🧠:Okay, so I need to prove that the number 0.a₁a₂a₃...aₙ..., which is formed by taking the units digit of the sum 1² + 2² + ... + n² for each n, is a rational number. Hmm, rational numbers have decimal expansions that either terminate or eventually repeat. So if I can show that this sequence of units digits eventually becomes periodic, then the number would be rational. That makes sense because a repeating decimal is rational.First, let me recall the formula for the sum of squares. The sum 1² + 2² + ... + n² is equal to n(n + 1)(2n + 1)/6. Right? Let me confirm that formula. Yeah, I think that's correct. So the sum up to n is given by that cubic polynomial. But we're interested in the units digit of this sum for each n. So for each n, compute S(n) = 1² + 2² + ... + n² mod 10, and then the decimal 0.a₁a₂a₃... is formed by these digits aₙ = S(n) mod 10.To show that 0.a₁a₂a₃... is rational, we need to show that the sequence {aₙ} is eventually periodic. Because if the sequence of digits after some point repeats every period k, then the decimal expansion is periodic, hence rational.So the key is to prove that the units digit of S(n) is periodic. Since we're dealing with modular arithmetic mod 10, perhaps there is a period after which the units digits repeat. Let's see.But first, let me see if the sequence aₙ is periodic. Let's compute some initial terms to check for a pattern.Compute S(n) mod 10 for n from 1 to, say, 20.Let me compute each term step by step:n=1: 1² = 1 → sum=1 → a₁=1n=2: 1² + 2² = 1 + 4 = 5 → a₂=5n=3: 5 + 9 = 14 → a₃=4n=4: 14 + 16 = 30 → a₄=0n=5: 30 + 25 = 55 → a₅=5n=6: 55 + 36 = 91 → a₆=1n=7: 91 + 49 = 140 → a₇=0n=8: 140 + 64 = 204 → a₈=4n=9: 204 + 81 = 285 → a₉=5n=10: 285 + 100 = 385 → a₁₀=5n=11: 385 + 121 = 506 → a₁₁=6n=12: 506 + 144 = 650 → a₁₂=0n=13: 650 + 169 = 819 → a₁₃=9n=14: 819 + 196 = 1015 → a₁₄=5n=15: 1015 + 225 = 1240 → a₁₅=0n=16: 1240 + 256 = 1496 → a₁₆=6n=17: 1496 + 289 = 1785 → a₁₇=5n=18: 1785 + 324 = 2109 → a₁₈=9n=19: 2109 + 361 = 2470 → a₁₉=0n=20: 2470 + 400 = 2870 → a₂₀=0So writing out the units digits from a₁ to a₂₀:1, 5, 4, 0, 5, 1, 0, 4, 5, 5, 6, 0, 9, 5, 0, 6, 5, 9, 0, 0Hmm, that seems a bit irregular. Let me check if I made any mistakes in calculation.Wait, let me verify some of these:At n=5: sum up to 5 is 55, which is correct. 55 mod 10 is 5.n=6: 55 + 36 = 91. 91 mod 10 is 1. Correct.n=7: 91 + 49 = 140. 140 mod 10 is 0. Correct.n=8: 140 + 64 = 204. 204 mod 10 is 4. Correct.n=9: 204 + 81 = 285. 285 mod 10 is 5. Correct.n=10: 285 + 100 = 385. 385 mod 10 is 5. Correct.n=11: 385 + 121 = 506. 506 mod 10 is 6. Correct.n=12: 506 + 144 = 650. 650 mod 10 is 0. Correct.n=13: 650 + 169 = 819. 819 mod 10 is 9. Correct.n=14: 819 + 196 = 1015. 1015 mod 10 is 5. Correct.n=15: 1015 + 225 = 1240. 1240 mod 10 is 0. Correct.n=16: 1240 + 256 = 1496. 1496 mod 10 is 6. Correct.n=17: 1496 + 289 = 1785. 1785 mod 10 is 5. Correct.n=18: 1785 + 324 = 2109. 2109 mod 10 is 9. Correct.n=19: 2109 + 361 = 2470. 2470 mod 10 is 0. Correct.n=20: 2470 + 400 = 2870. 2870 mod 10 is 0. Correct.Okay, so the calculations are correct. The sequence of units digits is as above. Now, looking for a pattern or periodicity here. Let's see.Looking at the first 20 terms:1, 5, 4, 0, 5, 1, 0, 4, 5, 5, 6, 0, 9, 5, 0, 6, 5, 9, 0, 0Is there a repeating cycle here? Let's see.Looking from n=1 to n=10: 1,5,4,0,5,1,0,4,5,5n=11 to n=20: 6,0,9,5,0,6,5,9,0,0Not immediately obvious. Maybe the period is longer? Let me check up to n=30.But calculating manually up to 30 might take time. Alternatively, perhaps we can look at the formula for S(n) mod 10.Given that S(n) = n(n + 1)(2n + 1)/6 mod 10.So if we can compute n(n + 1)(2n + 1)/6 mod 10, that gives us aₙ.But since we are dealing with mod 10, perhaps we can find a period in n such that n, n + 1, 2n + 1, and the division by 6 all repeat with some period.But division by 6 modulo 10 requires that 6 and 10 are coprime? Wait, gcd(6,10)=2, so division isn't always possible. Hmm, but S(n) is an integer, so when we compute n(n + 1)(2n + 1)/6, it must be an integer, so n(n + 1)(2n + 1) is divisible by 6. Therefore, modulo 10, it's equivalent to (n(n + 1)(2n + 1)/6) mod 10.But since n(n + 1)(2n + 1) is divisible by 6, then n(n + 1)(2n + 1)/6 is an integer, so when we take that integer mod 10, we get the unit digit.So perhaps instead of computing the sum each time, we can compute the formula mod 10. Let's see.But how to compute n(n + 1)(2n + 1)/6 mod 10?Let me note that 10 = 2 * 5. Maybe using the Chinese Remainder Theorem, compute modulo 2 and modulo 5, then combine.But first, let me note that if we can find a period modulo m, then the sequence will have period dividing m. Since mod 10, the period would be related to the least common multiple of the periods modulo 2 and modulo 5.Alternatively, compute S(n) mod 10 directly using the formula.Alternatively, notice that n, n+1, and 2n+1 are three consecutive numbers? Not exactly. Wait, n and n+1 are consecutive, so one of them is even, and among three consecutive numbers, one is divisible by 3. But here we have n, n+1, and 2n + 1. Hmm.But perhaps, instead of that, consider that the formula n(n + 1)(2n + 1)/6 is an integer. So for each n, n(n + 1)(2n + 1) is divisible by 6, so when divided by 6, gives an integer. Then, modulo 10, we can compute this as:First compute the product n(n + 1)(2n + 1) mod 60, then divide by 6, since 60 is 10 * 6. Because 6 * 10 = 60, so if we compute the product modulo 60, then divide by 6, we get the result modulo 10. Let me check.Wait, if we have S(n) = n(n + 1)(2n + 1)/6. Then, to compute S(n) mod 10, we can compute [n(n + 1)(2n + 1) mod 60] / 6 mod 10. Because 60 is the modulus such that when we divide by 6, the result mod 10 is the same as S(n) mod 10. Let me verify with an example.Take n=2:Product = 2*3*5 = 30. 30 mod 60 is 30. 30 /6 =5. 5 mod10=5. Which matches a₂=5.n=3:Product=3*4*7=84. 84 mod60=24. 24 /6=4. 4 mod10=4. Correct, a₃=4.n=4:4*5*9=180. 180 mod60=0. 0/6=0. 0 mod10=0. Correct, a₄=0.n=5:5*6*11=330. 330 mod60=30. 30 /6=5. 5 mod10=5. Correct, a₅=5.Okay, seems to work. So in general, to compute aₙ = S(n) mod10, compute n(n +1)(2n +1) mod60, then divide by6, then take mod10. Because 60 is the least common multiple of 6 and 10? Wait, 6 and 10 have lcm 30. But since we need to compute n(n +1)(2n +1)/6 mod10, which is equivalent to (n(n +1)(2n +1) mod60)/6 mod10. Let me check why 60.Because 10*6=60. So when we compute the product mod60, then divide by6, the result mod10 is the same as the original division by6 mod10. Because:Suppose x ≡ y mod60. Then x/6 ≡ y/6 mod10. Because x = y +60k, so x/6 = y/6 +10k, so x/6 ≡ y/6 mod10. Therefore, yes, computing the product mod60, dividing by6, gives the same result mod10.Therefore, aₙ = [ (n(n +1)(2n +1) mod60) /6 ] mod10.Therefore, the sequence aₙ is determined by n mod60, because the product n(n +1)(2n +1) mod60 depends only on n mod60. Because:If n ≡ m mod60, then n = m +60k. Then:n ≡ m mod60,n +1 ≡ m +1 mod60,2n +1 ≡ 2m +1 +120k ≡ 2m +1 mod60.Therefore, n(n +1)(2n +1) ≡ m(m +1)(2m +1) mod60.Therefore, the product depends only on n mod60, hence the term aₙ depends only on n mod60. Therefore, the sequence aₙ is periodic with period dividing60. Therefore, the sequence aₙ is periodic with period at most60. Therefore, the decimal expansion 0.a₁a₂a₃... is eventually periodic, hence rational.Therefore, the number is rational.Wait, but does that mean the period is 60? Or could it be shorter?In the example above, up to n=20, we didn't see a repetition, but the period could be 60. However, even if the period is 60, the decimal expansion would still be periodic, hence rational.Therefore, since aₙ is periodic with period 60, the decimal 0.a₁a₂a₃... is a repeating decimal, hence rational. Therefore, the number is rational.Alternatively, even if the period is less than60, the conclusion still holds.Therefore, the key idea is that the unit digit aₙ depends on n mod60, hence the sequence is periodic with period60, which makes the decimal expansion periodic, hence rational.Thus, the number is rational.Final AnswerThe number is rational, so the final answer is boxed{0.a_1a_2a_3ldots} text{ is rational} or in boxed form as per instruction. Wait, the question says "Prove that the number 0.a₁a₂⋯aₙ⋯ is a rational number." So the answer is simply stating that it's rational, hence boxed as boxed{0.a_1a_2a_3ldots} is a rational number. But the standard way is to present the conclusion as boxed{text{rational}} or in the problem's context, since they ask to prove it, but the answer should be boxed as per instruction. Maybe the user expects the boxed word "rational".But checking the problem statement again: "Prove that the number 0.a₁a₂⋯aₙ⋯ is a rational number." So the answer is that it's rational, and in the box, we put boxed{0.a_1a_2a_3ldots text{ is a rational number}}. But maybe in the standard math competition format, the answer is just boxed{text{rational}}.But in the instructions, it says "put your final answer within boxed{}". So perhaps the user expects the statement to be boxed. Maybe:boxed{0.a_1a_2a_3ldots text{ is a rational number}}But in Chinese problems, sometimes they expect just boxed{text{rational}}.But given the original problem states "the number 0.a₁a₂⋯aₙ⋯ is a rational number", so the answer to be proven is that it's rational. So the final box is just boxed{text{rational}}.Wait, let me check previous examples. For example, if the question is "Prove that ... is rational", then the answer is boxed{text{rational}} or sometimes boxed{0} if it's zero, but here since it's a statement to be proven, perhaps the answer is simply boxed{text{rational}}.Alternatively, maybe the problem expects the decimal itself to be identified as rational, which is done by proving periodicity, so the answer is boxed as rational. So I will go with boxed{text{rational}}.

✅:To prove that the number (0.a_1a_2a_3ldots) formed by the units digits of the sums (1^2 + 2^2 + cdots + n^2) for (n = 1, 2, 3, ldots) is rational, we need to show that the sequence ({a_n}) is eventually periodic. The sum (1^2 + 2^2 + cdots + n^2) is given by the formula (frac{n(n+1)(2n+1)}{6}). We are interested in the units digit of this sum, which is equivalent to computing (frac{n(n+1)(2n+1)}{6} mod 10).To analyze the periodicity, we consider the expression modulo 60. Since (60) is the least common multiple of (6) and (10), we compute (n(n+1)(2n+1) mod 60). This product modulo 60 depends only on (n mod 60). Dividing the result by 6 and taking modulo 10 gives the units digit (a_n).Because the product (n(n+1)(2n+1) mod 60) is periodic with period 60, the sequence (a_n) is also periodic with period dividing 60. Therefore, the decimal expansion (0.a_1a_2a_3ldots) is eventually periodic, hence rational.Thus, the number (0.a_1a_2a_3ldots) is rational, and the final answer is (boxed{text{rational}}).

🔑:1. Define the position digits: Let ( a_{n} ) be the unit's digit of the sum ( 1^{2} + 2^{2} + cdots + n^{2} ), for ( n = 1, 2, 3, ldots ).2. Examine the pattern in the unit's place of the squares: Define ( b_{n} ) to be the unit’s place digit of ( n^{2} ). Notice the relationship between units digits of squares with ( n ) and ( n+10 ): [ (n+10)^{2} = n^{2} + 20n + 100. ] Since ( 100 ) is a multiple of ( 10 ), the units place of ( (n+10)^2 ) is the same as that of ( n^{2} ), i.e., ( b_{n+10} = b_{n} ).3. Sum of units digits: The unit's digit of the sum ( sum_{k=1}^{n} k^{2} ) is given by the units digit of ( sum_{k=1}^{n} b_{k} ). Thus, the units digit of ( a_{n} ) is: [ a_n equiv b_1 + b_2 + cdots + b_n pmod{10}. ]4. Finding the units digits of the first 20 terms: Calculate the units digits of the first few terms of ( 1^2, 2^2, ldots, 20^2 ): | ( n ) | ( n^2 ) | Unit's digit (( b_n )) | |---------|----------|---------------------------| | 1 | 1 | 1 | | 2 | 4 | 4 | | 3 | 9 | 9 | | 4 | 16 | 6 | | 5 | 25 | 5 | | 6 | 36 | 6 | | 7 | 49 | 9 | | 8 | 64 | 4 | | 9 | 81 | 1 | | 10 | 100 | 0 | | 11 | 121 | 1 | | 12 | 144 | 4 | | 13 | 169 | 9 | | 14 | 196 | 6 | | 15 | 225 | 5 | | 16 | 256 | 6 | | 17 | 289 | 9 | | 18 | 324 | 4 | | 19 | 361 | 1 | | 20 | 400 | 0 | Summarize these as ( b_k ) values: [ a_{20} = sum_{k=1}^{20} b_k pmod{10} = 0. ]5. Detect periodicity: Using the periodicity in ( b_n ): [ a_{n+20} = sum_{k=1}^{n+20} b_k pmod{10} = left( sum_{k=1}^{20} b_k + sum_{k=21}^{n+20} b_k right) pmod{10}. ] Given ( a_{20} = 0 ), it simplifies to: [ a_{n+20} equiv sum_{k=21}^{n+20} b_k equiv sum_{k=1}^{n} b_k equiv a_n pmod{10}. ]6. Conclusion: Since (a_n) is periodic with a period of 20, the decimal (0.a_1a_2a_3 ldots) forms a repeating, hence rational, decimal representation.Therefore, we conclude that (0.a_1a_2 cdots a_n cdots) is a rational number.[boxed{text{Rational}}]

🧠:Okay, let's tackle this problem step by step. The question says that an even perfect square in the decimal system is of the form (overline{a b 1 a b}). We need to find what this perfect square is. First, let's understand the structure of the number. The number is a 5-digit number written as (overline{a b 1 a b}). Breaking this down, the digits are: - First digit: (a)- Second digit: (b)- Third digit: (1)- Fourth digit: (a)- Fifth digit: (b)So the number looks like (ab1ab) in digits. For example, if (a=2) and (b=3), the number would be 23123. But since it's a perfect square and even, the last digit must be even. The digits (b) is the last digit, so (b) must be even (0, 2, 4, 6, 8). Also, the number is a perfect square, so we need to find a 5-digit number that fits this pattern and is the square of some integer. Since it's a 5-digit number, the square root must be between 100 (100^2 = 10000) and 316 (316^2 = 99856). So we're looking for a number N where 100 ≤ N ≤ 316, such that N^2 has the form ab1ab.Let me note down the key points:1. The number is 5 digits: (overline{a b 1 a b}).2. It's an even perfect square, so the last digit (b) must be even.3. The third digit is 1.4. The first digit is (a), the second (b), third 1, fourth (a), fifth (b).So our number is structured as:- Position 1: a- Position 2: b- Position 3: 1- Position 4: a- Position 5: bLet me think about how to approach this. Since the number is a square, maybe we can narrow down possible values for (a) and (b) by considering properties of squares.First, since the third digit is 1, the number when written as N^2 must have 1 in the hundreds place (third digit from the right). Let's denote the number as (N^2 = 10000a + 1000b + 100*1 + 10a + b = 10000a + 1000b + 100 + 10a + b = (10000a + 10a) + (1000b + b) + 100 = 10010a + 1001b + 100).So, (N^2 = 10010a + 1001b + 100). Let's factor this expression a bit: 10010a = 1001*10*a, 1001b = 1001*b, and 100. So the expression can be written as 1001*(10a + b) + 100. Hmm, not sure if that's helpful yet.Alternatively, maybe we can look at possible values of (a) and (b). Since (a) is the first digit of a 5-digit number, (a) can be from 1 to 9. But since the number is even, the last digit (b) must be even (0,2,4,6,8). Also, (b) is the second digit, so it can be from 0 to 9, but with the constraint that the last digit is (b), so (b) must be even. Moreover, the square's last digit is (b), so the square N^2 ends with (b). Since (b) is even, possible values for (b) are 0, 2, 4, 6, 8. Let's recall that the last digit of a square number depends on the last digit of N. For even squares:- If N ends with 0, then N^2 ends with 0.- If N ends with 2 or 8, then N^2 ends with 4.- If N ends with 4 or 6, then N^2 ends with 6.- If N ends with 1, 3, 7, 9, the square ends with 1, 9, 9, 1 respectively (but since N is even, these are not considered here).Wait, since N is even, it must end with 0, 2, 4, 6, or 8. Therefore, the square N^2 will end with 0, 4, 6, or 6, 4 respectively. So possible (b) values (last digit of square) are 0,4,6. Because:- If N ends with 0: N^2 ends with 0.- If N ends with 2 or 8: N^2 ends with 4.- If N ends with 4 or 6: N^2 ends with 6.So possible (b)s are 0,4,6. So (b in {0,4,6}).That narrows down possible (b) to 0,4,6. Let's keep that in mind.Also, the second digit is (b), so the tens place of the square is (b). Wait, actually, in the number (overline{a b 1 a b}), the second digit is (b), which is the ten-thousands place? Wait, no: in a 5-digit number, the digits are:- Position 1: Ten thousands place (digit a)- Position 2: Thousands place (digit b)- Position 3: Hundreds place (1)- Position 4: Tens place (digit a)- Position 5: Units place (digit b)So the number is:a*10^4 + b*10^3 + 1*10^2 + a*10 + b.So the square N^2 must equal this number.Since the third digit (hundreds place) is 1, the hundreds digit of N^2 is 1.So, if we can find N such that when squared, the hundreds digit is 1, the tens digit is a, the units digit is b, the thousands digit is b, and the ten-thousands digit is a.Given that N is between 100 and 316, as the square is 5 digits.Now, since the last digit of N^2 is b (0,4,6), the last digit of N must be such that when squared, it gives b. Let's consider possible endings:Case 1: b=0. Then N must end with 0 (since 0^2=0). So N is a multiple of 10. Let's consider N=...0. Then N^2 ends with 00, but wait, the last digit is b=0, and the tens digit is a. But in this case, N^2 would have last two digits as 00. Therefore, the tens digit (position 4) would be 0. But position 4 is a. So a=0. But a is the first digit of a 5-digit number, which can't be 0. Therefore, b cannot be 0. So we can eliminate b=0.Case 2: b=4. Then N must end with 2 or 8, since 2^2=4 and 8^2=64. So N ends with 2 or 8. Then the last digit of N^2 is 4, and the tens digit (position 4) is a. Let's note that if N ends with 2 or 8, the last two digits of N^2 would be something like x4, where x is the tens digit. For example, 12^2=144, so last two digits 44. 18^2=324, last two digits 24. So the tens digit (position 4) is determined by the square's tens digit. So in this case, a must equal the tens digit of N^2. For example, if N ends with 2, then N=...k2, and N^2=...k^2*? Wait, maybe a better approach is to note that if N ends with 2 or 8, then the last two digits of N^2 can be calculated.Let me recall that for numbers ending with 2:(10k + 2)^2 = 100k^2 + 40k + 4. So the last two digits are 40k + 4. So the tens digit is (40k + 4)/10 mod 10, which is (4k + 0.4) but since k is an integer, 40k + 4 will have the tens digit as (4k + 0) mod 10. Wait, 40k is a multiple of 10, so 40k mod 100 is 40k mod 100. Then 40k +4 mod 100 is (40k +4). Therefore, the tens digit is (40k +4)/10 mod 10 = (4k + 0.4) but since we are dealing with integers, it's actually (40k +4)/10 = 4k + 0.4, but the tens digit is the integer part of (40k +4)/10 divided by 10? Wait, maybe better to take specific examples.Let's take N=12: 12^2=144. Last two digits 44. So tens digit is 4. Then a would have to be 4. Similarly, N=22: 22^2=484. Last two digits 84. Tens digit 8. a=8. N=32: 32^2=1024. Last two digits 24. Tens digit 2. a=2. N=42: 42^2=1764. Last two digits 64. Tens digit 6. a=6. Similarly, N=52: 52^2=2704. Last two digits 04. Tens digit 0. a=0. But a can't be 0. N=62: 3844. Last two digits 44. Tens digit 4. a=4. N=72: 5184. Last two digits 84. a=8. N=82: 6724. Last two digits 24. a=2. N=92: 8464. Last two digits 64. a=6. N=102: 10404. Last two digits 04. a=0. Not valid. N=112: 12544. Last two digits 44. a=4. Etc.Similarly, if N ends with 8, let's compute (10k +8)^2 = 100k^2 + 160k + 64. The last two digits are 60k + 64. For example, N=18: 18^2=324. Last two digits 24. Tens digit 2. So a=2. N=28: 784. Last two digits 84. a=8. N=38: 1444. Last two digits 44. a=4. N=48: 2304. Last two digits 04. a=0. N=58: 3364. Last two digits 64. a=6. N=68: 4624. Last two digits 24. a=2. N=78: 6084. Last two digits 84. a=8. N=88: 7744. Last two digits 44. a=4. N=98: 9604. Last two digits 04. a=0. N=108: 11664. Last two digits 64. a=6. Etc.So when N ends with 2 or 8, the tens digit (a) cycles through 4,8,2,6,0,4,8,2,6,0,... depending on k. However, a cannot be 0. So possible values for a when b=4 are 2,4,6,8.Similarly, if b=6, then N must end with 4 or 6, since 4^2=16 and 6^2=36. Therefore, if N ends with 4 or 6, the last digit of N^2 is 6. Let's analyze this case.For N ending with 4: (10k +4)^2 = 100k^2 +80k +16. The last two digits are 80k +16. So the tens digit is (80k +16)/10 mod 10 = (8k +1.6) => integer part 8k +1, so the tens digit is (8k +1) mod 10. Let's test with examples.N=14: 14^2=196. Last two digits 96. Tens digit 9. a=9. But since the number is even, a must be even? Wait, no, the number itself is even, but the digits can be any except that the last digit is even. The other digits can be any. Wait, the problem states it's an even perfect square, which means it's divisible by 2, so the last digit is even. But the other digits can be any. So a can be from 1-9 (first digit), and in the case of N ending with 4, a could be 9 here, but in our pattern, the tens digit is a, which must be a digit from 0-9, but also the fourth digit is a, which is the tens digit. However, the first digit is a as well, so a can't be 0, but can be 1-9. So in this case, if a=9, that's allowed. But wait, the number is of the form ab1ab. So if a=9 and b=6, the number would be 9 6 1 9 6, which is 96196. Let's check if that's a square. Maybe, but let's see.Wait, let's take N=14: 196. But that's a 3-digit number. Let's take a larger N. Let's take N=104: 104^2=10816. Last two digits 16. So a=1 (tens digit is 1), but the number is 1 0 1 1 0? Wait, no. Wait, N=104: 10816. The digits are 1,0,8,1,6. So the third digit is 8, not 1. Doesn't fit.Wait, maybe better to take N= 224: 224^2=50176. Last two digits 76. So a=7 (tens digit). The number is 5 0 1 7 6. But the third digit is 1, so maybe? Wait, 50176: digits are 5,0,1,7,6. So the third digit is 1. So that fits the pattern a=5, b=0, but here a=5, b=0, but the tens digit is 7, which should be a=5. Wait, inconsistency here. Wait, the number is 50176. The digits are:Position 1: 5 (a)Position 2: 0 (b)Position 3: 1Position 4: 7 (a)Position 5: 6 (b)But according to the pattern, position 4 should be a=5, but here it's 7. So that doesn't fit.Wait, so perhaps when b=6, a is determined by the tens digit of the square. Let's see. For N ending with 4:Take N= 34: 34^2=1156. Last two digits 56. So a=5. But N=34 is 1156, which is a 4-digit number. Doesn't fit.N= 44: 44^2=1936. Last two digits 36. So a=3. But 1936 is a 4-digit number.N= 54: 54^2=2916. Last two digits 16. a=1. The number 2916: digits 2,9,1,6. But it's a 4-digit number. Doesn't fit.N= 64: 4096. 4-digit number. Last two digits 96. a=9.N= 74: 5476. 4-digit. Last two digits 76. a=7.N=84: 7056. Last two digits 56. a=5. Number is 7056: 7,0,5,6. 4-digit.N=94: 8836. Last two digits 36. a=3.N=104: 10816. 5-digit. Last two digits 16. a=1. The number is 1 0 8 1 6. Third digit is 8, not 1. Doesn't fit.N=114: 12996. Wait, 114^2=12996. Let's check: 100^2=10000, 110^2=12100, 114^2=12100 + 2*110*4 + 4^2 = 12100 + 880 + 16= 12996. So digits 1,2,9,9,6. Third digit is 9. Doesn't fit.N=124: 124^2=15376. Last two digits 76. a=7. The number is 1 5 3 7 6. Third digit 3. Doesn't fit.Hmm, seems challenging. Let's check N= 224: as before, 50176. Third digit is 1. Wait, 50176: digits 5,0,1,7,6. So third digit is 1. Wait! That does have a 1 in the third position. Let's verify:224 squared is 224*224. Let me calculate:224*200=44800224*24=5376So total 44800 +5376=50176. Correct. So 50176 is 5 0 1 7 6. So according to the pattern, a should be 5 (first digit), b should be 0 (second digit), third digit 1, fourth digit a=5, fifth digit b=0. But the actual number is 5 0 1 7 6. So the fourth digit is 7 instead of 5, which contradicts. So this is not matching.But wait, the problem says the number is of the form ab1ab. So first digit a, second b, third 1, fourth a, fifth b. So in the number 50176, the digits are 5,0,1,7,6. So a=5, b=0, fourth digit should be a=5, but it's 7. Therefore, this is not a match.So even though the third digit is 1, the fourth digit doesn't match a. Therefore, this is not the number we're looking for.Similarly, let's check other numbers where the third digit is 1.Wait, perhaps we can narrow down the candidates by considering that the number is of the form ab1ab. So, the number can be written as 10000a + 1000b + 100*1 + 10a + b = 10000a + 1000b + 100 + 10a + b = (10000a + 10a) + (1000b + b) + 100 = 10010a + 1001b + 100.So N^2 = 10010a + 1001b + 100.Factorizing 1001, we know 1001 = 7*11*13. So 1001b is 7*11*13*b. Similarly, 10010a = 10*1001a. So:N^2 = 10*1001a + 1001b + 100 = 1001*(10a + b) + 100.Not sure if that helps directly, but maybe this can be factored or give some modular constraints.Alternatively, perhaps we can set up the equation N^2 = 10010a + 1001b + 100, where a ∈ {1,...,9}, b ∈ {0,4,6}.But maybe instead of algebra, we can approach this by considering possible a and b values.Given that the number is ab1ab, let's list possible values for a (1-9) and b (0,4,6), construct the number, and check if it's a perfect square.There are 9*3=27 possible combinations. Let's see:Start with b=4:For b=4, the number is a41a4. So positions: a,4,1,a,4.So numbers like 14114, 24124, ..., 94194.Similarly, for b=6: a61a6. So 16116, 26126,...,96196.For b=0: a01a0. But since b=0, the second digit is 0, last digit 0. The number is a01a0. For example, 10110, 20120,...,90190.But earlier we saw that if b=0, then the tens digit (a) would have to be 0, but a is the first digit, which can't be 0. So b=0 is invalid. Wait, no. Wait, if b=0, the number is a01a0. The tens digit is a, which is the fourth digit. But the first digit is a, so a can be from 1-9. So for example, 10110: a=1, b=0. The tens digit is a=1, which matches. But the number ends with 0, which is even, so it's a possible candidate. But wait, the square would have to end with 0, so N must be a multiple of 10. Let's check if such a number exists.Take a=1, b=0: 10110. Is 10110 a perfect square? The square root of 10110 is approximately 100.55, which is not an integer. 100^2=10000, 101^2=10201. So 10110 isn't a square.Similarly, a=2: 20120. Square root is about 141.84. 141^2=19881, 142^2=20164. 20164-20120=44. Not a square.a=3:30130. Square root ~173.6. 173^2=29929, 174^2=30276. Not a square.a=4:40140. Square root ~200.35. 200^2=40000, 201^2=40401. Not a square.Similarly, continuing up to a=9:90190. Square root ~300.316. 300^2=90000, 301^2=90601. Not a square.Therefore, b=0 gives no solutions. So focus on b=4 and b=6.Now let's check b=4:Numbers are a41a4. For a=1 to 9:1:141142:241243:341344:441445:541546:641647:741748:841849:94194Check which of these are perfect squares.Compute square roots:14114: sqrt(14114) ≈ 118.8. 118^2=13924, 119^2=14161. Not a square.24124: sqrt ≈ 155.3. 155^2=24025, 156^2=24336. Not a square.34134: sqrt ≈ 184.7. 184^2=33856, 185^2=34225. Not a square.44144: sqrt ≈ 210.1. 210^2=44100, 211^2=44521. Not a square.54154: sqrt ≈ 232.7. 232^2=53824, 233^2=54289. Not a square.64164: sqrt ≈ 253.3. 253^2=64009, 254^2=64516. Not a square.74174: sqrt ≈ 272.3. 272^2=73984, 273^2=74529. Not a square.84184: sqrt ≈ 290.1. 290^2=84100, 291^2=84681. Not a square.94194: sqrt ≈ 306.9. 306^2=93636, 307^2=94249. Not a square.So none of the b=4 cases work. Now check b=6:Numbers are a61a6. For a=1 to 9:1:161162:261263:361364:461465:561566:661667:761768:861869:96196Check each:16116: sqrt ≈ 127.0. 127^2=16129. Close. 126^2=15876. 127^2=16129. So 16116 is not a square.26126: sqrt ≈ 161.6. 161^2=25921, 162^2=26244. Not a square.36136: sqrt ≈ 190.1. 190^2=36100, 191^2=36481. Not a square.46146: sqrt ≈ 214.8. 214^2=45796, 215^2=46225. Not a square.56156: sqrt ≈ 237.0. 237^2=56169, 236^2=55696. Not a square.66166: sqrt ≈ 257.2. 257^2=66049, 258^2=66564. Not a square.76176: sqrt ≈ 276.0. 276^2=76176. Wait, 276*276=?Let me calculate 276^2:200^2=4000070^2=49006^2=36Cross terms: 2*200*70=28000, 2*200*6=2400, 2*70*6=840So total: 40000 + 4900 +36 +28000 +2400 +840 = 40000 + 4900 = 44900 +36=44936 +28000=72936 +2400=75336 +840=76176. Yes! So 276^2=76176. Which is the number 76176. Let's check the digits:7 (a=7), 6 (b=6), 1, 7 (a=7), 6 (b=6). So indeed, the number is 7 6 1 7 6, which fits the pattern (overline{a b 1 a b}) with a=7 and b=6. Wait, that seems to fit perfectly. So 76176 is a perfect square (276^2), it's even (ends with 6, which is even), and has the form a b 1 a b with a=7 and b=6. But let's verify this. 276 squared is indeed 76176. Let's confirm:276 * 276:Break it down:270 * 270 = 72900270 * 6 = 16206 * 270 = 16206 * 6 = 36So total: 72900 + 1620 + 1620 +36 = 72900 + 3240 +36 = 72900 + 3276 = 76176. Correct.So 276^2=76176, which is 76176. The digits are 7,6,1,7,6. So a=7, b=6. Therefore, this fits the pattern (overline{a b 1 a b}) = 76176.Thus, the answer is 76176.But wait, let's check the other numbers in case there are multiple solutions.Continuing with b=6:After 76176, check 86186:sqrt ≈ 293.4. 293^2=85849, 294^2=86436. Not a square.96196: sqrt ≈310.1. 310^2=96100, 311^2=96721. Not a square.Therefore, only 76176 fits the criteria. Therefore, the even perfect square of the form ab1ab is 76176, which is 276 squared.

✅:The even perfect square of the form (overline{a b 1 a b}) is 76176. Step-by-Step Explanation:1. Structure Analysis: The number is a 5-digit palindrome-like form (overline{a b 1 a b}), where (a) and (b) are digits. Key constraints: - It is an even perfect square, so the last digit (b) must be even (0, 2, 4, 6, 8). - The third digit is fixed as 1.2. Range Determination: Since it's a 5-digit number, its square root (N) must satisfy (100 leq N leq 316).3. Digit Constraints for (b): - (b) is the units digit of the square, so (b in {0, 4, 6}) (since even squares end in 0, 4, or 6). - Eliminating (b = 0) (as it forces the first digit (a = 0), invalid for a 5-digit number), focus shifts to (b = 4) and (b = 6).4. Formulating the Number: - For (b = 4), the number is (a41a4). Testing all (a) values (1-9) reveals none are perfect squares. - For (b = 6), the number is (a61a6). Testing (a = 7) gives (76176).5. Verification: - Calculate (276^2 = 76176), confirming it fits the structure (overline{7 6 1 7 6}). - Check that all digits align: (a = 7), (b = 6), third digit 1, and the number is even.Answer: (boxed{76176})

🔑:We are given that the number is an even square number in the form overline{ab1ab} in the decimal system. We need to find which square number fits this description.1. Define the number: Let the required number be 2x, where x is some integer. Therefore, the square of this number will be: [ (2x)^2 = 4x^2 ] According to the given problem: [ 4x^2 = 10000a + 1000b + 100 + 10a + b = 10000a + 1000b + 100 + 10a + b ] Simplifying: [ 4x^2 = 10010a + 1001b + 100 ] [ 4x^2 - 100 = 10010a + 1001b ]2. Factors of 1001: Recognize that: [ 1001 = 7 times 11 times 13 ]3. Transform equation: Since 1001 = 7 times 11 times 13, factor the equation as follows: [ 4 left( x^2 - 25 right) = 7 times 11 times 13 times (10a + b) ] Let: [ 10a + b = 4y quad text{(since x is a multiple of 2)} ]4. Condition on y: Given: [ 10a + b < 100 quad text{(since 10a + b is a two-digit number)} ] It follows: [ y < 25 ]5. Form equation in x: We know: [ left( x + 5 right) left( x - 5 right) = 7 times 11 times 13 times y ] The task now is to find factors that satisfy the equation, distinguishing the contributions of 7, 11, 13, and y.6. Possible factorizations: Consider possible products of 7, 11, 13 with y < 25 - If one part is 7 times 11 = 77, the other should be 91y rightarrow text{invalid as 91 is not within defined range} - If one part is 7 times 13 = 91, the other should be 77y rightarrow text{invalid as 77 is not prime} - If one part is 11 times 13 = 143, then: [ 7y = 133 quad implies quad y = 19 ]7. Verifying with the value of y: Therefore, [ y = 19 quad Rightarrow quad 10a + b = 4y = 76 ] Given a and b such that 10a + b = 76: [ a = 7, quad b = 6 ]8. Final calculations: [ begin{aligned} x^2 - 25 &= 7 times 11 times 13 times 19 = 19019 x^2 &= 19044 quad Rightarrow quad x = 138 2x &= 276 end{aligned} ] Now: [ 4x^2 = 276^2 = 76176 ]In conclusion, we verify that:[4x^2 = 276^2 = 76176]which fits the form overline{ab1ab}, where a=7 and b=6.boxed{76176}

🧠:Okay, let's see. The problem is asking two things: first, how many six-digit numbers have all distinct digits, and second, of those, how many have exactly four odd digits. Hmm. Let me start with the first part.So, six-digit numbers. That means the number can't start with zero, right? Because if the first digit is zero, it's not a six-digit number anymore. So, the first digit has to be from 1 to 9. Then the rest can be from 0 to 9, but all digits need to be distinct. So, for the first digit, we have 9 choices. Then the second digit can be any of the remaining 9 digits (since one is already used, and now zero is allowed), the third digit would have 8 choices, and so on, decreasing by one each time. Wait, let me write that out.First digit: 9 options (1-9)Second digit: 9 options (0-9 except the first digit)Third digit: 8 optionsFourth digit: 7 optionsFifth digit: 6 optionsSixth digit: 5 optionsSo, multiplying those together: 9 * 9 * 8 * 7 * 6 * 5. Let me compute that. Let's see: 9*9 is 81, then 81*8=648, 648*7=4536, 4536*6=27216, 27216*5=136080. So, 136,080 six-digit numbers with all distinct digits. That seems right. Let me check another way. The formula for permutations of n things taken k at a time is n! / (n-k)!. But here, the first digit is restricted. So, total numbers = 9 * 9 * 8 * 7 * 6 * 5, which is 9 * 9P5. Wait, 9P5 is 9*8*7*6*5, so yes, 9*9P5 = 9*15120 = 136080. Yep, same result. Okay, so the first part is 136,080.Now, the second part: how many of these six-digit numbers have exactly four odd digits. All digits are distinct, so we need to ensure that exactly four of the six digits are odd, and the remaining two are even. Also, digits must not repeat.First, let's recall that the digits are 0-9. Odd digits are 1,3,5,7,9 (five digits), and even digits are 0,2,4,6,8 (five digits). So, there are five odd and five even digits.But since the first digit cannot be zero, we have to be careful when considering positions. The four odd digits can be placed in any of the six positions, but we have to ensure that the first digit is not zero. However, since odd digits don't include zero, if we choose to have an even digit in the first position, it can't be zero. Wait, but the problem is that even digits include zero. So, if the first digit is even, it must be one of 2,4,6,8. So, the even digits for the first position are only four options (excluding zero). But other even digits (positions 2-6) can include zero.Therefore, when constructing numbers with exactly four odd digits, we need to consider two cases:Case 1: The first digit is odd. Then, we have three more odd digits in the remaining five positions, and two even digits in the remaining positions.Case 2: The first digit is even (but not zero). Then, we have four odd digits in the remaining five positions, and one even digit in the remaining positions.Wait, hold on. The total number of odd digits is four. So, depending on whether the first digit is odd or even, the distribution changes.If the first digit is odd, then we need three more odd digits in the remaining five digits. If the first digit is even, then all four odd digits must be in the remaining five digits. Since the first digit can't be zero, if it's even, it's one of 2,4,6,8. So, let's break it down into these two cases.Case 1: First digit is odd. Then, three of the remaining five digits are odd, and two are even.Case 2: First digit is even. Then, four of the remaining five digits are odd, and one is even.We can compute the number of possibilities for each case and sum them.Let's start with Case 1: First digit is odd.First digit: 5 choices (1,3,5,7,9).Now, we need three more odd digits and two even digits in the remaining five positions. But all digits must be distinct.So, first, how many ways to choose the three odd digits and two even digits.But wait, since digits have to be distinct, we need to choose three different odd digits from the remaining four odd digits (since one odd digit is already used in the first position), and two different even digits from all five even digits (since even digits haven't been used yet).Wait, is that right? Let's think. If the first digit is odd, say we picked 1, then the remaining odd digits available are 3,5,7,9 (four left). So, for the three remaining odd digits, we need to choose three distinct ones from these four. The number of ways is C(4,3) for choosing which three, then permuting them in the five positions. Wait, but actually, we need to assign digits to specific positions. Hmm. Maybe a better approach is:Total process:1. Choose the first digit: 5 options (odd).2. Choose three positions out of the remaining five for the odd digits. The number of ways to choose positions: C(5,3).3. Assign the three remaining odd digits to these positions. Since we already used one odd digit, there are 4 left. So, permutations: P(4,3) = 4*3*2 = 24.4. Assign the two even digits to the remaining two positions. Since even digits include 0, but these positions can be any even digits not already used. Wait, but we haven't used any even digits yet, right? Because the first digit is odd, so all even digits (0,2,4,6,8) are available. However, in these two positions (which are not the first digit), we can use any even digits, including zero, but they must be distinct. So, we need to choose two distinct even digits and arrange them in the two positions. The number of ways is P(5,2) = 5*4 = 20.So, multiplying these together for Case 1:5 (first digit) * C(5,3) * P(4,3) * P(5,2)Let me compute that:C(5,3) is 10.P(4,3) is 24.P(5,2) is 20.So, 5 * 10 * 24 * 20.First, 5*10=50, 50*24=1200, 1200*20=24,000.Case 1: 24,000.Case 2: First digit is even (non-zero).First digit: 4 choices (2,4,6,8).Then, we need four odd digits in the remaining five positions and one even digit. Wait, no. Wait, the total number of odd digits is four. Since the first digit is even, the four odd digits must be in the remaining five positions. So, four of the five remaining digits are odd, and one is even.But all digits must be distinct. So:1. Choose first digit: 4 options (even, non-zero).2. Choose four positions out of the remaining five for the odd digits. Wait, but since we need four odd digits in the remaining five positions, the number of ways to choose positions is C(5,4) = 5.3. Assign the four odd digits to these positions. Since no odd digits have been used yet (first digit was even), we have all five odd digits (1,3,5,7,9) available, but we need to choose four distinct ones. The number of permutations is P(5,4) = 5*4*3*2 = 120.4. Assign the one even digit to the remaining position. But wait, we already used one even digit (the first digit). So, the remaining even digits are 0 and the other three even digits not used in the first position. Wait, no: first digit is even (2,4,6,8). So, the remaining even digits are 0 and the remaining three even digits (since one was used in the first digit). For example, if first digit is 2, the remaining even digits are 0,4,6,8. So, four options. But we need to choose one even digit for the remaining position, which is distinct from the first digit. Therefore, the number of options is 4 (since one even digit was used, and zero is allowed here).Wait, let me clarify. The even digits are 0,2,4,6,8. If the first digit is one of 2,4,6,8, then the remaining even digits are 0 and the other four minus one. So, for example, if first digit is 2, remaining even digits are 0,4,6,8. So, four choices. Similarly, if first digit is 4, remaining even digits are 0,2,6,8. So, in general, for the remaining even digit (since we need exactly one even digit in the remaining five positions), we have 4 choices (since one even digit is already used, and zero is allowed here). But we have to place this remaining even digit in one of the five positions, but since we have already allocated four positions to odd digits, the remaining position is fixed. Wait, no: in step 2, we choose four positions for odd digits, leaving one position for the even digit. Therefore, the even digit is assigned to that specific position.But we need to make sure that this even digit is distinct from the first digit and from other digits. Since we are using distinct digits, the even digit in the remaining position can't be the same as the first digit, but can be zero. So, the number of choices is 5 even digits (0,2,4,6,8) minus the one used in the first digit. So, 4 choices.Therefore, the number of ways for Case 2:4 (first digit) * C(5,4) * P(5,4) * 4Compute that:C(5,4) is 5.P(5,4) is 120.So, 4 * 5 * 120 * 4.First, 4*5=20, 20*120=2400, 2400*4=9600.Case 2: 9,600.Therefore, total numbers with exactly four odd digits: 24,000 + 9,600 = 33,600.Wait, let me verify that again.Case 1:First digit: 5 options (odd).Positions for remaining three odds: C(5,3) = 10.Permutations for odds: P(4,3) = 24 (since one odd already used).Permutations for evens: P(5,2) = 20 (all evens available, including zero, since first digit is odd and no evens used yet).So, 5 * 10 * 24 * 20 = 24,000.Case 2:First digit: 4 options (even non-zero).Positions for four odds: C(5,4) = 5.Permutations for odds: P(5,4) = 120 (since no odds used yet).Permutations for the remaining even digit: 4 options (since one even is used in first digit, and remaining evens are 5-1=4, including zero).But wait, in this case, since we have only one even digit to place, it's just 4 choices for that digit, multiplied by the number of arrangements. Wait, but once we choose the position (which is fixed once we choose the four positions for odds), the even digit is placed in the remaining position. So, it's 4 choices for the even digit.Therefore, total for Case 2: 4 * 5 * 120 * 4. Wait, but hold on, why is there a *4 at the end?Wait, no. Let's clarify:First digit: 4 choices.Choosing positions for the four odd digits: C(5,4) = 5. So, 5 ways to choose where the odds go.Then, assign the four odd digits: P(5,4) = 120. Because we have five odd digits (1,3,5,7,9) and we need to choose four distinct ones and arrange them in the four chosen positions.Then, the remaining one position (which was not chosen for the odd digits) must be an even digit. Since the first digit is already an even digit (2,4,6,8), we cannot reuse that. So, available even digits are 0 and the remaining four even digits minus the one used in the first digit. Wait, the first digit is one of 2,4,6,8, so the remaining even digits are 0 and the other three. For example, if first digit is 2, remaining evens are 0,4,6,8. That's four options. Similarly, for any first digit, the remaining evens are 0 and three others. So, 4 options.Therefore, for the remaining even digit, we have 4 choices.Therefore, total for Case 2 is 4 (first digit) * [C(5,4) * P(5,4)] * [1 * 4].Breaking it down:4 * [5 * 120] * [4] = 4 * 600 * 4 = 4 * 2400 = 9600. Yes, that's right.Therefore, total is 24,000 + 9,600 = 33,600.Hmm. Let me check if there's another way to compute this.Alternative approach:Total six-digit numbers with all distinct digits: 136,080.But maybe we can compute the number of such numbers with exactly four odd digits directly.First, decide how many odd digits are in the number. Exactly four. So, we need four odd digits and two even digits.But we have to consider the first digit cannot be zero. So, we can divide into two cases based on whether the first digit is odd or even, as before.Wait, but maybe another way: calculate the total number without considering position first, then adjust for the first digit.But perhaps not. Let's see.Alternatively, think of it as:Choose four positions to be odd digits and two positions to be even digits. However, the first digit cannot be zero, so if an even digit is in the first position, it must be 2,4,6,8.So, split into two scenarios:1. The first digit is odd: Then, three of the remaining five digits are odd, and two are even (but the even digits can include zero).2. The first digit is even: Then, the remaining five digits include four odd and one even (which can be zero).This is similar to before.But maybe let's compute it using combinations.Total number of ways:Number of ways = (number of ways where first digit is odd and three more odds) + (number of ways where first digit is even and four more odds).Which is exactly what we did before.So, seems like 24,000 + 9,600 = 33,600 is the answer.Wait, let me confirm with another method.Suppose we ignore the first digit restriction for a moment. Then, the number of six-digit numbers with all distinct digits and exactly four odd digits would be:Choose four positions for odd digits, two for even. But first digit cannot be zero.Wait, but even digits include zero, so if one of the even digits is in the first position, it can't be zero. Hmm.Alternatively, total number of numbers (allowing leading zero) with four odd digits: C(6,4) * P(5,4) * P(5,2). But wait, no. Wait, if we allow leading zero, first digit can be zero. But we need to exclude those cases where the first digit is zero. Hmm. Maybe this is more complicated.Alternatively, total numbers with four odd digits, all distinct, allowing leading zeros:First, choose four positions for odd digits. For each choice:- Assign four distinct odd digits: P(5,4) = 120.- Assign two distinct even digits: P(5,2) = 20.Number of ways: C(6,4) * 120 * 20 = 15 * 120 * 20 = 15*2400=36,000.But this includes numbers where the first digit is zero. So, how many of these have first digit zero?If the first digit is zero, then we have five remaining digits with exactly four odd digits. So:First digit is zero (even digit). Then, we need four odd digits and one even digit in the remaining five positions.Number of such numbers:Choose four positions out of the remaining five for odd digits: C(5,4)=5.Assign four distinct odd digits: P(5,4)=120.Assign one even digit (since first digit is zero, which is even, remaining even digits are 2,4,6,8. So, four choices. Assign to the remaining position: 4 options.So, total numbers with first digit zero: 1 (for zero) * 5 (positions) * 120 (odds) * 4 (evens) = 1*5*120*4=2400.Therefore, total valid numbers (without leading zero) would be 36,000 - 2,400 = 33,600. Which matches our previous result. So, that's a good check.Therefore, the answer for the second part is 33,600.So, summarizing:Total six-digit numbers with all distinct digits: 136,080.Among these, numbers with exactly four odd digits: 33,600.I think that's correct. Let me just recap the steps to ensure no oversight.First part: 9 options for first digit, then 9,8,7,6,5 for the rest: 9*9*8*7*6*5=136080. Correct.Second part: calculated using two cases, got 24k +9.6k=33.6k, verified by an alternative method subtracting invalid numbers from total allowing leading zeros. Got same result. Seems solid.Final AnswerThe total number of six-digit numbers with all distinct digits is boxed{136080}, and the number of those with exactly four odd digits is boxed{33600}.

✅:To determine the number of six-digit numbers with all distinct digits, we consider the following:1. The first digit cannot be zero, so there are 9 choices (1-9).2. The remaining five digits can be any of the remaining 9, 8, 7, 6, and 5 digits respectively (since each subsequent digit must be distinct).Calculating this, we get:[9 times 9 times 8 times 7 times 6 times 5 = 136080]Next, we need to find how many of these six-digit numbers have exactly four odd digits. We consider two cases:Case 1: The first digit is odd.1. Choose the first digit (5 choices: 1, 3, 5, 7, 9).2. Choose 3 positions out of the remaining 5 for odd digits ((binom{5}{3} = 10) ways).3. Permute the remaining 4 odd digits in these 3 positions ((P(4,3) = 24)).4. Permute 2 distinct even digits (including 0) in the remaining 2 positions ((P(5,2) = 20)).Calculating this, we get:[5 times 10 times 24 times 20 = 24000]Case 2: The first digit is even (non-zero).1. Choose the first digit (4 choices: 2, 4, 6, 8).2. Choose 4 positions out of the remaining 5 for odd digits ((binom{5}{4} = 5) ways).3. Permute 5 odd digits in these 4 positions ((P(5,4) = 120)).4. Choose 1 even digit (excluding the one used in the first digit) for the remaining position (4 choices).Calculating this, we get:[4 times 5 times 120 times 4 = 9600]Summing both cases, we get the total number of six-digit numbers with exactly four odd digits:[24000 + 9600 = 33600] Final AnswerThe total number of six-digit numbers with all distinct digits is boxed{136080}, and the number of those with exactly four odd digits is boxed{33600}.

🔑:To find the number of all six-digit numbers composed of unique digits and then the number of such numbers that contain exactly four odd digits, let's break down the problem into two parts:1. Total Number of Six-Digit Numbers with All Unique Digits:We start by determining the total number of six-digit numbers with unique digits from the set ( {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ).First, calculate the number of ways to select 6 different digits from these 10 digits. This can be achieved via permutations since the order matters. Using the formula for permutations without repetition: [ P(10, 6) = frac{10!}{(10-6)!} = 10 times 9 times 8 times 7 times 6 times 5 ]Next, consider that if 0 is one of these digits, it cannot be the first digit. We remove the invalid cases by accounting for the fact that ( frac{1}{10} ) of all arrangements will incorrectly place a zero at the first position. Correct the count by multiplying by ( frac{9}{10} ):[frac{9}{10} P(10, 6) = frac{9}{10} times 10 times 9 times 8 times 7 times 6 times 5 = 9 times 9 times 8 times 7 times 6 times 5 = 136080]Thus, the total number of valid six-digit numbers with unique digits is:[boxed{136080}]2. Number of Six-Digit Numbers with Exactly Four Odd Digits:- Step 1: Choose 4 odd digits from the set ( {1, 3, 5, 7, 9} ). This can be achieved using combinations since the order doesn't matter here:[binom{5}{4} = 5]- Step 2: Choose 2 even digits from the set ( {0, 2, 4, 6, 8} ):[binom{5}{2} = 10]However, in 4 of these configurations, we include a 0, while in 6 configurations, we exclude the 0. Therefore, for combinations that do and don’t include 0, we handle permutations differently:- Case 1: Combinations containing 0: - We must choose 4 odd digits and 1 other even digit out of 4 remaining. - We get ( 5 times 4 ) such groups since we have ( 5 ) choices for the odd digits and ( 4 ) valid setups for the even one that includes 0. - Permutations of these 6 digits (given the restriction on 0) and correcting for 0 not being in the first place, multiply non-zero positions:[5 times 4 times 6! times frac{5}{6} = 5 times 4 times 720 times frac{5}{6} = 5 times 4 times 600 = 12000]- Case 2: Combinations excluding 0: - We select 4 odd digits and 2 valid even digits. - Permutation calculation:[5 times 6 times 6! = 5 times 6 times 720 = 21600]Summing both situations for final total:[12000 + 21600 = 33600]Thus, the total number of six-digit numbers containing exactly four odd digits:[boxed{33600}]

🧠:Alright, so I need to prove these three statements about the geometric configuration involving a circle, lines, and tangents. Let me try to visualize this first.We have circle O with horizontal line m passing through its center. Line l is perpendicular to m and intersects m at point M, which is to the right of the center. Points A, B, C are on line l, outside the circle, above line m. A is farthest from M, then B, then C closest to M. From each of these points, tangents are drawn to the circle: AP, BQ, CR with P, Q, R as points of tangency. First, when l is tangent to the circle, so line l touches the circle at exactly one point. Then, in the other cases, l either intersects the circle (two points) or is entirely outside (no intersection). The three parts of the problem require proving different inequalities or equalities involving the lengths AB, BC, AC multiplied by the lengths of the tangents CR, AP, BQ.Let me start with part 1: When l is tangent to circle O, prove that AB·CR + BC·AP = AC·BQ.Hmm. Since l is tangent to the circle, that means the distance from the center of the circle to line l is equal to the radius. Since line l is perpendicular to m, which passes through the center, the point of tangency must lie on line l. Wait, but in the problem statement, points A, B, C are on line l and outside the circle. If l is tangent, then the only point on l that is on the circle is the tangency point. However, the points A, B, C are outside the circle, so they must lie above the tangency point (since they are above m). Wait, but if l is tangent to the circle, then the tangency point is the closest point on l to the circle's center. So A, B, C are above this tangency point. But the tangents from A, B, C to the circle would each have their own points of tangency. Hmm.Wait, let me make sure I have the configuration correct. If line l is tangent to the circle, then the point of tangency must be on line l. But since points A, B, C are on l and outside the circle, then the tangency point of line l must be between M and the circle's center? Wait, no. If line l is tangent to the circle, then the distance from the center to line l is equal to the radius. Since line l is perpendicular to m, which passes through the center, the tangency point would be at the foot of the perpendicular from the center to line l. Wait, but line m is horizontal and passes through the center, and line l is perpendicular to m, so line l is vertical. Therefore, the foot of the perpendicular from the center to line l is the point where line l meets line m, which is point M. Wait, but if line l is tangent to the circle, then the distance from the center to line l must be equal to the radius. However, point M is the intersection of l and m, and if the center is on m, then the distance from the center to line l is the distance from the center to point M. Therefore, if line l is tangent to the circle, then OM (the distance from center O to M) must be equal to the radius. But M is on the right side of the center, so if OM is equal to the radius, then the circle is tangent to line l at point M. But in the problem statement, points A, B, C are on line l, outside the circle, above line m. If the circle is tangent to line l at M, then M is on the circle, but since M is the intersection of l and m, which is also on the center's line. So points A, B, C are above M on line l, which is vertical. So the circle is tangent to line l at M, and points A, B, C are above M. Therefore, the tangents from A, B, C to the circle will have lengths equal to the square root of (OA² - r²), where OA is the distance from A to the center. But since A, B, C are colinear on l, which is vertical, their distances to the center can be calculated accordingly.Wait, perhaps using power of a point. The power of a point with respect to a circle is equal to the square of the length of the tangent from the point to the circle. So for point A, the power is AP² = OA² - r², similarly for BQ² and CR². But how does that help with the given equation?We need to relate AB·CR + BC·AP with AC·BQ.Given that A, B, C are colinear on line l, with A above B above C. So AB is the distance from A to B, BC from B to C, AC from A to C. Since they are colinear, AC = AB + BC.So maybe we can express everything in terms of coordinates. Let me set up a coordinate system.Let me place the center of the circle O at (h, 0) on line m, which is horizontal. Line l is perpendicular to m, so it's vertical, and intersects m at M. Since M is on the right side of the center, the x-coordinate of M is greater than h. Let me set coordinates such that the center O is at (0, 0) for simplicity. Wait, but the problem says M is on the right side of the center. If we place the center at (0, 0), then line m is the x-axis. Line l is perpendicular to m, so it's the vertical line passing through M. Since M is on the right side of the center, M is at (d, 0) where d > 0. Then line l is x = d. The circle has center (0, 0) and radius r. If line l is tangent to the circle, then the distance from the center to line l must be equal to the radius. The distance from (0,0) to x = d is |d|, so if line l is tangent, then d = r. Therefore, M is at (r, 0). Points A, B, C are on line l (x = r), above line m (y > 0), outside the circle. Since the circle is tangent to line l at (r, 0), the points A, B, C are at (r, y_A), (r, y_B), (r, y_C) with y_A > y_B > y_C > 0.The tangent from A to the circle will touch the circle at some point P. Similarly for BQ and CR. The lengths AP, BQ, CR can be calculated using the power of the point. The power of point A with respect to the circle is AP² = (distance from A to center)^2 - radius^2. The distance from A (r, y_A) to center (0,0) is sqrt(r² + y_A²). Therefore, AP² = (r² + y_A²) - r² = y_A², so AP = y_A. Similarly, BQ = y_B, CR = y_C. Wait, is that correct?Wait, if the center is at (0,0), line l is x = r, and point A is (r, y_A). The distance from A to the center is sqrt(r² + y_A²). The power of point A is OA² - r² = (r² + y_A²) - r² = y_A², so the length of the tangent from A to the circle is sqrt(y_A²) = y_A. Similarly, for B and C, the lengths are y_B and y_C. So AP = y_A, BQ = y_B, CR = y_C.Therefore, in the case where l is tangent to the circle (d = r), the lengths of the tangents are just the y-coordinates of the points A, B, C. That simplifies things.Now, we need to prove AB·CR + BC·AP = AC·BQ.First, compute AB, BC, AC. Since all points are on the vertical line x = r, their coordinates are (r, y_A), (r, y_B), (r, y_C) with y_A > y_B > y_C. Therefore:AB = y_A - y_BBC = y_B - y_CAC = y_A - y_CCR = y_CAP = y_ABQ = y_BSo substituting into the equation:AB·CR + BC·AP = (y_A - y_B)·y_C + (y_B - y_C)·y_A= y_A y_C - y_B y_C + y_B y_A - y_C y_ASimplify: The first term is y_A y_C, then - y_B y_C, then + y_B y_A, then - y_C y_A. The y_A y_C and - y_C y_A cancel each other. Then we have - y_B y_C + y_B y_A = y_B (y_A - y_C)On the other hand, AC·BQ = (y_A - y_C)·y_BTherefore, both sides are equal: y_B (y_A - y_C) = y_B (y_A - y_C). Therefore, the equation holds. So part 1 is proven.That was straightforward once I set up coordinates and realized that the lengths of the tangents are just the y-coordinates. So part 1 is done.Now part 2: When l intersects the circle, prove AB·CR + BC·AP < AC·BQ.Similarly, when line l intersects the circle, the distance from the center to line l is less than the radius. In coordinate terms, using the previous setup, if center is at (0,0), line l is x = d, and the distance from center to line l is |d|. If line l intersects the circle, then |d| < r. Therefore, the points of intersection are at (d, y) where x = d and the circle equation x² + y² = r². Therefore, y = ±sqrt(r² - d²). So line l intersects the circle at (d, sqrt(r² - d²)) and (d, -sqrt(r² - d²)). However, in the problem, points A, B, C are on l, outside the circle, above line m (which is the x-axis). Therefore, points A, B, C are above y = sqrt(r² - d²). Since l intersects the circle, d < r.In this case, the length of the tangent from A (d, y_A) to the circle is sqrt( (d)^2 + y_A^2 - r^2 ). Similarly for BQ and CR. So AP = sqrt(d² + y_A² - r²), BQ = sqrt(d² + y_B² - r²), CR = sqrt(d² + y_C² - r²).So now, we need to compute AB·CR + BC·AP and compare it with AC·BQ.First, AB = y_A - y_B, BC = y_B - y_C, AC = y_A - y_C.Therefore, the left-hand side (LHS) is:AB·CR + BC·AP = (y_A - y_B) sqrt(d² + y_C² - r²) + (y_B - y_C) sqrt(d² + y_A² - r²)The right-hand side (RHS) is:AC·BQ = (y_A - y_C) sqrt(d² + y_B² - r²)We need to show that LHS < RHS when d < r.This seems more complicated. Maybe we can use some inequality or algebraic manipulation.Alternatively, perhaps consider the previous case where d = r (part 1) as a boundary case and see how the expressions change when d decreases (since in part 2, d < r). Maybe the terms involving sqrt(d² + y² - r²) increase or decrease accordingly.Wait, when d decreases (since d < r now), the term under the sqrt for each tangent becomes d² + y² - r². Since d is smaller, d² is smaller, so for a given y, the value under the sqrt is smaller, so the tangent lengths are shorter? Wait, but when d decreases, the distance from the center to line l is less, so the points A, B, C are closer to the circle. Therefore, the tangent lengths should be longer. Wait, perhaps I need to check.Wait, the power of point A is OA² - r², where OA is the distance from A to the center. OA² = d² + y_A². So the power is d² + y_A² - r². Therefore, the tangent length squared is this. If d decreases, then OA² decreases, so the power decreases, which would mean the tangent length decreases. Wait, but if d decreases, the point A is moving left towards the center, so even though it's moving horizontally, its distance to the center is decreasing, hence the tangent length should decrease. So if in part 1, when d = r, the tangent lengths were y_A, y_B, y_C, but when d < r, the tangent lengths are sqrt(d² + y² - r²), which is sqrt( (y² + d²) - r² ). So compared to part 1, where d = r, we have sqrt( y² + r² - r² ) = y. But when d < r, sqrt( y² + d² - r² ). Since d² - r² is negative (because d < r), the expression becomes sqrt( y² - (r² - d²) ). Therefore, the tangent lengths are sqrt( y² - k² ), where k = sqrt(r² - d²). So k is the y-coordinate where line l intersects the circle.So points A, B, C have y-coordinates greater than k. Therefore, the tangent lengths are sqrt(y² - k²). For example, AP = sqrt(y_A² - k²), BQ = sqrt(y_B² - k²), CR = sqrt(y_C² - k²).Therefore, the equation to prove is:(y_A - y_B) sqrt(y_C² - k²) + (y_B - y_C) sqrt(y_A² - k²) < (y_A - y_C) sqrt(y_B² - k²)This seems like an inequality that might be provable using the Cauchy-Schwarz inequality or some other algebraic inequality. Let me consider substituting variables to simplify.Let me denote:Let a = y_A, b = y_B, c = y_C, with a > b > c > k.We need to prove:(a - b) sqrt(c² - k²) + (b - c) sqrt(a² - k²) < (a - c) sqrt(b² - k²)Let me square both sides to eliminate the square roots. But before that, note that all terms are positive because a > b > c > k and k > 0. Therefore, squaring should preserve the inequality.Left-hand side (LHS):[(a - b) sqrt(c² - k²) + (b - c) sqrt(a² - k²)]²Right-hand side (RHS):[(a - c) sqrt(b² - k²)]²Expand LHS:= (a - b)² (c² - k²) + 2(a - b)(b - c) sqrt(c² - k²) sqrt(a² - k²) + (b - c)² (a² - k²)RHS:= (a - c)² (b² - k²)So we need to show that:(a - b)² (c² - k²) + 2(a - b)(b - c) sqrt((c² - k²)(a² - k²)) + (b - c)² (a² - k²) < (a - c)² (b² - k²)Let me expand both sides and see if we can compare term by term.First, note that (a - c)² = (a - b + b - c)² = (a - b)² + 2(a - b)(b - c) + (b - c)²Therefore, RHS expanded is:[(a - b)² + 2(a - b)(b - c) + (b - c)²] (b² - k²)= (a - b)² (b² - k²) + 2(a - b)(b - c)(b² - k²) + (b - c)² (b² - k²)Therefore, subtract LHS from RHS and show that the result is positive:RHS - LHS = [(a - b)² (b² - k²) + 2(a - b)(b - c)(b² - k²) + (b - c)² (b² - k²)] - [(a - b)² (c² - k²) + 2(a - b)(b - c) sqrt((c² - k²)(a² - k²)) + (b - c)² (a² - k²)] > 0Factor terms:= (a - b)^2 [b² - k² - c² + k²] + 2(a - b)(b - c)[b² - k² - sqrt((c² - k²)(a² - k²))] + (b - c)^2 [b² - k² - a² + k²]Simplify each bracket:First term: (a - b)^2 (b² - c²)Second term: 2(a - b)(b - c) [b² - k² - sqrt((c² - k²)(a² - k²))]Third term: (b - c)^2 (b² - a²) = -(b - c)^2 (a² - b²)So overall:= (a - b)^2 (b² - c²) - (b - c)^2 (a² - b²) + 2(a - b)(b - c)[b² - k² - sqrt((c² - k²)(a² - k²))]Factor the first two terms:Note that (a - b)^2 (b² - c²) - (b - c)^2 (a² - b²) can be written as:(a - b)^2 (b - c)(b + c) - (b - c)^2 (a - b)(a + b)Factor out (a - b)(b - c):= (a - b)(b - c)[(a - b)(b + c) - (b - c)(a + b)]Compute inside the brackets:(a - b)(b + c) - (b - c)(a + b)Expand both:= ab + ac - b² - bc - [ab + b² - ac - bc]= ab + ac - b² - bc - ab - b² + ac + bcSimplify:= (ab - ab) + (ac + ac) + (-b² - b²) + (-bc + bc)= 2ac - 2b²Therefore, the first two terms become:(a - b)(b - c)(2ac - 2b²) = 2(a - b)(b - c)(ac - b²)So putting back into RHS - LHS:2(a - b)(b - c)(ac - b²) + 2(a - b)(b - c)[b² - k² - sqrt((c² - k²)(a² - k²))] > 0Factor out 2(a - b)(b - c):2(a - b)(b - c)[ac - b² + b² - k² - sqrt((c² - k²)(a² - k²))] > 0Simplify inside the brackets:ac - k² - sqrt((c² - k²)(a² - k²))Therefore, RHS - LHS > 0 is equivalent to:2(a - b)(b - c)[ac - k² - sqrt((c² - k²)(a² - k²))] > 0Since a > b > c, (a - b) > 0 and (b - c) > 0, so the sign depends on the term inside the brackets: [ac - k² - sqrt((c² - k²)(a² - k²))]We need to show that ac - k² > sqrt((c² - k²)(a² - k²))Square both sides (since both sides are positive):(ac - k²)^2 > (c² - k²)(a² - k²)Expand left side:a²c² - 2ac k² + k⁴Right side:c²a² - c²k² - a²k² + k⁴Subtract right side from left side:(a²c² - 2ac k² + k⁴) - (a²c² - c²k² - a²k² + k⁴) = (-2ac k²) - (-c²k² - a²k²) = -2ac k² + c²k² + a²k² = k²(a² + c² - 2ac) = k²(a - c)^2 ≥ 0But equality holds only when a = c, which is not the case since a > c. Therefore, (ac - k²)^2 > (c² - k²)(a² - k²), which implies ac - k² > sqrt((c² - k²)(a² - k²))Thus, the term inside the brackets is positive, and since 2(a - b)(b - c) is positive, the entire expression is positive. Therefore, RHS - LHS > 0, so LHS < RHS. Hence, part 2 is proven.Moving on to part 3: When l is outside the circle, prove AB·CR + BC·AP > AC·BQ.When line l is outside the circle, the distance from the center to line l is greater than the radius. In coordinates, with center at (0,0), line l is x = d, so |d| > r. Points A, B, C are on line l (x = d), above the x-axis, with y_A > y_B > y_C > 0. The power of a point A with respect to the circle is AP² = d² + y_A² - r², so AP = sqrt(d² + y_A² - r²). Similarly for BQ and CR.Similarly to part 2, we need to show:AB·CR + BC·AP > AC·BQUsing the same substitutions as before, with a = y_A, b = y_B, c = y_C, but now d > r. However, the key difference is that when l is outside the circle, the expression under the sqrt for the tangent lengths is d² + y² - r², which is now greater than y² (since d² > r²). Therefore, the tangent lengths are longer than in the case where l is tangent (part 1). However, the structure of the inequality is similar to part 2 but reversed.Following the same approach as part 2, let's set up the inequality:(a - b) sqrt(c² + k²) + (b - c) sqrt(a² + k²) > (a - c) sqrt(b² + k²)Wait, here, if d > r, then k² = d² - r² (since the power of point M (d,0) would be d² - r², which is the square of the tangent length from M to the circle). Therefore, the tangent lengths from A, B, C are sqrt(d² + y² - r²) = sqrt(y² + (d² - r²)) = sqrt(y² + k²), where k = sqrt(d² - r²). So AP = sqrt(a² + k²), BQ = sqrt(b² + k²), CR = sqrt(c² + k²).Therefore, we need to show:(a - b) sqrt(c² + k²) + (b - c) sqrt(a² + k²) > (a - c) sqrt(b² + k²)This is similar to part 2, but now with k² added inside the square roots instead of subtracted.Again, square both sides to compare.Left-hand side (LHS):[(a - b) sqrt(c² + k²) + (b - c) sqrt(a² + k²)]²Right-hand side (RHS):[(a - c) sqrt(b² + k²)]²Expand LHS:= (a - b)² (c² + k²) + 2(a - b)(b - c) sqrt((c² + k²)(a² + k²)) + (b - c)² (a² + k²)RHS:= (a - c)² (b² + k²)Similar to part 2, expand RHS:(a - c)^2 (b² + k²) = [(a - b) + (b - c)]² (b² + k²) = (a - b)^2 (b² + k²) + 2(a - b)(b - c)(b² + k²) + (b - c)^2 (b² + k²)Now compute RHS - LHS:= (a - b)^2 (b² + k²) + 2(a - b)(b - c)(b² + k²) + (b - c)^2 (b² + k²) - [(a - b)^2 (c² + k²) + 2(a - b)(b - c) sqrt((c² + k²)(a² + k²)) + (b - c)^2 (a² + k²)]Factor terms:= (a - b)^2 [b² + k² - c² - k²] + 2(a - b)(b - c)[b² + k² - sqrt((c² + k²)(a² + k²))] + (b - c)^2 [b² + k² - a² - k²]Simplify each bracket:First term: (a - b)^2 (b² - c²)Second term: 2(a - b)(b - c)[b² + k² - sqrt((c² + k²)(a² + k²))]Third term: (b - c)^2 (b² - a²) = - (b - c)^2 (a² - b²)Therefore:= (a - b)^2 (b² - c²) - (b - c)^2 (a² - b²) + 2(a - b)(b - c)[b² + k² - sqrt((c² + k²)(a² + k²))]Factor the first two terms as before:= (a - b)(b - c)[2ac - 2b²] + 2(a - b)(b - c)[b² + k² - sqrt((c² + k²)(a² + k²))]Factor out 2(a - b)(b - c):= 2(a - b)(b - c)[ac - b² + b² + k² - sqrt((c² + k²)(a² + k²))]Simplify inside the brackets:ac + k² - sqrt((c² + k²)(a² + k²))We need to show that this term is positive, i.e., ac + k² > sqrt((c² + k²)(a² + k²))Square both sides:(ac + k²)^2 > (c² + k²)(a² + k²)Expand left side:a²c² + 2ac k² + k⁴Right side:c²a² + c²k² + a²k² + k⁴Subtract right side from left side:(a²c² + 2ac k² + k⁴) - (a²c² + c²k² + a²k² + k⁴) = 2ac k² - c²k² - a²k² = k²(2ac - c² - a²) = -k²(a - c)^2But this is negative unless a = c, which they are not. Therefore, we have:(ac + k²)^2 - (c² + k²)(a² + k²) = -k²(a - c)^2 < 0Which implies that (ac + k²)^2 < (c² + k²)(a² + k²), hence ac + k² < sqrt((c² + k²)(a² + k²))Therefore, the term inside the brackets is negative. Since 2(a - b)(b - c) is positive, the entire expression RHS - LHS is negative, meaning LHS > RHS. Therefore, AB·CR + BC·AP > AC·BQ. Hence, part 3 is proven.So summarizing:1. When l is tangent, the tangent lengths are the y-coordinates, leading to equality.2. When l intersects the circle, after manipulating the inequality, we found LHS < RHS.3. When l is outside, similar steps showed LHS > RHS.Thus, all three parts are proven using coordinate geometry and algebraic manipulation with inequalities.

✅:Step-by-Step Explanation and Proof:1. When ( l ) is tangent to circle ( O ):- Setup: Let the circle ( O ) have center at ( (0,0) ) and radius ( r ). Line ( m ) is the x-axis, and line ( l ) (vertical) is tangent at ( M ), which is ( (r, 0) ). Points ( A, B, C ) on ( l ) have coordinates ( (r, y_A) ), ( (r, y_B) ), ( (r, y_C) ) with ( y_A > y_B > y_C > 0 ).- Tangent Lengths: Since ( l ) is tangent, the power of each point gives ( AP = y_A ), ( BQ = y_B ), ( CR = y_C ).- Compute Segments: [ AB = y_A - y_B,quad BC = y_B - y_C,quad AC = y_A - y_C ] - Substitute into Equation: [ AB cdot CR + BC cdot AP = (y_A - y_B)y_C + (y_B - y_C)y_A = y_B(y_A - y_C) = AC cdot BQ ] Hence, equality holds.2. When ( l ) intersects circle ( O ):- Setup: Let line ( l ) be ( x = d ) with ( d < r ). The circle intersects ( l ) at ( (d, pm sqrt{r^2 - d^2}) ). Points ( A, B, C ) are above ( y = sqrt{r^2 - d^2} ).- Tangent Lengths: Power of point gives ( AP = sqrt{y_A^2 + d^2 - r^2} ), similarly for ( BQ ), ( CR ).- Inequality Setup: Define ( k = sqrt{r^2 - d^2} ). Show: [ (y_A - y_B)sqrt{y_C^2 - k^2} + (y_B - y_C)sqrt{y_A^2 - k^2} < (y_A - y_C)sqrt{y_B^2 - k^2} ] - Square Both Sides and Simplify: After expansion and algebraic manipulation, using ( ac - k^2 > sqrt{(c^2 - k^2)(a^2 - k^2)} ), the inequality reduces to a positive expression, proving ( LHS < RHS ).3. When ( l ) is outside circle ( O ):- Setup: Line ( l ) is ( x = d ) with ( d > r ). Points ( A, B, C ) have tangents ( AP = sqrt{y_A^2 + d^2 - r^2} ).- Inequality Setup: Show: [ (y_A - y_B)sqrt{y_C^2 + k^2} + (y_B - y_C)sqrt{y_A^2 + k^2} > (y_A - y_C)sqrt{y_B^2 + k^2} ] where ( k = sqrt{d^2 - r^2} ).- Square Both Sides and Simplify: Similar to part 2, but due to the reversed sign in ( k ), the inequality results in ( LHS > RHS ).Conclusion:Through coordinate analysis and algebraic manipulation, all three statements are proven by evaluating the expressions under different geometric conditions of line ( l ) relative to circle ( O ).Final Answer1. (boxed{AB cdot CR + BC cdot AP = AC cdot BQ})2. (boxed{AB cdot CR + BC cdot AP < AC cdot BQ})3. (boxed{AB cdot CR + BC cdot AP > AC cdot BQ})

🔑:Given: The circle (O) has radius (r), the horizontal line (m) passes through the center of circle (O), the line (l) is perpendicular to (m) at point (M). Points (A, B, C) lie on line (l) outside the circle, above line (m) with (A) being the farthest from (M), (C) being the closest, and (B) in between. The segments (AP, BQ, CR) are secants of the circle (O) with (P, Q, R) as the corresponding points of intersection.Let:- (OM = x)- (AM = a)- (BM = b)- (CM = c)Given that (a > b > c > 0), we derive the equations for the lengths (AP, BQ, CR):From the Pythagorean theorem applied to ( triangle AMP ), we have:[ AM^2 + OM^2 - OP^2 = AP^2 ][ a^2 + x^2 - r^2 = AP^2 ]Let ( t = x^2 - r^2 ). Then,[ AP^2 = a^2 + t ][ AP = sqrt{a^2 + t} ]Similarly:[ BQ = sqrt{b^2 + t} ][ CR = sqrt{c^2 + t} ]We are to prove:1. When ( l ) is tangent to the circle (O),[ AB cdot CR + BC cdot AP = AC cdot BQ ]2. When ( l ) intersects the circle (O),[ AB cdot CR + BC cdot AP < AC cdot BQ ]3. When ( l ) is outside the circle (O),[ AB cdot CR + BC cdot AP > AC cdot BQ ]To proceed, define:[ G = left( AB cdot CR + BC cdot AP right)^2 - left( AC cdot BQ right)^2 ]First, compute:[ G = left( (a - b) sqrt{c^2 + t} + (b - c) sqrt{a^2 + t} right)^2 - (a - c)^2 (b^2 + t) ]Next, expand and simplify:[ G = left( (a - b) sqrt{c^2 + t} + (b - c) sqrt{a^2 + t} right)^2 - left( (a - b) + (b - c) right)^2 (b^2 + t) ][ = (a - b)^2 (c^2 + t) + (b - c)^2 (a^2 + t) + 2(a - b)(b - c) sqrt{c^2 + t} sqrt{a^2 + t} ][ - (a - b)^2 (b^2 + t) - (b - c)^2 (b^2 + t) - 2(a - b)(b - c) (b^2 + t) ][ = 2(a - b)(b - c)left[ sqrt{(c^2 + t)(a^2 + t)} - (ac + t) right] ] Case 1: (l) is tangent to circle (O):When ( l ) is tangent to the circle (O), (x = r) thus (t = x^2 - r^2 = 0):[ G = 2(a - b)(b - c) left[sqrt{(c^2)(a^2)} - (ac) right] = 0 ]Hence, [ AB cdot CR + BC cdot AP = AC cdot BQ ][ boxed{1} ] is proved. Case 2: (l) intersects circle (O):When ( l ) intersects the circle, (0 < x < r), thus (0 < t < r^2). Since (c) is outside the circle:[ x^2 + c^2 > r^2 ][ t = x^2 - r^2 > -c^2 > -a^2 ]As a result,[ 2(a - b)(b - c)[ sqrt{(c^2 + t)(a^2 + t)} - ac - t ] < 0 ]So,[ AB cdot CR + BC cdot AP < AC cdot BQ ][ boxed{2} ] is proved. Case 3: (l) is outside the circle:When ( l ) is outside the circle, (x > r), thus (t > 0). Consequently:[ 2(a - b)(b - c)[ sqrt{(c^2 + t)(a^2 + t)} - ac - t ] > 0 ]So,[ AB cdot CR + BC cdot AP > AC cdot BQ ][ boxed{3} ] is proved.Thus, the proofs for all three cases are concluded.

🧠:Okay, so I need to find the value of the function f at 2016/2017, given this functional equation: (x - 1)f(x) + f(1/x) = 1/(x - 1) for all x not equal to 0 or 1. Hmm, functional equations can sometimes be tricky, but maybe I can substitute specific values for x to create a system of equations and solve for f(x). Let me try to think through this step by step.First, the equation given is for all x ≠ 0, 1. So, I can choose any x except those two values. The function involves both f(x) and f(1/x), so maybe if I substitute x with 1/x, I can get another equation that relates f(1/x) and f(x). Then, I can solve the two equations together to find f(x).Let me write down the original equation:(x - 1)f(x) + f(1/x) = 1/(x - 1) ...(1)Now, substitute x with 1/x in equation (1). That would give:( (1/x) - 1 )f(1/x) + f(x) = 1/( (1/x) - 1 )Let me simplify both sides. Starting with the left side:( (1 - x)/x )f(1/x) + f(x)And the right side:1/( (1 - x)/x ) = x/(1 - x)So the second equation becomes:( (1 - x)/x )f(1/x) + f(x) = x/(1 - x) ...(2)Now, I have two equations:Equation (1): (x - 1)f(x) + f(1/x) = 1/(x - 1)Equation (2): ( (1 - x)/x )f(1/x) + f(x) = x/(1 - x)Let me denote these as two equations with variables f(x) and f(1/x). Let's rewrite them for clarity.Let me write equation (1) as:A * f(x) + B * f(1/x) = Cand equation (2) as:D * f(1/x) + E * f(x) = FSo, in equation (1):A = (x - 1)B = 1C = 1/(x - 1)In equation (2):D = (1 - x)/xE = 1F = x/(1 - x)Now, we can solve this system of equations for f(x) and f(1/x) using substitution or elimination. Let's try elimination.From equation (1):f(1/x) = C - A f(x) = [1/(x - 1)] - (x - 1)f(x)Substitute this into equation (2):D * [C - A f(x)] + E f(x) = FLet me compute each term.First, D * C = ( (1 - x)/x ) * (1/(x - 1)) )Simplify this:(1 - x)/x * 1/(x - 1) = ( - (x - 1)/x ) * 1/(x - 1) = -1/xThen, D * (-A f(x)) = ( (1 - x)/x ) * ( - (x - 1) f(x) )Simplify:(1 - x)/x * ( - (x - 1) ) f(x) = ( (1 - x)( - (x - 1) ) ) / x * f(x)Note that (1 - x)( - (x - 1) ) = (1 - x)( - (x - 1) ) = (1 - x)( -x + 1 ) = (1 - x)(1 - x) = (1 - x)^2Wait, hold on:Wait, (x - 1) = -(1 - x), so (1 - x) * ( - (x - 1) ) = (1 - x) * ( - ( - (1 - x) ) ) = (1 - x)(1 - x) = (1 - x)^2So, D * (-A f(x)) = (1 - x)^2 / x * f(x)Therefore, equation (2) becomes:-1/x + ( (1 - x)^2 / x ) f(x) + f(x) = FBut F is x/(1 - x). Let me write the entire equation:-1/x + [ ( (1 - x)^2 / x ) + 1 ] f(x) = x/(1 - x)Let me combine the terms involving f(x). First, factor out f(x):[ ( (1 - x)^2 / x ) + 1 ] f(x) = x/(1 - x) + 1/xCompute the coefficient:( (1 - x)^2 / x ) + 1 = ( (1 - 2x + x^2) / x ) + 1 = (1 - 2x + x^2 + x)/x = (1 - x + x^2)/xWait, let me check:Wait, (1 - x)^2 is 1 - 2x + x^2, correct. Then divided by x: (1 - 2x + x^2)/x. Adding 1 is same as adding x/x, so total is (1 - 2x + x^2 + x)/x = (1 - x + x^2)/x. Correct.So, coefficient is (1 - x + x^2)/x.The right side is x/(1 - x) + 1/x. Let's combine these two terms:x/(1 - x) + 1/x = [x^2 + (1 - x)] / [x(1 - x)] = [x^2 + 1 - x] / [x(1 - x)]So, numerator is x^2 - x + 1, denominator is x(1 - x)Therefore, the equation becomes:( (x^2 - x + 1)/x ) f(x) = (x^2 - x + 1)/[x(1 - x)]Multiply both sides by x/(x^2 - x + 1):f(x) = [ (x^2 - x + 1)/[x(1 - x)] ] * [ x/(x^2 - x + 1) ) ] = 1/(1 - x)Wait, that's interesting. Let me verify that step.Left side: (x^2 - x + 1)/x * f(x)Right side: (x^2 - x + 1)/[x(1 - x)]Therefore, dividing both sides by (x^2 - x + 1)/x gives:f(x) = [ (x^2 - x + 1)/[x(1 - x)] ] / [ (x^2 - x + 1)/x ) ]Which simplifies to:f(x) = [ (x^2 - x + 1)/x(1 - x) ) ] * [ x/(x^2 - x + 1) ) ] = 1/(1 - x)So f(x) = 1/(1 - x) ?Wait, but let's check if this works. Let's substitute back into original equation.If f(x) = 1/(1 - x), then f(1/x) = 1/(1 - 1/x) = x/(x - 1)Then, original equation:(x - 1)f(x) + f(1/x) = (x - 1)/(1 - x) + x/(x - 1)Simplify:(x - 1)/(1 - x) is -1, and x/(x - 1) is positive. So:-1 + x/(x - 1) = (- (x - 1) + x ) / (x - 1) = (-x + 1 + x ) / (x - 1) = 1/(x - 1)Which is the right side. So yes, this works. Therefore, f(x) = 1/(1 - x) is the solution.Wait, but that seems too straightforward. Let me check with equation (2) as well, just to be sure.Equation (2) was:( (1 - x)/x )f(1/x) + f(x) = x/(1 - x)If f(x) = 1/(1 - x), then f(1/x) = 1/(1 - 1/x) = x/(x - 1)Left side:( (1 - x)/x ) * (x/(x - 1)) + 1/(1 - x)Simplify:( (1 - x)/x ) * (x/(x - 1)) = (1 - x)/(x - 1) = -1Then, adding 1/(1 - x):-1 + 1/(1 - x) = - (1 - x)/(1 - x) + 1/(1 - x) = (-1 + x + 1)/(1 - x) = x/(1 - x)Which is equal to the right side. So yes, it works for equation (2) as well. Therefore, the solution is indeed f(x) = 1/(1 - x).Wait, but then the problem is solved? So f(2016/2017) = 1/(1 - 2016/2017) = 1/(1/2017) = 2017. But let me verify again.Wait, the functional equation is (x - 1)f(x) + f(1/x) = 1/(x - 1). If f(x) = 1/(1 - x), then:(x - 1)*[1/(1 - x)] + f(1/x) = (x - 1)/(- (x - 1)) + f(1/x) = -1 + f(1/x). And according to the equation, this should equal 1/(x - 1). But according to f(1/x) = x/(x - 1). So -1 + x/(x - 1) = [ - (x - 1) + x ] / (x - 1) = 1/(x - 1), which works. So yes, correct.Therefore, the function f(x) is 1/(1 - x). Therefore, f(2016/2017) is 1/(1 - 2016/2017) = 1/(1/2017) = 2017. So the answer is 2017. That seems straightforward. Wait, but the problem is presented as a functional equation, which might lead one to think it's more complex, but maybe in this case, substitution works directly.Alternatively, perhaps there's a mistake in my reasoning? Let me check again.Suppose f(x) = 1/(1 - x). Then f(1/x) = 1/(1 - 1/x) = x/(x - 1). Then:Left-hand side of original equation: (x - 1)/(1 - x) + x/(x - 1) = (-1) + x/(x - 1) = (- (x - 1)/(x - 1)) + x/(x - 1) = (-x + 1 + x)/(x - 1) = 1/(x - 1), which is the right-hand side. Correct.Therefore, the solution is f(x) = 1/(1 - x). Hence, f(2016/2017) = 2017. Therefore, the answer is boxed{2017}Wait, but this seems almost too simple. Did I miss something here? Let me think again. The problem says "the function f satisfies the equation for each x not equal to 0 and 1." So, if f(x) = 1/(1 - x), then it's defined for x ≠ 1, which is consistent with the domain given (x ≠ 0,1). So as long as x ≠ 1, which is true for x = 2016/2017, then f(x) is defined. So yes, this seems correct.Alternatively, maybe the problem is designed to test the ability to solve such functional equations, and this is a standard approach. So, in conclusion, after solving the system of equations obtained by substituting x and 1/x, we find that f(x) = 1/(1 - x), hence the required value is 2017.Final Answerboxed{2017}

✅:Given the functional equation ((x - 1)f(x) + fleft(frac{1}{x}right) = frac{1}{x - 1}) for all (x neq 0) and (x neq 1), we need to find (fleft(frac{2016}{2017}right)).First, we substitute (x) with (frac{1}{x}) in the original equation to obtain a second equation:[left(frac{1}{x} - 1right)fleft(frac{1}{x}right) + f(x) = frac{1}{frac{1}{x} - 1}]Simplifying both sides of this new equation, we get:[left(frac{1 - x}{x}right)fleft(frac{1}{x}right) + f(x) = frac{x}{1 - x}]Now we have two equations:1. ((x - 1)f(x) + fleft(frac{1}{x}right) = frac{1}{x - 1})2. (left(frac{1 - x}{x}right)fleft(frac{1}{x}right) + f(x) = frac{x}{1 - x})We solve these equations as a system. From the first equation, express (fleft(frac{1}{x}right)):[fleft(frac{1}{x}right) = frac{1}{x - 1} - (x - 1)f(x)]Substitute this into the second equation:[left(frac{1 - x}{x}right)left(frac{1}{x - 1} - (x - 1)f(x)right) + f(x) = frac{x}{1 - x}]Simplify the left-hand side:[left(frac{1 - x}{x}right)left(frac{1}{x - 1}right) - left(frac{1 - x}{x}right)(x - 1)f(x) + f(x)]Simplifying each term:[frac{1 - x}{x} cdot frac{1}{x - 1} = -frac{1}{x}][-left(frac{1 - x}{x}right)(x - 1)f(x) = frac{(1 - x)^2}{x}f(x)]Combining these terms:[-frac{1}{x} + frac{(1 - x)^2}{x}f(x) + f(x) = frac{x}{1 - x}]Combine the coefficients of (f(x)):[left(frac{(1 - x)^2 + x}{x}right)f(x) - frac{1}{x} = frac{x}{1 - x}]Simplify the coefficient:[frac{1 - x + x^2}{x}f(x) - frac{1}{x} = frac{x}{1 - x}]Solve for (f(x)):[frac{x^2 - x + 1}{x}f(x) = frac{x}{1 - x} + frac{1}{x}]Combine the right-hand side terms:[frac{x^2 - x + 1}{x}f(x) = frac{x^2 - x + 1}{x(1 - x)}]Thus, we find:[f(x) = frac{1}{1 - x}]Therefore, substituting (x = frac{2016}{2017}):[fleft(frac{2016}{2017}right) = frac{1}{1 - frac{2016}{2017}} = frac{1}{frac{1}{2017}} = 2017]The final answer is (boxed{2017}).

🔑:1. Start with the given functional equation: [ (x-1) f(x) + fleft(frac{1}{x}right) = frac{1}{x-1} ]2. Substitute ( x ) with ( frac{1}{x} ) in the equation: [ left(frac{1}{x} - 1right) fleft(frac{1}{x}right) + f(x) = frac{x}{1-x} ]3. Now we have a system of two equations: [ begin{cases} (x-1) f(x) + fleft(frac{1}{x}right) = frac{1}{x-1} left(frac{1}{x} - 1right) fleft(frac{1}{x}right) + f(x) = frac{x}{1-x} end{cases} ]4. Multiply the first equation by ( frac{1-x}{x} ): [ frac{(x-1)(1-x)}{x} f(x) + frac{1-x}{x} fleft(frac{1}{x}right) = frac{(1-x)}{(x-1)x} ] Simplify the equation: [ frac{(1-x)^2}{x} f(x) + frac{1-x}{x} fleft(frac{1}{x}right) = frac{1-x}{(1-x)x} ] Simplify further: [ frac{(1-x)^2}{x} f(x) + frac{1-x}{x} fleft(frac{1}{x}right) = frac{1}{1-x} cdot frac{1-x}{x} ] Simplify to get: [ frac{(1-x)^2}{x} f(x) + frac{1-x}{x} fleft(frac{1}{x}right) = frac{1}{x} ]5. Subtract this equation from the second equation in the system: [ left(frac{1}{x} - 1right) fleft(frac{1}{x}right) + f(x) = frac{x}{1-x} ] [ - left[ frac{(1-x)^2}{x} f(x) + frac{1-x}{x} fleft(frac{1}{x}right) = frac{1}{x} right] ] Simplify the subtraction step-by-step: [ left(frac{1}{x} - 1right) fleft(frac{1}{x}right) + f(x) - frac{(1-x)^2}{x} f(x) - frac{1-x}{x} fleft(frac{1}{x}right) = frac{x}{1-x} - frac{1}{x} ] Combine like terms: [ left[ left(frac{1}{x} - 1right) - frac{1-x}{x} right] fleft(frac{1}{x}right) + left[ 1-frac{(1-x)^2}{x} right] f(x) = frac{x}{1-x} - frac{1}{x} ] Simplify and solve for ( f(x) ): [ frac{x^2-x+1}{x} f(x) = frac{x^2-x+1}{x(1-x)} ] [ f(x) = frac{1}{1-x} ]6. Determine ( fleft(frac{2016}{2017}right) ): [ fleft(frac{2016}{2017}right) = frac{1}{1 - frac{2016}{2017}} = frac{1}{frac{2017 - 2016}{2017}} = 2017 ] Conclusion:[boxed{2017}]

🧠:Okay, let me try to figure out how to prove that the sum of the distances from 45 points to point A is not equal to the sum of the distances to point B, given that all these points are outside the segment AB on a line. Hmm, first, I need to visualize the problem. So, we have a line with points A and B. The segment AB is part of this line, and there are 45 points marked outside of AB. That means all these points are either to the left of A or to the right of B on the line. The problem states that the sum of their distances to A is not equal to the sum of their distances to B. I need to prove that these two sums can't be the same.Let me start by considering the positions of the points. If all points are outside AB, then each point is either in the region extending to the left of A or to the right of B. Let's denote the coordinate system such that point A is at position 0 and point B is at position d, where d is the distance between A and B. Then, all the 45 points will have coordinates either less than 0 (left of A) or greater than d (right of B).Let me assign variables. Suppose there are m points to the left of A, with coordinates x₁, x₂, ..., xₘ, each less than 0. And n points to the right of B, with coordinates y₁, y₂, ..., yₙ, each greater than d. Since there are 45 points in total, m + n = 45.Now, the sum of distances from all points to A would be the sum of |x_i - A| + |y_j - A|. But since A is at 0, this simplifies to sum |x_i| + sum |y_j|. Similarly, the sum of distances to B would be sum |x_i - d| + sum |y_j - d|. Because B is at d.But wait, since all x_i are to the left of A (i.e., negative), |x_i| is just -x_i. Similarly, |x_i - d| would be d - x_i because x_i < 0 < d. For the points y_j to the right of B (i.e., y_j > d), |y_j - A| is |y_j - 0| = y_j, and |y_j - B| is |y_j - d| = y_j - d.So, the sum of distances to A (S_A) is sum_{i=1 to m} (-x_i) + sum_{j=1 to n} y_j.The sum of distances to B (S_B) is sum_{i=1 to m} (d - x_i) + sum_{j=1 to n} (y_j - d).Let me compute the difference S_A - S_B to see if it can be zero.Calculating:S_A - S_B = [sum(-x_i) + sum y_j] - [sum(d - x_i) + sum(y_j - d)]Let's distribute the sums:= [sum(-x_i) + sum y_j] - [sum d - sum x_i + sum y_j - sum d]Wait, sum over m terms of d is m*d, and sum over n terms of d is n*d? Wait, in S_B, the first sum is over m terms (each x_i), each term is (d - x_i), so sum_{i=1 to m} (d - x_i) = m*d - sum x_i.Similarly, the second sum in S_B is sum_{j=1 to n} (y_j - d) = sum y_j - n*d.Therefore, S_B = m*d - sum x_i + sum y_j - n*d.So, S_A = - sum x_i + sum y_j.Therefore, S_A - S_B = (- sum x_i + sum y_j) - (m*d - sum x_i + sum y_j - n*d)Let me compute that:= (- sum x_i + sum y_j) - m*d + sum x_i - sum y_j + n*dSimplify terms:- sum x_i and + sum x_i cancel out.sum y_j and - sum y_j cancel out.So we have: - m*d + n*d = d*(n - m)Therefore, S_A - S_B = d*(n - m)But the problem states that all 45 points are outside AB. So m + n = 45. Therefore, n = 45 - m.Substituting back into S_A - S_B:= d*( (45 - m) - m ) = d*(45 - 2m)Therefore, S_A - S_B = d*(45 - 2m)Now, the original problem claims that S_A is not equal to S_B. Which would mean that S_A - S_B ≠ 0. Therefore, we need to show that d*(45 - 2m) ≠ 0.Given that d is the distance between A and B, which is a positive number (since A and B are distinct points), d ≠ 0. Therefore, for S_A - S_B to be zero, we need 45 - 2m = 0. But 45 is an odd number, and 2m is an even number (since m is an integer). The difference between an odd number and an even number cannot be zero. Therefore, 45 - 2m ≠ 0 for any integer m.Therefore, S_A - S_B = d*(45 - 2m) ≠ 0, which means S_A ≠ S_B.Hence, proved that the sum of the distances from the points to A cannot equal the sum of the distances to B.Wait, let me check that again. If m is the number of points left of A, and n = 45 - m. Then 45 - 2m must not be zero. Since 45 is odd, 2m is even. So 45 - even = odd, which can never be zero. Therefore, yes, 45 - 2m ≠ 0 for any integer m. Therefore, S_A - S_B is a non-zero multiple of d, so it cannot be zero. Therefore, the two sums cannot be equal.Therefore, the proof is complete. The key idea was to express the difference between the two sums and realize that due to the parity (odd vs even), the difference can't be zero.Let me check with an example. Suppose all 45 points are on one side, say left of A. Then m=45, n=0. Then S_A - S_B = d*(0 - 45) = -45d ≠ 0. If half are on each side, but 45 is odd, so m=22, n=23. Then 45 - 2m = 45 - 44 =1 ≠0. Similarly, m=23, n=22, 45 -46= -1≠0. So in all cases, the difference is non-zero. Hence, proved.

✅:To prove that the sum of distances from 45 points outside segment ( AB ) to ( A ) is not equal to the sum of distances to ( B ), follow this reasoning:1. Coordinate System Setup: Place ( A ) at ( 0 ) and ( B ) at ( d > 0 ). The 45 points lie either to the left of ( A ) (coordinates ( x_i < 0 )) or to the right of ( B ) (coordinates ( y_j > d )).2. Sum of Distances: - Sum to ( A ): [ S_A = sum_{i=1}^m |x_i| + sum_{j=1}^n y_j = -sum_{i=1}^m x_i + sum_{j=1}^n y_j ] (since ( x_i < 0 ), ( |x_i| = -x_i ), and ( y_j > 0 )). - Sum to ( B ): [ S_B = sum_{i=1}^m (d - x_i) + sum_{j=1}^n (y_j - d) = md - sum_{i=1}^m x_i + sum_{j=1}^n y_j - nd ]3. Difference Between Sums: [ S_A - S_B = left(-sum x_i + sum y_jright) - left(md - sum x_i + sum y_j - ndright) ] Simplifying: [ S_A - S_B = d(n - m) ] Since ( m + n = 45 ), substitute ( n = 45 - m ): [ S_A - S_B = d(45 - 2m) ]4. Parity Argument: The term ( 45 - 2m ) is critical. Since ( 45 ) is odd and ( 2m ) is even for any integer ( m ), their difference ( 45 - 2m ) is always odd, hence never zero. Therefore: [ S_A - S_B neq 0 implies S_A neq S_B ]Conclusion: The difference ( S_A - S_B ) is a non-zero multiple of ( d ), proving the sums cannot be equal. This holds regardless of how the 45 points are distributed between the two sides of ( AB ).boxed{S_A neq S_B}

🔑:1. Define the Segment Length: Let the length of the segment ( AB ) be 1 unit.2. Consider Points Outside the Segment: Consider 45 points marked on the line, all lying outside the segment ( AB ).3. Distance Difference Calculation: For each point ( P ) outside the segment ( AB ), denote the distances from ( P ) to ( A ) and from ( P ) to ( B ) as ( d_A ) and ( d_B ) respectively.4. Relation Between Distances: As the segment ( AB ) has a length of 1, the relationship between ( d_A ) (distance from the point to ( A )) and ( d_B ) (distance from the point to ( B )) is given by: [ |d_A - d_B| = 1. ] This difference is always ( pm 1 ).5. Sum of Differences: To find the difference between the sum of distances from the points to ( A ) and the sum of distances from the points to ( B ), we need to sum up the individual differences. Denote the points as ( P_1, P_2, ldots, P_{45} ). Therefore, the total difference between the sums is: [ sum_{i=1}^{45} (|d_{A_i} - d_{B_i}|) ] Since ( |d_{A_i} - d_{B_i}| ) is ( pm 1 ) for each point, this sum becomes: [ sum_{i=1}^{45} (pm 1). ]6. Odd Sum Argument: Since there are 45 points, which is an odd number, summing an odd quantity of (pm 1) results in an odd number. An odd sum cannot be zero: [ sum_{i=1}^{45} (pm 1) = text{odd number} neq 0. ] Hence, the sum of the distances from these points to ( A ) is not equal to the sum of the distances from these points to ( B ). Conclusion: [ boxed{text{The sum of the distances from the points to } A text{ is not equal to the sum of the distances from the points to } B.} ]