inclusion-exclusion step by step.First, compute the total number of solutions without any restrictions: C(20 + 4 - 1, 4 - 1) = C(23, 3). Let's compute that. C(23, 3) is 23*22*21/(3*2*1) = (23*22*21)/6. Let me calculate that:23*22 = 506, 506*21 = 10,626. Then 10,626 divided by 6 is 1,771. So there are 1,771 solutions without restrictions.But now we need to subtract the cases where any digit is greater than 9. So, for each digit that could be 10 or more, we subtract those cases. However, since there are 4 digits, we have to consider the overcounts here.First, subtract the cases where one particular digit is ≥10. Let's fix one digit, say a, and set a' = a - 10, so a' ≥ 0. Then the equation becomes a' + 10 + b + c + d = 20 ⇒ a' + b + c + d = 10. The number of solutions for this is C(10 + 4 - 1, 4 - 1) = C(13, 3) = 286. Since there are 4 digits, each of which could be the one exceeding 9, we multiply by 4: 4*286 = 1,144.But now we have subtracted too much because if two digits are ≥10, we subtracted those cases twice. So we need to add them back in. How many solutions have two digits ≥10?Let’s fix two digits, say a and b, and set a' = a - 10, b' = b - 10, so a', b' ≥ 0. Then the equation becomes a' + 10 + b' + 10 + c + d = 20 ⇒ a' + b' + c + d = 0. The number of non-negative solutions to this is C(0 + 4 - 1, 4 - 1) = C(3, 3) = 1. So for each pair of digits, there's 1 solution. How many pairs are there? C(4, 2) = 6. So we add back 6*1 = 6.Now, moving to inclusion-exclusion, the next term would be subtracting cases where three digits are ≥10. Let's check if that's possible. If three digits are each at least 10, then their total would be at least 30, but the sum needed is 20, which is impossible. So the number of solutions where three or more digits are ≥10 is zero. Therefore, we can stop here.Putting it all together: total solutions = unrestricted solutions - solutions with one digit ≥10 + solutions with two digits ≥10.So that would be 1,771 - 1,144 + 6 = let's compute that.1,771 - 1,144 = 627; 627 + 6 = 633.Wait, so according to this, there are 633 possible 4-digit combinations where the digits sum to 20 and each digit is between 0 and 9.But hold on, let me verify this because sometimes when applying inclusion-exclusion, mistakes can happen.First, check the unrestricted solutions: C(23,3)=1771. Correct. Then, subtract the cases where any one digit is over 9. For each digit, we set it to at least 10, then the remaining sum is 10. The number of solutions for each digit is C(10 + 4 -1, 4 -1)=C(13,3)=286. Multiply by 4 gives 1144. Then, add back the cases where two digits are over 9. For each pair, the remaining sum is 20 -10 -10=0, so solutions C(0 +4 -1,4-1)=C(3,3)=1. Multiply by C(4,2)=6 gives 6. So total is 1771 -1144 +6=633.But wait, let's think: if two digits are over 9, then each is at least 10, so two digits take away 20, so the remaining two digits must sum to 0, which means both are 0. So the only solution in that case is 10,10,0,0. But since each digit can be arranged, how many such combinations are there? For each pair of digits set to 10 and the other two set to 0. The number of such combinations is the number of ways to choose which two digits are 10: C(4,2)=6. Each of these contributes one solution, so 6 total. So that's correct.Similarly, for the case where one digit is over 9, say 10, then the remaining three digits sum to 10. The number of solutions is C(13,3)=286, which is correct. So 4 digits, each can be the one over 9, so 4*286=1144.But wait, when we subtract these 1144, we subtract all cases where a digit is over 9. But if two digits are over 9, they were subtracted twice, once for each digit. Hence, adding them back once via inclusion-exclusion.So total solutions: 1771 -1144 +6=633. That seems right.But let's check with another method. For example, generating functions. The generating function for each digit is (x^0 + x^1 + ... +x^9). So the generating function for four digits is (x^0 +x^1 +...+x^9)^4. We need the coefficient of x^20 in this expansion.Alternatively, we can compute this coefficient using inclusion-exclusion as well.But perhaps it's easier to compute via generating functions. Let's see.The generating function for a single digit is (1 - x^10)/(1 - x). Therefore, the generating function for four digits is (1 - x^10)^4 / (1 - x)^4.We need the coefficient of x^20 in this product. Expanding this, we can use the binomial theorem:(1 - x^10)^4 = 1 - 4x^10 + 6x^20 - 4x^30 + x^40Then, multiplying by 1/(1 - x)^4, which is the generating function for the number of non-negative integer solutions (i.e., coefficients are C(n + 3, 3)).Therefore, the coefficient of x^20 in (1 - x^10)^4 / (1 - x)^4 is:C(20 + 3, 3) - 4*C(20 -10 + 3, 3) + 6*C(20 -20 + 3, 3) - 4*C(20 -30 + 3, 3) + C(20 -40 +3, 3)But note that C(n,3) is zero if n <3.Compute each term:First term: C(23,3) = 1771Second term: -4*C(13,3) = -4*286 = -1144Third term: +6*C(3,3) = +6*1 = +6Fourth term: -4*C(-7,3)=0 (since negative)Fifth term: +C(-17,3)=0So total is 1771 -1144 +6 = 633. Which matches our previous result. So that confirms it.Therefore, the number of valid passwords is 633. Therefore, Xiaoqiang would need to try at most 633 times to unlock the phone.Wait, but let me just think again: are there any possible mistakes here? For example, in generating functions, we have (1 - x^10)^4 multiplied by 1/(1 -x)^4. Then, the coefficient of x^20 is indeed the same as calculated. So this seems correct.Alternatively, maybe we can try to compute this manually for a smaller case to check.Suppose, for example, if we have a 2-digit password, sum to 10, each digit 0-9. How many solutions?By stars and bars, total solutions without restriction: C(10 + 2 -1, 2 -1)=C(11,1)=11. Subtract the cases where a digit is ≥10. If a digit is ≥10, then set a' = a -10, so equation becomes a' + b = 0. So a'=0 and b=0. So only one solution where a=10, b=0 and a=0, b=10. So 2 solutions to subtract. So total solutions:11 -2=9. But let's list them:Digits can be (0,10) invalid, (1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), (9,1), (10,0) invalid. So indeed 9 valid solutions. Which matches 11 -2=9. So that works.Similarly, applying inclusion-exclusion here: C(11,1) - 2*C(1,1)=11-2=9. Correct.So the method works here. So returning to our problem, 633 seems correct.But let me check with another approach. For example, enumerating the number of solutions where all digits are between 0 and 9 and sum to 20. Maybe we can think of this as a constrained integer composition problem.But another way: Let's consider the digits as variables a, b, c, d with 0 ≤ a,b,c,d ≤9 and a + b + c + d =20.We can rephrase each variable as 9 - a', 9 - b', etc., but I don't know if that helps.Alternatively, since the maximum sum for four digits is 4*9=36, so 20 is somewhere in the middle. But maybe not helpful.Alternatively, think of the problem as distributing 20 units into 4 digits, each digit can take 0-9.Alternatively, since each digit is between 0 and 9, we can model this as a constrained stars and bars problem. But the inclusion-exclusion approach seems the standard way here, and we already verified with generating functions.Another way to check is to compute the number using recursive counting.For example, for each digit position, the number of ways to choose a digit from 0 to 9 such that the remaining digits sum to the remaining total.But that might be cumbersome for a 4-digit password, but let's try to sketch it.Define f(n, k) as the number of ways to have n digits summing to k.We need f(4,20).Base cases:f(1, k) = 1 if 0 ≤k ≤9, else 0.Recurrence:f(n, k) = sum_{i=0}^9 f(n-1, k -i) for n ≥2.So compute f(4,20):Compute f(1, k) for k=0-9: all 1s.Then compute f(2, k) for k=0-18.f(2, k) = sum_{i=0}^9 f(1, k -i). Since f(1, k -i) is 1 if 0 ≤k -i ≤9, else 0.Therefore, f(2, k) = number of integers i in [0,9] such that k -i is in [0,9]. That is, i ∈ [k-9, k] ∩ [0,9].So for each k from 0 to 18:If k ≤9, then i can be from 0 to k, so k+1 terms.If k ≥10, then i can be from k-9 to 9, so 9 - (k -9) +1 = 19 -k terms.So f(2,k):k=0:1k=1:2...k=9:10k=10:9k=11:8...k=18:1Now compute f(3, k) for k=0-27.f(3,k) = sum_{i=0}^9 f(2, k -i). So for each k, sum the f(2, k -i) where i=0 to9.Similarly, this would require knowing the values of f(2, j) for j from k-9 to k.But let's compute f(3, k) up to k=20.But maybe this will take too long, but let's try to compute f(3, k) for k up to 20.Wait, but f(3, k) is needed for k up to 20, but since each digit is up to9, the maximum sum for 3 digits is 27. But since we need f(4,20), which is sum_{i=0}^9 f(3,20 -i). So we need f(3, j) where j ranges from 20 -9=11 to 20.Therefore, compute f(3,11) to f(3,20).First, compute f(2, k) for k from 0 to 18 as above.Then, compute f(3,11) = sum_{i=0}^9 f(2,11 -i) = sum_{i=0}^9 f(2,11 -i)Which is f(2,11) + f(2,10) + f(2,9) + ... + f(2,2)Wait, when i=0: f(2,11 -0)=f(2,11)=8i=1: f(2,10)=9i=2: f(2,9)=10i=3: f(2,8)=9Wait, hold on:Wait, for f(3,11):i ranges from 0 to9, so 11 -i ranges from 11 down to 2. So:f(3,11) = f(2,11) + f(2,10) + f(2,9) + f(2,8) + f(2,7) + f(2,6) + f(2,5) + f(2,4) + f(2,3) + f(2,2)From earlier, f(2,11)=8, f(2,10)=9, f(2,9)=10, f(2,8)=9, f(2,7)=8, f(2,6)=7, f(2,5)=6, f(2,4)=5, f(2,3)=4, f(2,2)=3.Adding these up: 8 +9 +10 +9 +8 +7 +6 +5 +4 +3.Let's compute:8+9=1717+10=2727+9=3636+8=4444+7=5151+6=5757+5=6262+4=6666+3=69So f(3,11)=69.Similarly, compute f(3,12):sum_{i=0}^9 f(2,12 -i) = sum_{i=0}^9 f(2,12 -i) where 12 -i ranges from 12 down to 3.Thus, f(2,12)=7, f(2,11)=8, f(2,10)=9, f(2,9)=10, f(2,8)=9, f(2,7)=8, f(2,6)=7, f(2,5)=6, f(2,4)=5, f(2,3)=4.Adding these:7+8=1515+9=2424+10=3434+9=4343+8=5151+7=5858+6=6464+5=6969+4=73So f(3,12)=73.f(3,13):sum_{i=0}^9 f(2,13 -i) = f(2,13)=6, f(2,12)=7, f(2,11)=8, f(2,10)=9, f(2,9)=10, f(2,8)=9, f(2,7)=8, f(2,6)=7, f(2,5)=6, f(2,4)=5.Adding:6+7=1313+8=2121+9=3030+10=4040+9=4949+8=5757+7=6464+6=7070+5=75So f(3,13)=75.f(3,14):sum_{i=0}^9 f(2,14 -i) = f(2,14)=5, f(2,13)=6, f(2,12)=7, f(2,11)=8, f(2,10)=9, f(2,9)=10, f(2,8)=9, f(2,7)=8, f(2,6)=7, f(2,5)=6.Adding:5+6=1111+7=1818+8=2626+9=3535+10=4545+9=5454+8=6262+7=6969+6=75So f(3,14)=75.f(3,15):sum_{i=0}^9 f(2,15 -i) = f(2,15)=4, f(2,14)=5, f(2,13)=6, f(2,12)=7, f(2,11)=8, f(2,10)=9, f(2,9)=10, f(2,8)=9, f(2,7)=8, f(2,6)=7.Adding:4+5=99+6=1515+7=2222+8=3030+9=3939+10=4949+9=5858+8=6666+7=73So f(3,15)=73.f(3,16):sum_{i=0}^9 f(2,16 -i) = f(2,16)=3, f(2,15)=4, f(2,14)=5, f(2,13)=6, f(2,12)=7, f(2,11)=8, f(2,10)=9, f(2,9)=10, f(2,8)=9, f(2,7)=8.Adding:3+4=77+5=1212+6=1818+7=2525+8=3333+9=4242+10=5252+9=6161+8=69So f(3,16)=69.f(3,17):sum_{i=0}^9 f(2,17 -i) = f(2,17)=2, f(2,16)=3, f(2,15)=4, f(2,14)=5, f(2,13)=6, f(2,12)=7, f(2,11)=8, f(2,10)=9, f(2,9)=10, f(2,8)=9.Adding:2+3=55+4=99+5=1414+6=2020+7=2727+8=3535+9=4444+10=5454+9=63So f(3,17)=63.f(3,18):sum_{i=0}^9 f(2,18 -i) = f(2,18)=1, f(2,17)=2, f(2,16)=3, f(2,15)=4, f(2,14)=5, f(2,13)=6, f(2,12)=7, f(2,11)=8, f(2,10)=9, f(2,9)=10.Adding:1+2=33+3=66+4=1010+5=1515+6=2121+7=2828+8=3636+9=4545+10=55So f(3,18)=55.f(3,19):sum_{i=0}^9 f(2,19 -i) = f(2,19)=0, f(2,18)=1, f(2,17)=2, f(2,16)=3, f(2,15)=4, f(2,14)=5, f(2,13)=6, f(2,12)=7, f(2,11)=8, f(2,10)=9.Adding:0+1=11+2=33+3=66+4=1010+5=1515+6=2121+7=2828+8=3636+9=45So f(3,19)=45.f(3,20):sum_{i=0}^9 f(2,20 -i) = f(2,20)=0 (since maximum for two digits is 18), f(2,19)=0, f(2,18)=1, f(2,17)=2, f(2,16)=3, f(2,15)=4, f(2,14)=5, f(2,13)=6, f(2,12)=7, f(2,11)=8, f(2,10)=9.Wait, for i=0: f(2,20)=0i=1: f(2,19)=0i=2: f(2,18)=1i=3: f(2,17)=2i=4: f(2,16)=3i=5: f(2,15)=4i=6: f(2,14)=5i=7: f(2,13)=6i=8: f(2,12)=7i=9: f(2,11)=8Adding these up:0+0=00+1=11+2=33+3=66+4=1010+5=1515+6=2121+7=2828+8=36So f(3,20)=36.Now, compute f(4,20) = sum_{i=0}^9 f(3,20 -i) = sum_{i=0}^9 f(3,20 -i) where 20 -i ranges from 20 down to 11.Therefore, f(4,20) = f(3,20) + f(3,19) + f(3,18) + f(3,17) + f(3,16) + f(3,15) + f(3,14) + f(3,13) + f(3,12) + f(3,11)From above:f(3,20)=36f(3,19)=45f(3,18)=55f(3,17)=63f(3,16)=69f(3,15)=73f(3,14)=75f(3,13)=75f(3,12)=73f(3,11)=69So adding these up:Start from the smallest:36 +45 =8181 +55=136136 +63=199199 +69=268268 +73=341341 +75=416416 +75=491491 +73=564564 +69=633So f(4,20)=633. Which matches our previous result. So this confirms that the inclusion-exclusion and generating functions methods are correct.Therefore, the answer is 633.Therefore, Xiaoqiang would need to try at most 633 times to unlock the phone.Final Answerboxed{633}</think>To determine the number of 4-digit passwords where the sum of the digits is 20, we need to find the number of non-negative integer solutions to the equation (a + b + c + d = 20) with each digit (a, b, c, d) between 0 and 9. Using the principle of inclusion-exclusion, we start by calculating the number of unrestricted solutions, then subtract the cases where any digit exceeds 9.1. Unrestricted solutions: The number of solutions to (a + b + c + d = 20) is given by the stars and bars method: [ binom{20 + 4 - 1}{4 - 1} = binom{23}{3} = 1771 ]2. Subtract cases where one digit is ≥ 10: For each digit, set it to (a' = a - 10) (so (a' geq 0)) and solve (a' + b + c + d = 10). The number of solutions is: [ binom{10 + 4 - 1}{4 - 1} = binom{13}{3} = 286 ] Since there are 4 digits, we multiply by 4: [ 4 times 286 = 1144 ]3. Add back cases where two digits are ≥ 10: For each pair of digits, set each to (a' = a - 10) and (b' = b - 10) (so (a', b' geq 0)) and solve (a' + b' + c + d = 0). The number of solutions is: [ binom{0 + 4 - 1}{4 - 1} = binom{3}{3} = 1 ] There are (binom{4}{2} = 6) pairs of digits, so we add back: [ 6 times 1 = 6 ]4. Combine results: Using inclusion-exclusion, the number of valid solutions is: [ 1771 - 1144 + 6 = 633 ]This result is confirmed by both the generating function method and recursive counting, ensuring the correctness of the solution.Thus, Xiaoqiang would need to try at most (boxed{633}) times to unlock the phone.]

🔑:【Analysis】Consider the case of 20 being split into 4 numbers, where:4 numbers are the same, there is 1 type: 55553 numbers are the same, there are 2 types: 6662, 44482 types of numbers each appear 2 times, there are 4 types: 9911, 8822, 7733, 66442 numbers are the same, the remaining are different, there are:If the same number is 9, then the sum of the remaining 2 numbers is 2, there is only 20 this 1 typeIf the same number is 8, then the sum of the remaining 2 numbers is 4, there are 40, 31 these 2 typesIf the same number is 7, then the sum of the remaining 2 numbers is 6, there are 60, 51, 42 these 3 typesIf the same number is 6, then the sum of the remaining 2 numbers is 8, there are 80, 71, 53 these 3 typesIf the same number is 5, then the sum of the remaining 2 numbers is 10, there are 91, 82, 73, 64 these 4 typesIf the same number is 4, then the sum of the remaining 2 numbers is 12, there are 93, 75 these 2 typesIf the same number is 3, then the sum of the remaining 2 numbers is 14, there are 95, 86 these 2 typesIf the same number is 2, then the sum of the remaining 2 numbers is 16, there is only 97 this 1 typeIf the same number is 1, then the sum of the remaining 2 numbers is 18, there are 0 types, a total of 18 types4 numbers are all different, there are 16 types:begin{aligned}20 & =9+8+3+0=9+8+2+1=9+7+4+0=9+7+3+1=9+6+5+0 & =9+6+4+1=9+6+3+2=9+5+4+2=8+7+5+0=8+7+4+1 & =8+7+3+2=8+6+5+1=8+6+4+2=8+5+4+3=7+6+5+2 & =7+6+4+3end{aligned}4 numbers are the same, there is only 1 arrangement3 numbers are the same, there are C_{4}^{1}=4 arrangements2 types of numbers each appear 2 times, there are C_{4}^{2}=6 arrangements2 numbers are the same, the remaining are different, there are P_{4}^{2}=12 arrangements4 numbers are all different, there are P_{4}^{4}=24 arrangementsIn summary, the password has 1 times 1+2 times 4+4 times 6+18 times 12+16 times 24=633 possibilities, meaning it can be unlocked with at most 633 attempts.

🧠:[<think>Okay, let's see. The problem says the first term of a sequence is 2005. Each next term is the sum of the cubes of the digits of the previous term. We need to find the 2005th term. The options are given from A to E. Hmm. Alright, let's break this down.First, the sequence starts at 2005. Then, each term after that is calculated by taking each digit of the previous term, cubing it, and summing those cubes. So, for example, the second term would be the sum of the cubes of 2, 0, 0, and 5. Let me compute that real quick.2 cubed is 8, 0 cubed is 0, another 0 cubed is 0, and 5 cubed is 125. So adding those up: 8 + 0 + 0 + 125 = 133. So the second term is 133.Then the third term would be the sum of the cubes of 1, 3, and 3. Let's check that. 1 cubed is 1, 3 cubed is 27, another 3 cubed is 27. So 1 + 27 + 27 = 55. That's the third term, 55.Fourth term would be the sum of the cubes of 5 and 5. 5 cubed is 125, another 5 cubed is 125. So 125 + 125 = 250. Fourth term is 250.Fifth term: sum of cubes of 2, 5, and 0. 2 cubed is 8, 5 cubed is 125, 0 cubed is 0. So 8 + 125 + 0 = 133. Wait, that's the same as the second term. So fifth term is 133, which was already term 2.So then the sixth term would be the same as term 3, which is 55. Then term 7 is 250 again, term 8 is 133, and so on. So it looks like we have a cycle here: 133, 55, 250, and then back to 133, repeating every three terms.Wait, let me verify that. Starting from term 2: 133, then 55, then 250, then 133 again. So the cycle is 133, 55, 250, 133, 55, 250,... So the cycle length is 3. So starting from term 2, every three terms, it repeats.So if that's the case, then terms 2, 5, 8, 11,... are 133. Terms 3, 6, 9, 12,... are 55. Terms 4, 7, 10, 13,... are 250. So the cycle is 133, 55, 250, repeating every three terms.Therefore, to find the 2005th term, we need to see where 2005 falls in this cycle. But wait, the cycle starts at term 2. So term 2 is the first term of the cycle. So term 2: cycle position 1 (133), term 3: cycle position 2 (55), term 4: cycle position 3 (250), term 5: cycle position 1 (133), etc.Therefore, starting from term 2, each term's position in the cycle can be found by (n - 2) mod 3. If (n - 2) mod 3 = 0, then it's the third term in the cycle (250), if it's 1, then first term (133), if it's 2, then second term (55).But we need term 2005. Let's calculate the position in the cycle. So term number 2005. Since the cycle starts at term 2, we subtract 1 to align it. Wait, maybe better to compute (2005 - 1) mod 3? Wait, no. Let's think.Starting from term 2 as the first term of the cycle. So term 2 corresponds to cycle index 1, term 3 to index 2, term 4 to index 3, term 5 to index 1, etc. So to find which index term 2005 is, we can compute (2005 - 2) mod 3. Wait, because term 2 is (2005 - 2) = 2003 terms ahead. Hmm, maybe not. Let's try an example.Let's take term 2: (2 - 2) = 0. 0 mod 3 is 0. But term 2 is cycle index 1. Wait, maybe my approach is off. Let me reindex.Alternatively, starting from term 2, the cycle is of length 3. So the term number can be calculated as term number = 2 + k*3 + m, where m is 0,1,2. So to find m, we can compute (2005 - 2) mod 3. Let's do that.2005 - 2 = 2003. 2003 divided by 3 is 667 * 3 = 2001, remainder 2. So 2003 mod 3 is 2. Therefore, m = 2. So the term is the third term in the cycle. Wait, but starting from term 2 as the first term. So if m=0, it's term 2 + 3k, which is the first term of the cycle (133). If m=1, term 2 + 3k +1, which is term 3 + 3k, which is the second term (55). If m=2, term 2 + 3k +2, which is term 4 + 3k, which is the third term (250). So since 2005 - 2 = 2003, which is 3k + 2, so m=2, so term is third in cycle, which is 250. But 250 is option E, but the answer choices are A)29, B)55, C)85, D)133, E)250. So E is 250. Wait, but hold on, let's check the cycle again because maybe I made a mistake.Wait, starting from term 2: term2=133, term3=55, term4=250, term5=133, term6=55, term7=250, term8=133, etc. So the cycle is indeed 133,55,250. So term numbers:2:1333:554:2505:1336:557:2508:1339:5510:250So the cycle is three terms: 133 (positions 2,5,8,...), 55 (positions3,6,9,...), 250 (positions4,7,10,...). So to get the 2005th term, we need to see which position 2005 is in this cycle.But 2005 is the term number. The cycle starts at term2, so term2, term3, term4 form one full cycle. Then term5, term6, term7 another cycle, etc.So the number of terms after term1 is 2004 terms (since term1 is 2005, term2 is next). Wait, term1 is 2005, term2 is calculated from term1, so term2 is the first term after the initial. So term2005 would be the 2004th term after term1. Wait, no. Wait, term1 is the first term, term2 is the second term. So term2005 is the 2005th term. So starting from term2 as the start of the cycle, term2 is cycle position1, term3 cycle position2, term4 cycle position3, term5 cycle position1, etc.Thus, to find the position within the cycle for term2005, we can compute (2005 - 1) mod 3. Wait, term2 corresponds to n=2. So term number minus 2 gives the number of terms after term2. Then mod 3. So (2005 -2) =2003. 2003 divided by 3 is 667*3=2001, remainder 2. So 2003 mod3=2. So cycle position is 2+1=3? Wait, maybe cycle positions are 0,1,2. So remainder 0 would be position3, remainder1 position1, remainder2 position2. Wait, this is getting confusing. Let me think again.Alternatively, list the term numbers and their cycle positions:Term2:1 (133)Term3:2 (55)Term4:3 (250)Term5:1 (133)Term6:2 (55)Term7:3 (250)So term number: cycle position2:13:24:35:16:27:3...So term number modulo 3 gives:Term2: 2 mod3=2 → cycle position1Term3:3 mod3=0 → cycle position2Term4:4 mod3=1 → cycle position3Term5:5 mod3=2 → cycle position1Term6:6 mod3=0 → cycle position2Term7:7 mod3=1 → cycle position3Wait, this is inconsistent. Maybe my mapping is wrong. Alternatively, perhaps to model this as term number -2, then mod3:Term2: (2-2)=0 → 0 mod3=0 → cycle position1?Wait, this is getting me confused. Maybe another approach. Let's note that term2 is cycle position1, term3 position2, term4 position3, term5 position1, term6 position2, term7 position3, etc. So the cycle positions repeat every 3 terms starting from term2. Therefore, term number =2 + 3k + m, where m=0,1,2. Then m=0 corresponds to cycle position1 (133), m=1 corresponds to cycle position2 (55), m=2 corresponds to cycle position3 (250). Therefore, for term number n, m=(n -2) mod3. So for term2005:m=(2005 -2) mod3=2003 mod3. Let's compute 2003 divided by3:3*667=2001, 2003-2001=2. So 2003 mod3=2. So m=2, which corresponds to cycle position3, which is 250. So the answer would be E)250. However, the answer choices include 250 as E. But wait, the answer isn't among the options? Wait, wait, the options are A)29, B)55, C)85, D)133, E)250. So E is 250. But in the problem statement, the answer options include E)250. So that's an option. So according to this reasoning, the 2005th term is 250. But let me cross-verify this because sometimes these cycle problems can be tricky.Wait, but let's check the cycle again manually for a few terms to make sure. Starting with term1:2005.Term1:2005Term2:2^3 +0^3 +0^3 +5^3=8+0+0+125=133Term3:1^3 +3^3 +3^3=1+27+27=55Term4:5^3 +5^3=125+125=250Term5:2^3 +5^3 +0^3=8+125+0=133Term6:1^3 +3^3 +3^3=55Term7:5^3 +5^3=250Term8:133Term9:55Term10:250Yes, the cycle is 133,55,250 repeating every three terms starting from term2. So term2, term5, term8,... term(2+3k)=133term3, term6, term9,... term(3+3k)=55term4, term7, term10,... term(4+3k)=250So for term number n, starting from term2, which is 2,5,8,..., the positions are 2,5,8,... So the general formula for term number in 133 is n=2+3k where k>=0.Similarly, term numbers in 55 are n=3+3k, and in 250 are n=4+3k.Therefore, for term2005, we need to see if 2005 is congruent to 2,3, or4 modulo3.Compute 2005 mod3:Sum of digits of 2005:2+0+0+5=7. 7 mod3=1. Therefore, 2005 mod3=1. So 2005=3k +1. Therefore, 2005=3k+1. So 2005-1=3k, so 2004=3k. So 2004 is divisible by3. Thus, 2005 mod3=1. So 2005 is congruent to1 modulo3.But our cycle positions for term numbers are:133: n≡2 mod3 (since term2,5,8,... which are 2 mod3)55: n≡3 mod3 (which is 0 mod3)250: n≡4 mod3 (which is 1 mod3)Wait, 4 mod3=1. So term numbers that are 1 mod3 correspond to 250. So 2005 mod3=1, so term2005 is 250. Which is optionE. So that's the answer.But let me verify again with term numbers:term2:133, 2 mod3=2term3:55,3 mod3=0term4:250,4 mod3=1term5:133,5 mod3=2term6:55,6 mod3=0term7:250,7 mod3=1term8:133,8 mod3=2term9:55,9 mod3=0term10:250,10 mod3=1Yes, so term numbers congruent to2 mod3:133term numbers congruent to0 mod3:55term numbers congruent to1 mod3:250Therefore, term2005:2005 mod3=1, so it's congruent to1 mod3, hence 250. So answer is E)250.But wait, the options given are:A)29B)55C)85D)133E)250So E is 250, which matches our conclusion. Therefore, the answer should be E)250.However, wait, let me check again if there's a mistake in the cycle. Because sometimes these sequences can have a different cycle. Let me compute a few more terms manually to ensure that the cycle is indeed 133,55,250.Starting from term1:2005Term1:2005Term2:133Term3:1^3 +3^3 +3^3=1 +27 +27=55Term4:5^3 +5^3=125 +125=250Term5:2^3 +5^3 +0^3=8 +125 +0=133Term6:1^3 +3^3 +3^3=55Term7:5^3 +5^3=250Term8:133So yes, it's repeating every three terms:133,55,250. So cycle length is3. Therefore, every term after term2 is in a cycle of3. So term2005, which is the 2005th term, would be in this cycle. Since term2005 is the 2005th term, and term2 is the start of the cycle, how many terms into the cycle is term2005?Number of terms in the cycle starting at term2: term2 is the first term of the cycle, term3 second, term4 third, term5 fourth? Wait no. Wait, term2 is the first term of the cycle, term3 second, term4 third, term5 first again. So the cycle is terms2,3,4, then repeats terms5,6,7, etc. So term2005 is term number2005.The position in the cycle can be found by (2005 -1) mod3. Wait, term2 is the first term in the cycle. So term2 corresponds to cycle position0? Or cycle position1? Maybe this is where confusion is happening.Alternative approach:Total number of terms in the cycle:3 (133,55,250)Starting from term2 as the first term of the cycle. So term2: index0, term3:index1, term4:index2, term5:index0, term6:index1, etc.Therefore, for term number n (where n >=2), the index in the cycle is (n -2) mod3.Thus, term2005: (2005 -2) mod3 =2003 mod3. As before, 2003 /3=667*3 +2, so remainder2. Therefore, index2 in the cycle, which is the third element, 250. Therefore, term2005 is250, which is optionE.Therefore, the answer is E)250.Final Answerboxed{E}</think>The first term of the sequence is (2005). Each succeeding term is the sum of the cubes of the digits of the previous term. We need to find the (2005^{text{th}}) term of the sequence.1. Calculating the first few terms: - The first term (a_1 = 2005). - The second term (a_2) is the sum of the cubes of the digits of (2005): [ 2^3 + 0^3 + 0^3 + 5^3 = 8 + 0 + 0 + 125 = 133 ] - The third term (a_3) is the sum of the cubes of the digits of (133): [ 1^3 + 3^3 + 3^3 = 1 + 27 + 27 = 55 ] - The fourth term (a_4) is the sum of the cubes of the digits of (55): [ 5^3 + 5^3 = 125 + 125 = 250 ] - The fifth term (a_5) is the sum of the cubes of the digits of (250): [ 2^3 + 5^3 + 0^3 = 8 + 125 + 0 = 133 ]2. Identifying the cycle: - The terms starting from (a_2) form a repeating cycle: (133, 55, 250). - The cycle length is (3).3. Determining the position of the (2005^{text{th}}) term in the cycle: - The cycle starts from the second term. Therefore, we need to find the position of the (2005^{text{th}}) term in the cycle. - Calculate ((2005 - 2) mod 3) to find the position in the cycle: [ 2005 - 2 = 2003 ] [ 2003 mod 3 = 2 ] - A remainder of (2) corresponds to the third term in the cycle (133, 55, 250), which is (250).Thus, the (2005^{text{th}}) term of the sequence is (boxed{E}).,<think>Okay, so I have this problem here about a sequence where the first term is 2005, and each next term is the sum of the cubes of the digits of the previous term. I need to find the 2005th term. The answer choices are 29, 55, 85, 133, and 250. Hmm, let me think through this step by step.First, let's make sure I understand the problem correctly. The sequence starts with 2005. Then, for each term after that, I take the previous term, look at each digit, cube those digits, and add them up to get the next term. So, for example, the second term would be the sum of the cubes of 2, 0, 0, and 5. Let me verify that.Calculating the second term: 2^3 + 0^3 + 0^3 + 5^3 = 8 + 0 + 0 + 125 = 133. So term 2 is 133. Then term 3 would be the sum of the cubes of 1, 3, and 3. That would be 1^3 + 3^3 + 3^3 = 1 + 27 + 27 = 55. Term 3 is 55. Term 4 would be the cubes of 5 and 5: 5^3 + 5^3 = 125 + 125 = 250. Term 4 is 250. Term 5: 2^3 + 5^3 + 0^3 = 8 + 125 + 0 = 133. Wait a minute, that's the same as term 2. So term 5 is 133, which we already saw. Term 6 would then be 55 again, right? So this seems to enter a loop here: 133 → 55 → 250 → 133 → 55 → 250...So starting from term 2, the sequence is 133, 55, 250, 133, 55, 250, etc. So the cycle here is of length 3: 133, 55, 250, repeating every three terms. So if we can figure out where the 2005th term falls in this cycle, we can determine which number it is.But first, let's make sure that this cycle really does repeat indefinitely and that there are no other numbers that might come up. Let's check a few more terms to confirm. Term 5 was 133, term 6 is 55, term 7 is 250, term 8 is 133, term 9 is 55, term 10 is 250... Yeah, it's definitely cycling through those three numbers. So from term 2 onwards, it's a cycle of 3.But let's check term 1 as well. Term 1 is 2005, term 2 is 133, term 3 is 55, term 4 is 250, term 5 is 133, and so on. So the cycle starts at term 2, and the cycle length is 3. Therefore, starting from term 2, every time you go three terms, you come back to the same number. So if we can figure out how many terms after term 1 we are when we get to term 2005, then we can take that number modulo 3 to see where in the cycle we land.Wait, term 1 is 2005. Then term 2 is the first term in the cycle. So term 2 corresponds to position 1 in the cycle (133), term 3 is position 2 (55), term 4 is position 3 (250), term 5 is position 1 (133), etc. So the cycle positions for terms after term 1 are as follows:Term 2: cycle position 1 (133)Term 3: cycle position 2 (55)Term 4: cycle position 3 (250)Term 5: cycle position 1 (133)Term 6: cycle position 2 (55)... and so on.Therefore, for term number n where n ≥ 2, the position in the cycle is (n - 2) mod 3. Then, depending on the result, we can assign the corresponding number. Let's test this with term 5: (5 - 2) = 3, 3 mod 3 is 0. Hmm, but cycle positions are 1, 2, 3. So maybe instead, (n - 2) mod 3 gives 0, 1, 2, which we can map to positions 3, 1, 2? Wait, maybe I need to adjust how I'm calculating this.Alternatively, since the cycle is of length 3 starting at term 2, the term number can be calculated as follows: starting from term 2, each subsequent term cycles through 133, 55, 250. So term 2: 133 (cycle index 0 if starting at 0), term 3: 55 (cycle index 1), term 4: 250 (cycle index 2), term 5: 133 (cycle index 0), etc. If we index the cycle starting at 0, then the cycle index is (n - 2) mod 3. So for term n:If (n - 2) mod 3 = 0 → 133If (n - 2) mod 3 = 1 → 55If (n - 2) mod 3 = 2 → 250Wait, but term 2: (2 - 2) mod 3 = 0 → 133, correct.Term 3: (3 - 2) mod 3 = 1 → 55, correct.Term 4: (4 - 2) mod 3 = 2 → 250, correct.Term 5: (5 - 2) mod 3 = 0 → 133, correct.Yes, that works. So in general, for term n (n ≥ 2), the value is:133 if (n - 2) mod 3 = 0,55 if (n - 2) mod 3 = 1,250 if (n - 2) mod 3 = 2.Therefore, to find the 2005th term, we need to compute (2005 - 2) mod 3, which is 2003 mod 3. Let's compute 2003 divided by 3.3 × 667 = 2001, so 2003 = 3×667 + 2, so 2003 mod 3 is 2. Therefore, (2005 - 2) mod 3 = 2, which corresponds to 250. But wait, 250 is not one of the answer choices. Wait, hold on, the answer choices are 29, 55, 85, 133, 250. Oh, 250 is option E. Wait, but the answer options include 250, which is E. But the options given are (A) 29, (B)55, (C)85, (D)133, (E)250. So 250 is an option. But the answer key might have different. Wait, but let me check again.Wait, in the problem statement, the answer choices are:(A) 29(B) 55(C) 85(D) 133(E) 250So 250 is an option (E). But according to the cycle we found, term 2005 would be 250. So is 250 the answer? But the original problem's answer choices include 250 as E, so maybe that's the answer. Wait, but let me double-check my calculations because sometimes these cycles can have a different starting point or maybe I made an error in calculating the cycle.Wait, let's recalculate the cycle. Starting from term 1: 2005.Term 1: 2005Term 2: 2^3 + 0^3 + 0^3 + 5^3 = 8 + 0 + 0 + 125 = 133Term 3: 1^3 + 3^3 + 3^3 = 1 + 27 + 27 = 55Term 4: 5^3 + 5^3 = 125 + 125 = 250Term 5: 2^3 + 5^3 + 0^3 = 8 + 125 + 0 = 133Term 6: 1^3 + 3^3 + 3^3 = 55Term 7: 5^3 + 5^3 = 250So yes, cycle is 133, 55, 250, repeating every three terms starting from term 2. Therefore, term 2: 133, term 3: 55, term 4: 250, term 5: 133, etc. So for term number n >= 2, the term is determined by (n - 2) mod 3.So term 2005: (2005 - 2) = 2003; 2003 mod 3. Let's compute 2003 / 3. 3*667 = 2001, so 2003 = 3*667 + 2. So 2003 mod 3 is 2. Therefore, (n - 2) mod 3 = 2, which corresponds to 250. So term 2005 is 250, which is option E.But wait, the answer choices given here include 250 as E. But looking back at the original problem, let me check again:The first term is 2005. The 2005th term. So the options are (A) 29, (B)55, (C)85, (D)133, (E)250.Wait, but according to my calculation, it's 250, which is E. But the problem is from an official source, perhaps an old AMC 10, and the answer might be different. Wait, maybe I made a miscalculation. Let's check again.Wait, maybe I made a mistake in calculating the cycle. Let me redo the first few terms.Term 1: 2005Term 2: 2³ + 0³ + 0³ + 5³ = 8 + 0 + 0 + 125 = 133. Correct.Term 3: 1³ + 3³ + 3³ = 1 + 27 + 27 = 55. Correct.Term 4: 5³ + 5³ = 125 + 125 = 250. Correct.Term 5: 2³ + 5³ + 0³ = 8 + 125 + 0 = 133. Correct.Term 6: same as term 3: 55Term 7: same as term 4: 250So cycle is indeed 133, 55, 250. So the cycle length is 3. So starting from term 2, every 3 terms it repeats.Therefore, term number 2: 133term 3:55term4:250term5:133term6:55term7:250...Therefore, for term n (n ≥ 2), the value is:133 when n ≡ 2 mod 3 (since 2, 5, 8,...)55 when n ≡ 0 mod 3 (since 3,6,9,...)250 when n ≡ 1 mod 3 (since 4,7,10,...)Wait, hold on. Wait, term 2:133, term3:55, term4:250, term5:133 (n=5), term6:55 (n=6), term7:250 (n=7). So term numbers and their modulo 3:Term2: n=2; 2 mod3=2 →133Term3: n=3;3 mod3=0 →55Term4:n=4;4 mod3=1→250Term5:n=5;5 mod3=2→133Term6:n=6;6 mod3=0→55Term7:n=7;7 mod3=1→250So the mapping is:n mod3=2 →133n mod3=0 →55n mod3=1 →250Therefore, to find term 2005, compute 2005 mod3.2005 divided by3: 3×668=2004, so 2005=3×668 +1. Therefore, 2005 mod3=1.Therefore, term2005 is 250, which is option E. So according to this, the answer is E)250.But wait, the original answer choices are given as:(A) 29(B)55(C)85(D)133(E)250So 250 is an option. But the problem is from an official source, and if I recall correctly, this problem might have a different answer. Wait, maybe I need to check if the cycle is actually different. Wait, perhaps I missed a step in the cycle?Wait, let's see. Starting from term1:2005term2:133term3:55term4:250term5:133term6:55term7:250Yes, so cycle is 133,55,250, repeating every 3 terms. So from term2 onwards, cycle length 3.Therefore, term numbers:2:1333:554:2505:1336:557:2508:133...So term number n:If n ≡2 mod3:133n≡0 mod3:55n≡1 mod3:250Therefore, term2005:2005 divided by3:3*668=2004, so remainder1. So 2005≡1 mod3, which maps to250. So answer is E)250.But the problem is given in a format where the answer is among the choices, and E is 250, so that's acceptable. Wait, but maybe the problem is different? Let me check the problem statement again."The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 2005th term of the sequence? (A)29, (B)55, (C)85, (D)133, (E)250"So according to my calculations, it's 250, which is E. But maybe I made a mistake in the cycle length? Let's confirm once more.Starting from term1:2005term2:133term3:55term4:250term5:133term6:55term7:250term8:133term9:55term10:250So the cycle is definitely 133,55,250. Therefore, every term starting from term2 is in a cycle of 3.Therefore, term2005 is 250. So the answer is E. However, in some versions of this problem, I remember the answer being different. Wait, maybe I miscalculated term2005's position?Wait, term1 is 2005, then term2 is 133, term3 is55, term4 is250, term5 is133, etc. So term2005 is the term after 2004 terms from term1. Since the cycle starts at term2, the number of terms in the cycle is 2005 -1 =2004 terms. Wait, no: term1 is the first term, term2 is the second, so term2005 is 2005th term, which is (2005 -1)=2004 terms after term1.But starting from term2, each cycle is 3 terms. So term2 is the first term in the cycle, term3 second, term4 third, term5 first, etc. So the number of terms in the cycle after term2 is 2005 -1 =2004 terms. Wait, term2 is the second term, so term2005 is (2005 -1)=2004 terms after term1, but since term2 is the start of the cycle, the number of terms in the cycle is 2005 -1 -1=2003 terms? Wait, maybe another approach.Alternatively, the number of terms after term1 is 2005 -1=2004 terms. Each cycle is 3 terms, so 2004 divided by3. 2004 ÷3=668, with remainder0. So 2004=3*668, so the 2004th term after term1 would be the last term in the cycle, which is250. Therefore, term2005=250. So that's consistent. So answer is250.Wait, but hold on. If there are 2004 terms after term1, which are part of the cycle, and 2004 is divisible by3 (since 2004=3*668), then the 2004th term after term1 would be the last term in the cycle, which is250. Therefore, term2005=250.But according to the problem, the answer is supposed to be one of the choices, and250 is an option. So unless there's a miscalculation here, I think the answer is250. But maybe in the problem's original source, the answer is different? Wait, let me check again the first few terms.Wait, term1:2005term2:2³ +0³+0³+5³=8+0+0+125=133term3:1³+3³+3³=1+27+27=55term4:5³+5³=125+125=250term5:2³+5³+0³=8+125+0=133term6:1³+3³+3³=55term7:5³+5³=250Yes, so the cycle is indeed133,55,250. So term numbers:2:1333:554:2505:1336:557:2508:133...So term numbers that are 2 mod3:133term numbers that are0 mod3:55term numbers that are1 mod3:250Term2005:2005 mod3.Since 2005 divided by3 is 668 with remainder1 (3*668=2004, 2005-2004=1). So2005 mod3=1. Therefore, term2005 is250.So answer isE)250. Hmm. But wait, perhaps there is a mistake in the problem statement? Or maybe I made a miscalculation. Wait, the problem says "the sum of the cubes of the digits". So, for term2, digits of term1 (2005) are2,0,0,5. Their cubes are8,0,0,125. Sum:8+0+0+125=133. Correct.Term3: digits of133 are1,3,3. Cubes:1,27,27. Sum:55. Correct.Term4: digits of55 are5,5. Cubes:125,125. Sum:250. Correct.Term5: digits of250 are2,5,0. Cubes:8,125,0. Sum:133. Correct.Yes, so cycle is confirmed.Alternatively, maybe the question is asking for the 2005th term starting from term1 as the first term. So term1:2005, term2:133,..., term2005: ?Yes, which is what I calculated. So term2005 is250. So answer isE.But in the original problem, answer might be different. Wait, maybe I have an old answer in mind. Let me check an external source. Wait, no, I should solve it myself. Alternatively, perhaps the cycle is longer? Let's see. Suppose there's another step beyond250. Let's check term5 again.Term5:133Term6:55Term7:250Term8:133Term9:55Term10:250No, it's definitely repeating every three terms. So cycle length3.Therefore, the answer is250, which is optionE.But the given options include55 and133 as well. But according to my calculation, it's250. Therefore, unless there's a miscalculation here, I think the answer isE. Alternatively, perhaps the problem is from a different source where the answer is55, but in this case, with the given terms,250 is correct.Wait, but let me check again term2005's position in the cycle. Starting from term2, cycle length3. Term2005 is term1 +2004 terms. Since term2 is the first term in the cycle, term2005 is term2 +2003 terms. 2003 divided by3 is667 cycles (3*667=2001) with remainder2. So term2005 is the third term in the cycle after667 full cycles. The third term in the cycle is250. Therefore, yes, term2005 is250. So answer isE. So I think the correct answer isE)250.But wait, in the original problem statement, the answer is one of the choices, and250 is an option. So unless I made a mistake in arithmetic, which I don't think I did, the answer should beE.Wait, but I just thought of something. Maybe there is a different cycle if you proceed further. Wait, term250: let's see what's after250.term8:133term9:55term10:250term11:133term12:55term13:250... So it's the same cycle. So no, the cycle doesn't change. So the answer is indeed250.Therefore, the correct answer isboxed{E}.Final Answerboxed{E}</think>The first term of the sequence is (2005). Each succeeding term is the sum of the cubes of the digits of the previous term. We need to find the (2005^{text{th}}) term of the sequence.1. Calculating the first few terms: - Term 1: (2005) - Term 2: (2^3 + 0^3 + 0^3 + 5^3 = 8 + 0 + 0 + 125 = 133) - Term 3: (1^3 + 3^3 + 3^3 = 1 + 27 + 27 = 55) - Term 4: (5^3 + 5^3 = 125 + 125 = 250) - Term 5: (2^3 + 5^3 + 0^3 = 8 + 125 + 0 = 133) - Term 6: (1^3 + 3^3 + 3^3 = 55) - Term 7: (5^3 + 5^3 = 250)2. Identifying the cycle: - The sequence enters a repeating cycle starting from term 2: (133, 55, 250). - The cycle length is 3.3. Determining the position of the (2005^{text{th}}) term in the cycle: - The cycle starts from term 2. Therefore, we need to find the position of term (2005) in the cycle. - Calculate ((2005 - 1) mod 3) to determine the position in the cycle (since term 2 is the first term in the cycle). - ((2005 - 1) = 2004) - (2004 mod 3 = 0), which corresponds to the third term in the cycle (since remainder 0 corresponds to the last term in the cycle).4. Mapping the result to the cycle: - The third term in the cycle (133, 55, 250) is (250).Thus, the (2005^{text{th}}) term of the sequence is (boxed{E}).]

🔑:Performing this operation several times yields the results of 133 for the second term, 55 for the third term, and 250 for the fourth term. The sum of the cubes of the digits of 250 equal 133, a complete cycle. The cycle is, excluding the first term, the 2^{text{nd}}, 3^{text{rd}}, and 4^{text{th}} terms will equal 133, 55, and 250, following the fourth term. Any term number that is equivalent to 1 (text{mod} 3) will produce a result of 250. It just so happens that 2005equiv 1 (text{mod} 3), which leads us to the answer of boxed{textbf{(E) } 250}.