🧠:Okay, so I need to find all positive integers x such that 2x + 1 is a perfect square. But also, in the range from 2x + 2 up to 3x + 2, there shouldn't be any perfect squares. Hmm, let's break this down step by step.First, the condition that 2x + 1 is a perfect square. Let me write that as 2x + 1 = n², where n is some positive integer. Then, solving for x, we get x = (n² - 1)/2. Since x has to be a positive integer, n² - 1 must be even. Since n² is either even or odd, n² - 1 will be odd if n is even, and even if n is odd. Wait, because if n is even, n² is even, so even minus 1 is odd. If n is odd, n² is odd, so odd minus 1 is even. Therefore, x is an integer only when n is odd. So n must be an odd integer. That makes sense. So n = 2k + 1 for some non-negative integer k, but since n is positive, k is non-negative. Then, substituting back, x = ((2k + 1)² - 1)/2. Let me compute that:(4k² + 4k + 1 - 1)/2 = (4k² + 4k)/2 = 2k² + 2k. So x = 2k² + 2k. So x is expressed in terms of k. So x is twice k(k + 1). That gives me all x such that 2x + 1 is a perfect square. So x must be of the form 2k² + 2k where k is a positive integer (k ≥ 0, but x is positive, so k ≥ 1? Let's check: if k=0, x=0, but x must be positive, so k ≥ 1). So k starts from 1. Therefore, possible x are 2(1)^2 + 2(1) = 4, 2(2)^2 + 2(2)= 12, 2(3)^2 + 2(3)= 24, etc.Now, the second part is that in the numbers from 2x + 2 up to 3x + 2, there are no perfect squares. So I need to ensure that between 2x + 2 and 3x + 2 inclusive, there are no squares. Let's express 2x + 2 and 3x + 2 in terms of k. Since x = 2k² + 2k, then 2x = 4k² + 4k, so 2x + 2 = 4k² + 4k + 2. Similarly, 3x + 2 = 3*(2k² + 2k) + 2 = 6k² + 6k + 2. So the interval we need to check is [4k² + 4k + 2, 6k² + 6k + 2]. There should be no perfect squares in this interval.So the problem reduces to finding all k ≥ 1 such that between 4k² + 4k + 2 and 6k² + 6k + 2 there are no perfect squares.Let me try to analyze this interval. Let's denote the lower bound as L(k) = 4k² + 4k + 2 and upper bound as U(k) = 6k² + 6k + 2.We need to check if there exists any perfect square m² such that L(k) ≤ m² ≤ U(k). If not, then such x is valid.So, to find for which k, the interval [L(k), U(k)] contains no perfect squares.First, let's find the approximate square roots of L(k) and U(k) to see which m's we need to check.Compute sqrt(L(k)) and sqrt(U(k)).sqrt(4k² + 4k + 2). Let's see, 4k² + 4k + 1 = (2k + 1)^2. So 4k² + 4k + 2 is just 1 more than a perfect square. Therefore, sqrt(4k² + 4k + 2) is between 2k + 1 and 2k + 2. Similarly, sqrt(6k² + 6k + 2). Let's see, 6k² + 6k + 2. Let's compare with (sqrt(6)k + something)^2. Maybe approximate sqrt(6)k + c. Let's see:( sqrt(6)k + a )² = 6k² + 2a sqrt(6)k + a². Comparing to 6k² + 6k + 2, so 2a sqrt(6)k ≈ 6k → a ≈ 6/(2 sqrt(6)) = 3/sqrt(6) = sqrt(6)/2 ≈ 1.2247. So approximate sqrt(6)k + sqrt(6)/2. Then, (sqrt(6)k + sqrt(6)/2)^2 = 6k² + 6k + 1.5. So 6k² + 6k + 1.5 is less than U(k) = 6k² + 6k + 2. Therefore, sqrt(U(k)) is between sqrt(6)k + sqrt(6)/2 and sqrt(6)k + sqrt(6)/2 + something. Hmm, maybe better to compute integers m such that m² is in [L(k), U(k)].Alternatively, since L(k) is just after (2k + 1)^2, perhaps the next square is (2k + 2)^2. Let's check:(2k + 2)^2 = 4k² + 8k + 4. Compare this to U(k) = 6k² + 6k + 2. For which k is 4k² + 8k + 4 ≤ 6k² + 6k + 2?Solving 4k² + 8k + 4 ≤ 6k² + 6k + 2→ 0 ≤ 2k² - 2k - 2→ 2k² - 2k - 2 ≥ 0Divide by 2: k² - k - 1 ≥ 0Solutions to k² - k - 1 = 0 are k = [1 ± sqrt(5)]/2. So positive root is (1 + sqrt(5))/2 ≈ 1.618. So for k ≥ 2, this inequality holds. Therefore, for k ≥ 2, (2k + 2)^2 ≤ U(k). So between L(k) = 4k² + 4k + 2 and U(k) = 6k² + 6k + 2, the next square after (2k + 1)^2 is (2k + 2)^2 = 4k² + 8k + 4. But we need to check if 4k² + 8k + 4 ≤ 6k² + 6k + 2. As we saw, that's true when k ≥ 2. So for k ≥ 2, the square (2k + 2)^2 is inside the interval [L(k), U(k)] only if 4k² + 8k + 4 ≤ 6k² + 6k + 2. Wait, but that's equivalent to 0 ≤ 2k² - 2k - 2, which, as we saw, holds for k ≥ 2. So for k ≥ 2, (2k + 2)^2 is less than or equal to U(k). Therefore, (2k + 2)^2 is inside the interval [4k² + 4k + 2, 6k² + 6k + 2] when 4k² + 8k + 4 ≥ 4k² + 4k + 2. Which simplifies to 4k + 2 ≥ 0, which is always true. Therefore, for k ≥ 2, (2k + 2)^2 is between L(k) and U(k). Therefore, for k ≥ 2, there exists at least one square in the interval, namely (2k + 2)^2. Therefore, such x would not satisfy the second condition. Therefore, only k = 1 might be valid?Wait, let's check for k = 1. Then x = 2(1)^2 + 2(1) = 4. Then 2x + 1 = 9, which is 3², a perfect square. Then, the interval from 2x + 2 = 10 to 3x + 2 = 14. So numbers 10, 11, 12, 13, 14. Are there any perfect squares here? The squares around this range: 3²=9, 4²=16. So between 10 and 14, there are no perfect squares. So for k=1, x=4, there are no squares in the interval. So x=4 is a solution.Now check k=2. x=2(4) + 4= 12. Then 2x + 1=25=5². Then the interval is from 2x+2=26 to 3x+2=38. Now check if there are squares between 26 and 38. 5²=25, 6²=36. 36 is within 26-38. So 36 is in the interval, so x=12 is invalid.Similarly, k=3. x=2(9)+6=24. 2x +1=49=7². Interval: 2x+2=50 to 3x+2=74. Squares in this range: 7²=49, 8²=64, 9²=81. 64 is between 50 and 74, so invalid.k=4. x=2(16)+8=40. 2x+1=81=9². Interval: 82 to 122. Squares: 9²=81, 10²=100, 11²=121. 100 and 121 are in the interval. So invalid.So seems like for k=1, x=4 is valid. For k ≥2, the next square (2k + 2)^2 falls into the interval. Wait, but when k=1, the next square after (2k +1)^2=9 is 16, which is outside 3x +2=14. So 16 is beyond 14. So k=1 is okay.But for k=2, (2k +1)^2=25, next square is 36, which is 36=6². 36 is within 26-38. So invalid. Similarly for higher k.Wait, so maybe only k=1 is valid. Therefore, x=4 is the only solution.But let's check k=0. Although x must be positive. If k=0, x=0. But x must be positive, so k starts from 1. So x=4 is the only solution? Wait, maybe check other possible squares in the interval.Wait, for example, maybe there are other squares in the interval apart from (2k + 2)^2. Let's check for k=1. The interval is 10-14. The squares are 16, which is outside. So no squares. For k=2, interval 26-38. Squares are 25 (excluded), 36 (included). So only 36. For k=3, interval 50-74. Squares are 64 and 81 (81 is outside). So 64 is in. For k=4, 82-122: 100, 121. Both in. So yes, the next square after (2k +1)^2 is (2k +2)^2, and for k ≥2, that square is inside the interval. For k=1, that square is 16, which is outside the interval. So x=4 is the only solution.But wait, maybe there are other squares in the interval. For example, between (2k +1)^2 and (2k +2)^2, sometimes there might be another square. Wait, squares grow as (m)^2, so between consecutive squares, there's no other square. So in the interval [L(k), U(k)] for k ≥2, (2k +2)^2 is the next square after (2k +1)^2, and since U(k) is 6k² +6k +2, and (2k +2)^2 =4k² +8k +4. So 4k² +8k +4 ≤6k² +6k +2 ?Which simplifies to 0 ≤2k² -2k -2. Which as before, holds when k ≥2. So for k=2, 4k² +8k +4= 16 +16 +4=36; 6k² +6k +2=24 +12 +2=38. So 36 ≤38, yes. So (2k +2)^2 is within the interval. For k=3: (2*3 +2)^2=8²=64. U(k)=6*9 +6*3 +2=54 +18 +2=74. 64 ≤74. So yes.Therefore, for k ≥2, the square (2k +2)^2 is inside the interval. Thus, making x invalid. Therefore, only k=1 gives x=4 as a valid solution.But let's check if there could be any other squares in the interval. For example, for some k, maybe (2k +3)^2 is less than U(k). Let's see.For k=1, U(k)=14. (2k +3)^2=5²=25 >14. So no. For k=2, U(k)=38. (2k +3)^2=7²=49 >38. For k=3, U(k)=74. (2k +3)^2=9²=81 >74. For k=4, U(k)=122. (2k +3)^2=11²=121 <122. Wait, here 11²=121 is less than 122. So in this case, 121 is within the interval [82,122]. So for k=4, 121 is in the interval. So (2k +3)^2=121=11² is in the interval. Therefore, even though (2k +2)^2=100 is also there, but (2k +3)^2=121 is also inside. So, in this case, there are two squares in the interval. So x=40 is invalid.Similarly, for k=5. x=2*25 +10=60. 2x +1=121=11². Interval: 122 to 182. Squares: 121 (excluded), 144, 169, 196 (excluded). So 144 and 169 are in [122,182]. So multiple squares. Hence invalid.Wait, so even when (2k +3)^2 is within U(k), which happens when (2k +3)^2 ≤6k² +6k +2. Let's solve for k:(2k +3)^2 ≤6k² +6k +24k² +12k +9 ≤6k² +6k +20 ≤2k² -6k -7Solving 2k² -6k -7 ≥0Quadratic equation: 2k² -6k -7=0Discriminant: 36 +56=92Solutions: [6 ± sqrt(92)]/4 = [6 ± 2*sqrt(23)]/4 = [3 ± sqrt(23)]/2 ≈ [3 ±4.796]/2. Positive root is (3 +4.796)/2≈3.898. So for k≥4, 2k² -6k -7 ≥0. Thus, (2k +3)^2 ≤6k² +6k +2 when k≥4. So for k≥4, (2k +3)^2 is also inside the interval. Hence, multiple squares. Therefore, for k≥2, (2k +2)^2 is inside, and for k≥4, even (2k +3)^2 is inside. Therefore, multiple squares in the interval, making x invalid.But even for k=3, (2k +3)^2=9²=81, while U(k)=74. So 81>74, so not in the interval. So for k=3, only 8²=64 is in the interval. For k=4, (2k +3)^2=11²=121, which is less than U(k)=122. So 121 is in the interval. Therefore, x=40 (k=4) would have two squares: 100 and 121. Hence, invalid.Therefore, the only k that works is k=1, leading to x=4.But wait, let's check k=1 again. x=4. 2x +1=9=3². Then the interval is 10 to14. No squares in there. Correct. What about x=0? x has to be positive, so x=0 is invalid. What about x=1? 2x +1=3, not a square. x=2: 2x +1=5, not square. x=3: 7, not square. x=4:9, which is square. So x=4 is the first candidate. Then x=5:2*5 +1=11, not square. x=6:13, not square. x=7:15, not square. x=8:17, not square. x=9:19, not square. x=10:21, not square. x=11:23, not square. x=12:25, which is square. So x=12 is another candidate. But as checked earlier, for x=12, the interval is 26-38, which includes 36=6². Hence invalid. So indeed, only x=4 is valid.Wait, but let's check x=24. 2x +1=49=7². Then interval 50-74. 64=8² is in there. So invalid.Wait, let's see if there are any other x beyond x=4 that could satisfy the condition. For example, maybe for some higher k, the interval [4k² +4k +2,6k² +6k +2] skips over the next few squares. Let's see. Suppose for larger k, the gap between consecutive squares is larger. Wait, the gap between m² and (m+1)² is 2m +1. As m increases, the gaps between squares increase. However, our interval [L(k), U(k)] is also expanding as k increases. Let's check for k=1: interval length is 14 -10=4. For k=2:38-26=12. For k=3:74-50=24. For k=4:122-82=40. So the interval length grows quadratically. Whereas the gaps between squares grow linearly. Therefore, for larger k, the interval will contain more squares. Thus, it's less likely for higher k to have intervals without squares. Hence, only k=1 (x=4) is possible.Alternatively, maybe another approach. Let's consider that 2x +1 =n². Then x=(n² -1)/2. The interval we need to check is from 2x +2 to 3x +2. Let's express this interval in terms of n. Since x=(n² -1)/2, then 2x +2= n² -1 +2 =n² +1. And 3x +2= (3(n² -1))/2 +2= (3n² -3)/2 +2= (3n² -3 +4)/2=(3n² +1)/2. So the interval is [n² +1, (3n² +1)/2]. We need to ensure there are no perfect squares in this interval.So the interval in terms of n is [n² +1, (3n² +1)/2]. Let's denote m² as a square in this interval. Then:n² +1 ≤m² ≤(3n² +1)/2We need to find n such that there is no integer m with m² in that interval.Let me analyze for n odd (since n must be odd for x to be integer, as we saw earlier). Let n=2k +1, k≥0. Then substituting back:n=2k +1. Then n²=4k² +4k +1. So the interval becomes:4k² +4k +1 +1=4k² +4k +2to(3*(4k² +4k +1) +1)/2=(12k² +12k +3 +1)/2=(12k² +12k +4)/2=6k² +6k +2So same as before. Therefore, the interval is [4k² +4k +2,6k² +6k +2]. As before.Now, we need to check if there exists m such that m² is in this interval. Let's find the possible m.sqrt(4k² +4k +2) ≈2k +1 + something. Since 4k² +4k +1=(2k +1)^2, so sqrt(4k² +4k +2) is slightly more than 2k +1. Similarly, sqrt(6k² +6k +2). Let's approximate sqrt(6)k + c.But maybe better to find m such that m >2k +1 and m² ≤6k² +6k +2.Let me check m=2k +2. Then m²=4k² +8k +4. Compare to 6k² +6k +2. So 4k² +8k +4 ≤6k² +6k +2?Which simplifies to 0 ≤2k² -2k -2. As before, for k≥2, this holds. So for k≥2, m=2k +2 gives a square within the interval. Hence, invalid. For k=1, m=2*1 +2=4, m²=16. Compare to upper bound 6(1)^2 +6(1) +2=14. 16>14, so m=4 is outside the interval. Hence, no squares in the interval for k=1. For k=0, but x=(n² -1)/2=(1 -1)/2=0, which is invalid as x must be positive. So only k=1 (n=3) gives x=4, which is valid.Therefore, the only solution is x=4.But to ensure thoroughness, let's check for m=2k +3. For k=4, m=11. 11²=121, which is within U(k)=122. So in that case, even m=2k +3 is inside. So for larger k, higher m might also be inside. But as established, even m=2k +2 is already inside for k≥2, so there's no need to check further. Hence, x=4 is the only solution.Another angle: suppose there exists m such that n² +1 ≤m² ≤(3n² +1)/2. Then:n² +1 ≤m² ≤(3n² +1)/2Multiply all parts by 2:2n² +2 ≤2m² ≤3n² +1Then, since 2m² must be an integer, but not sure if helpful.Alternatively, rearrange:n² +1 ≤m² ≤1.5n² +0.5Thus, m is between n and roughly sqrt(1.5)n. For example, sqrt(1.5)≈1.2247. So m is between n and ~1.2247n. Since m must be integer, the possible m's are n +1, n +2, ..., up to floor(1.2247n). Let's check for n=3 (k=1). 1.2247*3≈3.674. So m can be 4. But 4²=16, which is outside the upper bound (3*9 +1)/2=14. So no.For n=5 (k=2): sqrt(1.5)*5≈6.123. So m=6. 6²=36. Upper bound is (3*25 +1)/2=38. 36 ≤38. So yes, 36 is in.For n=7 (k=3): sqrt(1.5)*7≈8.573. So m=8 or 9. 8²=64, 9²=81. Upper bound is (3*49 +1)/2=74. 64 ≤74, so 64 is in.Therefore, for n≥5 (k≥2), m= next integer after n, which is n+1, gives a square within the interval. Hence, invalid. Only n=3 (k=1) works.Therefore, the only solution is x=4.

✅:The positive integers ( x ) satisfying the conditions are determined as follows:1. Condition 1: ( 2x + 1 ) must be a perfect square. Let ( 2x + 1 = n^2 ), where ( n ) is an odd integer. Solving for ( x ), we get: [ x = frac{n^2 - 1}{2}. ] Since ( n ) must be odd, let ( n = 2k + 1 ) for ( k geq 0 ). Substituting ( n ) gives: [ x = 2k^2 + 2k. ]2. Condition 2: The interval ( [2x + 2, 3x + 2] ) must contain no perfect squares. Substituting ( x = 2k^2 + 2k ), the interval becomes: [ [4k^2 + 4k + 2, 6k^2 + 6k + 2]. ]3. Analysis: - For ( k = 1 ), ( x = 4 ). The interval is ( [10, 14] ), which contains no perfect squares. - For ( k geq 2 ), the next perfect square after ( (2k + 1)^2 ) is ( (2k + 2)^2 = 4k^2 + 8k + 4 ). This square lies within the interval ( [4k^2 + 4k + 2, 6k^2 + 6k + 2] ) for ( k geq 2 ), violating the second condition.Conclusion: The only solution is ( x = 4 ).[boxed{4}]

🔑:Given the problem, we need to determine all positive integers (x) such that (2x + 1) is a perfect square, and there are no perfect squares in the numbers (2x + 2, 2x + 3, dots, 3x + 2).1. Step 1: Let (n) be an integer such that (2x + 1 = n^2).2. Step 2: From the equation (2x + 1 = n^2), we solve for (x): [ 2x + 1 = n^2 implies 2x = n^2 - 1 implies x = frac{n^2 - 1}{2} ]3. Step 3: We need to ensure that among the numbers (2x + 2, 2x + 3, dots, 3x + 2), there are no perfect squares. Substitute (x = frac{n^2 - 1}{2}) into (2x + 2) and (3x + 2):4. Step 4: Calculate the boundaries for these numbers: [ 2x + 2 = 2left(frac{n^2 - 1}{2}right) + 2 = n^2 + 1 ] [ 3x + 2 = 3left(frac{n^2 - 1}{2}right) + 2 = frac{3n^2 - 3}{2} + 2 = frac{3n^2 - 3 + 4}{2} = frac{3n^2 + 1}{2} ]5. Step 5: Ensure the range condition for (x): [ n^2 leq frac{3n^2 + 1}{2} < (n + 1)^2 ] Analyze this double inequality:6. Step 6: Analyze the first inequality: [ 2n^2 leq 3n^2 + 1 ] [ n^2 geq -1 ] This inequality holds for all integers (n).7. Step 7: Analyze the second inequality: [ frac{3n^2 + 1}{2} < n^2 + 2n + 1 ] [ 3n^2 + 1 < 2(n^2 + 2n + 1) ] [ 3n^2 + 1 < 2n^2 + 4n + 2 ] [ n^2 - 4n - 1 < 0 ]8. Step 8: Solve (n^2 - 4n - 1 < 0). This inequality can be solved by finding the roots of the corresponding equation: [ n^2 - 4n - 1 = 0 ] The roots are given by: [ n = frac{4 pm sqrt{16 + 4}}{2} = 2 pm sqrt{5} ]9. Step 9: The inequality (n^2 - 4n - 1 < 0) holds in the interval: [ 2 - sqrt{5} < n < 2 + sqrt{5} ]10. Step 10: Evaluate (2 - sqrt{5} < n < 2 + sqrt{5}) for integer values of (n). Since (2 - sqrt{5} approx -0.236) and (2 + sqrt{5} approx 4.236), the integers (n) must be within ({0, 1, 2, 3, 4}).11. Step 11: Test the valid integer values (n in {0, 1, 2, 3, 4}): - (n = 0): (x = frac{0^2 - 1}{2}) (not a positive integer) - (n = 1): (x = frac{1^2 - 1}{2} = 0) (not a positive integer) - (n = 2): (x = frac{2^2 - 1}{2} = frac{3}{2}) (not an integer) - (n = 3): (x = frac{3^2 - 1}{2} = frac{8}{2} = 4) (valid solution) - (n = 4): (x = frac{4^2 - 1}{2} = frac{15}{2}) (not an integer)12. Step 12: Verify the exclusion of perfect squares between (2x + 2) and (3x + 2) for (x = 4): [ 2x + 1 = 2(4) + 1 = 9 = 3^2 ] [ 2x + 2 = 2(4) + 2 = 10, quad 3x + 2 = 3(4) + 2 = 14 ] [ text{Numbers are } {10, 11, 12, 13, 14}; text{none of these are perfect squares} ] Conclusion:The only solution is (x = 4).[boxed{4}]

🧠:Okay, let me try to figure out the length of side AD in this quadrilateral ABCD. So, we have AB = 1, BC = 2, CD = √3, angle ABC is 120 degrees, and angle BCD is 90 degrees. Hmm, quadrilaterals can be tricky because they're not as straightforward as triangles. Maybe I should break this down into triangles and use some trigonometric laws like the Law of Cosines or maybe coordinate geometry? Let me think step by step.First, maybe I can draw the quadrilateral to visualize it. Let's start with point B, since we have two sides (AB and BC) and an angle at B. Then from point C, we have another side CD and a right angle at C. So, if I can figure out the coordinates of each point, I can then compute the distance between A and D.Coordinate geometry might be a good approach here. Let's assign coordinates to the points. Let me place point B at the origin (0,0) to simplify calculations. Then, since AB = 1, I need to place point A somewhere 1 unit away from B. But in which direction? Hmm, maybe along the x-axis for simplicity. Wait, but angle ABC is 120 degrees. If I place point B at (0,0), and point C somewhere, then angle at B is 120 degrees between AB and BC. Let me clarify.Let's set point B at (0,0). Let's let AB be along the positive x-axis. So point A would be at (1,0). Then, angle ABC is 120 degrees, so the direction from B to C makes a 120-degree angle with BA (which is along the negative x-axis, since BA is from B to A). Wait, angles are measured between two sides with the vertex at the common point. So angle at B is between BA and BC. If BA is from B to A (which is along the positive x-axis), and angle ABC is 120 degrees, then BC is making a 120-degree angle from BA. So the direction from B to C is 120 degrees from BA. Since BA is along the positive x-axis, the direction of BC would be 120 degrees counterclockwise from BA. Therefore, the coordinates of point C can be calculated using the length BC = 2 and the angle.Yes, so if we take BA as the reference (positive x-axis), then BC is at an angle of 120 degrees from BA. Therefore, point C has coordinates:C_x = BC * cos(120°)C_y = BC * sin(120°)Since BC = 2,cos(120°) is equal to cos(180°−60°) = -cos(60°) = -0.5sin(120°) is sin(180°−60°) = sin(60°) = √3/2Therefore,C_x = 2 * (-0.5) = -1C_y = 2 * (√3/2) = √3Wait, hold on. If BA is along the positive x-axis, then the angle between BA and BC is 120 degrees, which would mean that BC is going into the second quadrant (since 120 degrees from positive x-axis). But if we place point A at (1,0), then BA is from B(0,0) to A(1,0), which is the positive x-axis. Then, angle ABC of 120 degrees would mean that BC is going 120 degrees from BA, so 120 degrees counterclockwise from BA. That would put point C at (-1, √3) as calculated. But wait, BC is length 2. So coordinates of C would be ( -1, √3 ). Let me confirm that distance from B(0,0) to C(-1, √3):Distance = √[(-1 - 0)^2 + (√3 - 0)^2] = √[1 + 3] = √4 = 2. Correct. So point C is at (-1, √3).Now, from point C, we have CD = √3 and angle BCD = 90 degrees. So angle at C is 90 degrees between BC and CD. So we need to find the coordinates of point D such that CD = √3 and angle BCD is 90 degrees.Let's think about this. Since angle BCD is 90 degrees, the direction from C to D should be perpendicular to the direction from C to B. The direction from C to B is from C(-1, √3) to B(0,0), which is the vector (1, -√3). Therefore, the direction from C to D should be perpendicular to this vector. The perpendicular vectors to (1, -√3) are (√3, 1) and (-√3, -1). But we need to figure out which direction it is.Wait, angle BCD is 90 degrees. So at point C, the angle between CB and CD is 90 degrees. So CB is the vector from C to B, which is (1, -√3), as mentioned. Then CD is the vector from C to D, which needs to be perpendicular to CB. Therefore, the vector CD must satisfy the dot product with CB being zero.So let's denote the coordinates of D as (x, y). Then vector CD is (x - (-1), y - √3) = (x + 1, y - √3). Vector CB is (0 - (-1), 0 - √3) = (1, -√3). The dot product of CD and CB must be zero:(x + 1)(1) + (y - √3)(-√3) = 0Simplify:x + 1 - √3(y - √3) = 0x + 1 - √3 y + 3 = 0x - √3 y + 4 = 0So that's one equation. The other condition is that the length of CD is √3, so:√[(x + 1)^2 + (y - √3)^2] = √3Squaring both sides:(x + 1)^2 + (y - √3)^2 = 3So we have two equations:1) x - √3 y + 4 = 02) (x + 1)^2 + (y - √3)^2 = 3We can solve these equations to find the coordinates of D.From equation 1: x = √3 y - 4Plug into equation 2:(√3 y - 4 + 1)^2 + (y - √3)^2 = 3Simplify:(√3 y - 3)^2 + (y - √3)^2 = 3Let's compute each term:First term: (√3 y - 3)^2 = (√3 y)^2 - 2*√3 y*3 + 3^2 = 3 y² - 6√3 y + 9Second term: (y - √3)^2 = y² - 2 y √3 + (√3)^2 = y² - 2√3 y + 3Adding them together:3 y² - 6√3 y + 9 + y² - 2√3 y + 3 = 4 y² - 8√3 y + 12Set equal to 3:4 y² - 8√3 y + 12 = 3Subtract 3:4 y² - 8√3 y + 9 = 0Quadratic equation in y:4 y² - 8√3 y + 9 = 0Let's compute the discriminant:Δ = ( -8√3 )² - 4 *4*9 = 64*3 - 144 = 192 - 144 = 48Solutions:y = [8√3 ± √48]/(2*4) = [8√3 ± 4√3]/8 = [ (8√3 ± 4√3) ] /8So two possibilities:y = (8√3 + 4√3)/8 = 12√3 /8 = (3√3)/2ory = (8√3 - 4√3)/8 = 4√3 /8 = √3/2So two possible y-values: y = (3√3)/2 or y = √3/2Let's find the corresponding x for each.From equation 1: x = √3 y -4First, for y = (3√3)/2:x = √3*(3√3/2) -4 = (3*3)/2 -4 = 9/2 -4 = 9/2 -8/2 = 1/2So D is at (1/2, 3√3/2)Second, for y = √3/2:x = √3*(√3/2) -4 = (3/2) -4 = 3/2 -8/2 = -5/2So D is at (-5/2, √3/2)Now, we need to figure out which of these two points is the correct one. Since angle BCD is 90 degrees, and we are dealing with a quadrilateral, the position of D should be such that the quadrilateral doesn't intersect itself. Let's analyze both possibilities.First, let's consider point D at (1/2, 3√3/2). Let's see the direction from C(-1, √3) to D(1/2, 3√3/2). The vector CD is (1/2 - (-1), 3√3/2 - √3) = (3/2, √3/2). Then, vector CB is (1, -√3). Let's check the angle between CD and CB. The dot product is (3/2)(1) + (√3/2)(-√3) = 3/2 - (3/2) = 0. So that's correct, the angle is 90 degrees. Similarly, for the other point D(-5/2, √3/2), vector CD is (-5/2 - (-1), √3/2 - √3) = (-3/2, -√3/2). Then, the dot product with CB (1, -√3) is (-3/2)(1) + (-√3/2)(-√3) = -3/2 + (3/2) = 0. So both points satisfy the right angle condition. So both are valid, but we need to determine which one is part of the quadrilateral ABCD.Assuming the quadrilateral is convex, we can consider both possibilities, but likely, based on the given information, there is a unique solution. Let me check the positions.Point A is at (1,0), B at (0,0), C at (-1, √3). If D is at (1/2, 3√3/2), then the quadrilateral would go from A(1,0) to B(0,0) to C(-1, √3) to D(1/2, 3√3/2) and back to A. This might create a kite-shaped figure. Alternatively, if D is at (-5/2, √3/2), then D is further left. Let's see.Alternatively, perhaps the quadrilateral is constructed in such a way that the sides AB, BC, CD, DA are connected in order. So after CD, we connect back to A. So depending on where D is, the shape changes. Let me try to visualize both options.First, if D is (1/2, 3√3/2), which is above and to the right of point C. Then the quadrilateral would have points A(1,0), B(0,0), C(-1, √3), D(1/2, 3√3/2). Connecting these in order, it seems like a convex quadrilateral. The other point D(-5/2, √3/2) is to the left and below point C. So connecting C(-1, √3) to D(-5/2, √3/2), which is southwest direction, then back to A(1,0). That might create a concave quadrilateral or cross over other sides. Hmm. But without a diagram, it's hard to tell. However, since the problem doesn't specify whether the quadrilateral is convex or concave, both could be possible. However, since the problem asks for the exact length of AD, perhaps both possibilities yield the same length, or maybe only one is valid. Let's check the distances.First, if D is (1/2, 3√3/2):Distance from A(1,0) to D(1/2, 3√3/2):Δx = 1/2 - 1 = -1/2Δy = 3√3/2 - 0 = 3√3/2Distance AD = √[ (-1/2)^2 + (3√3/2)^2 ] = √[1/4 + 27/4] = √[28/4] = √[7] ≈ 2.6458Second, if D is (-5/2, √3/2):Distance from A(1,0) to D(-5/2, √3/2):Δx = -5/2 -1 = -7/2Δy = √3/2 -0 = √3/2Distance AD = √[ (-7/2)^2 + (√3/2)^2 ] = √[49/4 + 3/4] = √[52/4] = √[13] ≈ 3.6055So two different lengths. Hmm, so which one is correct? The problem must have a unique answer, so I must have made an incorrect assumption here.Wait, perhaps the direction of the angle is important. When angle BCD is 90 degrees, we have to consider the orientation. The right angle at C could be either on one side or the other. Let's recall that in the problem statement, quadrilateral ABCD is connected in order. So the angle at C is between BC and CD. The direction from C to D must be such that when moving from B to C to D, the turn is 90 degrees. Let's think about the orientation.The vector from B to C is (-1, √3). So from B(0,0) to C(-1, √3). At point C, we need to turn 90 degrees to get to D. Depending on the direction of the turn (left or right), the position of D changes.If we consider the right-hand rule, turning 90 degrees from BC towards the right would give one direction, and turning left would give the other. Let's compute the perpendicular vectors.The direction of BC is vector BC = C - B = (-1, √3). The perpendicular vectors to BC are (√3, 1) and (-√3, -1), since the dot product with BC must be zero: (-1)(√3) + (√3)(1) = -√3 + √3 = 0, and similarly for the other vector. Wait, no. Wait, perpendicular vectors to BC would satisfy (-1, √3) • (a, b) = 0 => -a + √3 b = 0 => a = √3 b. So any vector of the form (√3 b, b). For example, (√3, 1) and (-√3, -1). So, if we take the direction from C, moving in the direction of (√3, 1) or (-√3, -1). Since CD has length √3, which is the same as the vector (√3, 1) scaled appropriately. Let's check:The vector (√3, 1) has length √[(√3)^2 +1^2] = √[3+1] = √4 = 2. So if we scale it down by a factor of 1/2, we get (√3/2, 1/2), which has length 1. But CD is length √3. So scaling (√3,1) by √3/2 would give length √3. Let me see:Wait, to get a vector of length √3 in the direction of (√3,1), we can compute:The unit vector in the direction (√3,1) is (√3/2, 1/2). Multiply by √3: ( (√3)(√3)/2 , (√3)(1)/2 ) = (3/2, √3/2 ). So vector CD would be (3/2, √3/2). Therefore, adding this to point C(-1, √3):D = C + (3/2, √3/2) = (-1 + 3/2, √3 + √3/2 ) = (1/2, (3√3)/2 )Which is the first point we found. Similarly, if we take the direction (-√3, -1), the unit vector is (-√3/2, -1/2). Multiply by √3: (-√3*(√3)/2, -√3*(1)/2 ) = (-3/2, -√3/2 ). Adding this to point C(-1, √3):D = C + (-3/2, -√3/2 ) = (-1 - 3/2, √3 - √3/2 ) = (-5/2, √3/2 )Which is the second point. So both are valid, depending on the direction of the turn at point C.But in the problem statement, it's just given as quadrilateral ABCD with the given sides and angles. There's no information about the orientation (clockwise or counterclockwise). However, typically, quadrilaterals are considered in order, so ABCD is connected in order, which would imply a specific orientation. Let's check the order.Assuming the quadrilateral is ordered A-B-C-D-A. So moving from A to B to C to D to A. So the angle at B is between A-B-C, and the angle at C is between B-C-D.Given that, when moving from B to C, the direction towards D should be such that the angle at C is 90 degrees. Let's see. If we are moving from B to C, which is towards the left (from (0,0) to (-1, √3)), then the direction from C to D can be either upwards-right or downwards-left. The two options we have for D are (1/2, 3√3/2) which is up and right, and (-5/2, √3/2) which is down and left.But in a typical convex quadrilateral, the points are arranged such that the interior angles are less than 180 degrees. If we take D as (1/2, 3√3/2), then the quadrilateral would have points going from A(1,0) to B(0,0) to C(-1, √3) to D(1/2, 3√3/2). This seems to form a convex quadrilateral. The other point D(-5/2, √3/2) would make the quadrilateral go from C to D(-5/2, √3/2), which is further left, and then back to A(1,0). That might create a concave angle at D or intersect other sides. However, without a diagram, it's hard to be certain.But since both possibilities exist, and the problem is asking for the exact length of AD, which has two possible values, √7 and √13, but the problem states "the exact length", implying a unique answer. Therefore, I must have made a wrong assumption in setting up the coordinates.Wait, let's re-examine the initial placement. Maybe I placed point A incorrectly. I assumed point A is at (1,0), but depending on the angle at B, maybe the coordinate system can be set differently.Alternatively, perhaps using vectors or the Law of Cosines in triangles ABC and BCD, then combining the results.Wait, let's try another approach. Maybe break the quadrilateral into two triangles: ABC and BCD. Then find the coordinates or vectors accordingly.In triangle ABC, we know AB = 1, BC = 2, angle at B is 120 degrees. Then, in triangle BCD, BC = 2, CD = √3, angle at C is 90 degrees. Then, we can find coordinates accordingly.Wait, but we already tried coordinates. Alternatively, use vectors.Let me think. If we take point B as the origin, then vector BA is vector A - B = A - 0 = A. Similarly, vector BC is vector C - B = C. Given angle between BA and BC is 120 degrees, so the dot product:BA • BC = |BA||BC|cosθ = 1*2*cos(120°) = 2*(-1/2) = -1If we represent BA as vector a and BC as vector c, then a • c = -1.Assuming BA is along the x-axis, then vector BA = (1,0). Then vector BC can be represented as (x,y) such that x^2 + y^2 = 4 (since |BC| = 2) and (1,0) • (x,y) = x = -1. Therefore, x = -1, so y^2 = 4 - 1 = 3 => y = ±√3. But since angle ABC is 120 degrees, which is obtuse, and if we consider the standard position, the angle between BA and BC is 120 degrees, so the y-component should be positive. Therefore, vector BC is (-1, √3), so point C is at (-1, √3). Then, as before.Now, in triangle BCD, right-angled at C, so CD is perpendicular to BC. As before, vector CD is perpendicular to vector BC.Since vector BC is (-1, √3), then vector CD must satisfy (-1, √3) • (d_x, d_y) = 0, where (d_x, d_y) is the vector from C to D. Therefore, -d_x + √3 d_y = 0 => d_x = √3 d_y. And |CD| = √3, so sqrt(d_x^2 + d_y^2) = √3. Substituting d_x = √3 d_y:sqrt(3 d_y^2 + d_y^2) = sqrt(4 d_y^2) = 2|d_y| = √3 => |d_y| = √3 / 2. Therefore, d_y = ±√3/2, then d_x = ±√3*(√3/2) = ±3/2.Therefore, vector CD can be (3/2, √3/2) or (-3/2, -√3/2). Therefore, point D is either C + (3/2, √3/2) = (-1 + 3/2, √3 + √3/2) = (1/2, 3√3/2) or C + (-3/2, -√3/2) = (-1 - 3/2, √3 - √3/2) = (-5/2, √3/2). Same as before.So, we still have two possible positions for D. The problem is the problem statement doesn't specify which one, but the answer must be unique. Therefore, perhaps there's a property we're missing here.Wait, maybe the quadrilateral is closed? Wait, no. It's a quadrilateral with sides AB, BC, CD, DA. So once you fix A, B, C, D must be placed such that DA connects back to A. Depending on where D is, DA will have different lengths.But since we have two possibilities for D, resulting in AD being √7 or √13, but the problem asks for the exact length, implying a unique answer. Therefore, maybe I made a mistake in the direction of the angle at B?Wait, let's double-check the angle at B. When I placed point A at (1,0), point B at (0,0), and calculated point C at (-1, √3). Then angle ABC is the angle between BA and BC. Vector BA is (1,0), and vector BC is (-1, √3). The angle between them can be calculated using the dot product:cosθ = (BA • BC)/( |BA||BC| ) = ( (1)(-1) + (0)(√3) ) / (1*2) = (-1)/2 = -0.5Which corresponds to θ = 120 degrees, which is correct. So that part is correct.Alternatively, maybe the problem expects us to consider the quadrilateral as a three-dimensional figure, but that's unlikely as all sides and angles are given in a planar way.Alternatively, maybe using the Law of Cosines on the quadrilateral somehow. Wait, quadrilaterals can sometimes be solved by dividing into triangles and applying the Law of Cosines or Law of Sines.We have triangle ABC and triangle BCD. Then, if we can find the coordinates or the lengths of AC and BD, maybe we can find AD.But perhaps another approach is to compute the coordinates of D as above, leading to two possibilities, but then using the fact that in a convex quadrilateral, the order of the points matters. So if the quadrilateral is convex, then point D must be placed such that all interior angles are less than 180 degrees. Let's check both possibilities.First, D at (1/2, 3√3/2):Check the angles at A and D.Alternatively, compute the angles at D for both cases. But maybe that's complicated.Alternatively, compute the vectors of the sides and check the orientation.Alternatively, maybe the problem assumes a particular orientation. For example, in standard position, turning left rather than right. If we consider the right-hand rule, the positive perpendicular direction. Since angle BCD is 90 degrees, and if we consider the direction from BC to CD as a left turn, then D would be in one position, and a right turn would place it in the other.Given that angle ABC is 120 degrees, which is a left turn from AB to BC, if we follow the order A-B-C-D-A, then the turn at C from BC to CD would also be a left turn for the quadrilateral to be convex. Let's see.From point B to C is a movement to the left (from (0,0) to (-1, √3)). Then, a left turn at C would mean continuing to turn left, which would take us towards D(1/2, 3√3/2). Alternatively, a right turn at C would take us towards D(-5/2, √3/2). However, in standard quadrilaterals, the convexity usually follows consistent left or right turns, but without specific information, it's ambiguous.But since both options are possible, but the problem must have a unique answer, I need to check if there's another constraint I missed.Wait, perhaps the coordinates of point A. Wait, in my coordinate system, point A is at (1,0), B at (0,0), C at (-1, √3). If D is at (1/2, 3√3/2), then connecting D to A would cross over BC? Let's see. The line from D(1/2, 3√3/2) to A(1,0). The slope is (0 - 3√3/2)/(1 - 1/2) = (-3√3/2)/(1/2) = -3√3. The line from B(0,0) to C(-1, √3) has slope (√3 - 0)/(-1 -0) = -√3. So the lines DA and BC have different slopes, so they don't cross unless they intersect somewhere. But since DA goes from (1,0) to (1/2, 3√3/2), and BC goes from (0,0) to (-1, √3), they might not cross.Alternatively, if D is at (-5/2, √3/2), connecting to A(1,0) would cross BC? Let's see. The line from D(-5/2, √3/2) to A(1,0). The slope is (0 - √3/2)/(1 - (-5/2)) = (-√3/2)/(7/2) = -√3/7. The line BC has slope -√3. These are different, so they might not cross either.Alternatively, maybe using complex numbers.Let me represent the points as complex numbers. Let point B be 0, point A is 1 (on the real axis). Then, point C is 2∠120°, which is 2*(cos120° + i sin120°) = 2*(-1/2 + i*(√3/2)) = -1 + i√3. Then, angle BCD is 90°, so from point C, moving perpendicular to BC for length √3.The vector BC is C - B = (-1 + i√3) - 0 = -1 + i√3. To move perpendicular from C, we can multiply this vector by i or -i to get the perpendicular direction. Multiplying by i gives (-1 + i√3)*i = -i - √3. Which is a vector pointing in the direction (-√3, -1). Multiplying by -i gives (-1 + i√3)*(-i) = i + √3. Which is a vector pointing in the direction (√3, 1). These correspond to the two directions we found earlier.To get a vector of length √3, we need to scale these. The original vector BC has length 2, so multiplying by i gives a perpendicular vector of length 2. To get length √3, we scale by √3/2. So:Direction 1: ( (-√3, -1) )*(√3/2) = (-3/2, -√3/2 ). Adding to point C(-1, √3) gives D = (-1 - 3/2, √3 - √3/2 ) = (-5/2, √3/2 )Direction 2: ( (√3, 1) )*(√3/2) = (3/2, √3/2 ). Adding to point C gives D = (-1 + 3/2, √3 + √3/2 ) = (1/2, 3√3/2 )Same results as before. So again, two possibilities.But the problem states "the exact length of side AD". If both solutions are valid, then the answer would not be unique, which contradicts the problem's implication of a unique solution. Therefore, there must be a mistake in the assumption that both positions of D are valid.Wait, perhaps the problem defines the quadrilateral in a specific way, such that the order of the letters implies the orientation. For example, ABCD means that the quadrilateral is ordered such that the sides AB, BC, CD, DA are connected in that order without crossing. Therefore, depending on the position of D, the quadrilateral may intersect itself or not.Let's check both possibilities:Case 1: D is at (1/2, 3√3/2). The quadrilateral is A(1,0)-B(0,0)-C(-1, √3)-D(1/2, 3√3/2)-A. Plotting these points, the edges AB, BC, CD, DA. Connecting them in order, this seems to form a non-intersecting (simple) quadrilateral.Case 2: D is at (-5/2, √3/2). The quadrilateral is A(1,0)-B(0,0)-C(-1, √3)-D(-5/2, √3/2)-A. Connecting these, the edge DA would cross over the edge BC if we draw it? Let's check.From D(-5/2, √3/2) to A(1,0). Let's parametrize this line:x(t) = -5/2 + (1 - (-5/2))t = -5/2 + (7/2)ty(t) = √3/2 + (0 - √3/2)t = √3/2 - √3/2 tFor t from 0 to 1.The edge BC is from B(0,0) to C(-1, √3). Parametrize as:x(s) = 0 - sy(s) = 0 + √3 sFor s from 0 to 1.Check if these lines intersect for t and s in (0,1).Set x(t) = x(s):-5/2 + (7/2)t = -sAnd y(t) = y(s):√3/2 - √3/2 t = √3 sSimplify the x equation:(7/2)t + s = 5/2Multiply by 2:7t + 2s = 5From the y equation:√3/2 - √3/2 t = √3 sDivide both sides by √3:1/2 - (1/2)t = sSo s = (1 - t)/2Substitute into the x equation:7t + 2*( (1 - t)/2 ) = 5Simplify:7t + (1 - t) = 56t +1 =5 => 6t=4 => t=4/6=2/3Then s=(1 - 2/3)/2=(1/3)/2=1/6Since t=2/3 and s=1/6 are both in (0,1), the lines DA and BC intersect at this point. Therefore, the quadrilateral would intersect itself, making it a complex quadrilateral. Since the problem doesn't specify that it's simple, but generally, problems assume simple quadrilaterals unless stated otherwise. Therefore, the correct position of D is the one that makes the quadrilateral simple, i.e., non-intersecting, which is D at (1/2, 3√3/2).Therefore, the length AD is √[ (1/2 -1)^2 + (3√3/2 -0)^2 ] = √[ (-1/2)^2 + (3√3/2)^2 ] = √[ 1/4 + 27/4 ] = √[28/4] = √7Hence, the exact length of AD is √7.Final AnswerThe exact length of side (AD) is (boxed{sqrt{7}}).

✅:To find the exact length of side (AD) in the quadrilateral (ABCD) with given sides (AB = 1), (BC = 2), (CD = sqrt{3}), and angles (angle ABC = 120^circ) and (angle BCD = 90^circ), we can use coordinate geometry and vector analysis.1. Assign Coordinates: - Place point (B) at the origin ((0, 0)). - Place point (A) at ((1, 0)) along the positive x-axis. - Calculate coordinates of point (C) using the angle (120^circ) and length (BC = 2): [ C_x = 2 cos(120^circ) = -1, quad C_y = 2 sin(120^circ) = sqrt{3} ] Thus, (C) is at ((-1, sqrt{3})).2. Determine Coordinates of (D): - Since (angle BCD = 90^circ), vector (CD) must be perpendicular to vector (BC). Vector (BC) is ((-1, sqrt{3})). - Solve for coordinates of (D) such that (CD = sqrt{3}) and perpendicular to (BC): - Let (D = (x, y)). The vector (CD = (x + 1, y - sqrt{3})) must satisfy: [ (x + 1)(1) + (y - sqrt{3})(-sqrt{3}) = 0 quad text{(dot product)} ] [ sqrt{(x + 1)^2 + (y - sqrt{3})^2} = sqrt{3} ] - Solving these equations, we find two possible coordinates for (D): ((frac{1}{2}, frac{3sqrt{3}}{2})) and ((-frac{5}{2}, frac{sqrt{3}}{2})).3. Check for Non-Intersecting Quadrilateral: - The coordinates ((frac{1}{2}, frac{3sqrt{3}}{2})) result in a convex quadrilateral without intersecting sides. - The coordinates ((-frac{5}{2}, frac{sqrt{3}}{2})) result in a complex quadrilateral with intersecting sides.4. Calculate Distance (AD): - Using the valid coordinates (D(frac{1}{2}, frac{3sqrt{3}}{2})): [ AD = sqrt{left(1 - frac{1}{2}right)^2 + left(0 - frac{3sqrt{3}}{2}right)^2} = sqrt{left(frac{1}{2}right)^2 + left(frac{-3sqrt{3}}{2}right)^2} = sqrt{frac{1}{4} + frac{27}{4}} = sqrt{frac{28}{4}} = sqrt{7} ]Thus, the exact length of side (AD) is (boxed{sqrt{7}}).

🔑:# Method I:1. Place the Quadrilateral in the Coordinate System: - Place quadrilateral ABCD in a right-angled coordinate system such that segment CD lies on the x-axis and point C is at the origin. - The coordinates of the points are C(0, 0), D(sqrt{3}, 0), B(0, 2), and we need to find A.2. Calculate the Coordinates of Point A: - The triangle ABA' (where A' is the projection of A on y-axis) has A A' angle B A A' = 60^circ. - Using the sine and cosine values for angle BAA' which is 60^circ: [ A A' = AB sin 60^circ = 1 cdot frac{sqrt{3}}{2} = frac{sqrt{3}}{2} ] [ B A' = AB cos 60^circ = 1 cdot frac{1}{2} = frac{1}{2} ] - Thus, the coordinates of A become: [ A left( frac{sqrt{3}}{2}, frac{5}{2} right) ]3. Determine the Length of AD: - Using the distance formula between points A left( frac{sqrt{3}}{2}, frac{5}{2} right) and D(sqrt{3}, 0): [ AD = sqrt{left( frac{sqrt{3}}{2} - sqrt{3} right)^2 + left( frac{5}{2} right)^2} ] - Calculate the terms inside the square root: [ left( frac{sqrt{3}}{2} - sqrt{3} right)^2 = left( frac{sqrt{3}}{2} - frac{2sqrt{3}}{2} right)^2 = left( -frac{sqrt{3}}{2} right)^2 = frac{3}{4} ] [ left( frac{5}{2} right)^2 = frac{25}{4} ] - Adding these together: [ AD = sqrt{ frac{3}{4} + frac{25}{4} } = sqrt{ frac{28}{4} } = sqrt{7} ]4. Conclusion: [ boxed{sqrt{7}} ]# Method II:1. Find AC Using the Cosine Rule: - Using the law of cosines in triangle ABC: [ AC^{2} = AB^{2} + BC^{2} - 2 cdot AB cdot BC cdot cos 120^circ ] - Substitute the known values: [ AC^{2} = 1^{2} + 2^{2} - 2 cdot 1 cdot 2 cdot left( -frac{1}{2} right) = 1 + 4 + 2 = 7 ] [ AC = sqrt{7} ]2. Find DB Using Distance Formula: - Since triangle BCD is right-angled at C: [ DB^{2} = BC^{2} + CD^{2} = 2^{2} + (sqrt{3})^{2} = 4 + 3 = 7 ] [ DB = sqrt{7} ]3. Using the Law of Cosines to Find angle alpha: - Let alpha be angle BCA: [ cos alpha = frac{(AC)^{2} + (BC)^{2} - (AB)^{2}}{2 cdot AC cdot BC} = frac{7 + 4 - 1}{2 cdot 2 cdot sqrt{7}} = frac{10}{4 sqrt{7}} = frac{5 sqrt{7}}{14} ]4. Using Law of Cosines in Delta ADC: - cos (90^circ - alpha) = sin alpha: [ sin alpha = sqrt{1 - cos^{2} alpha} = sqrt{1 - left( frac{5 sqrt{7}}{14} right)^{2}} = sqrt{1 - frac{75}{196}} = sqrt{frac{121}{196}} = frac{11}{14} ] - Apply cosine rule in triangle ADC: [ AD^{2} = AC^{2} + CD^{2} - 2 cdot AC cdot CD cdot cos (90^circ - alpha) ] [ AD^{2} = 7 + 3 - (2 cdot sqrt{7} cdot sqrt{3} cdot frac{3 sqrt{7}}{14}) = 7 + 3 - 6 = 7 ] [ AD = sqrt{7} ]5. Conclusion: [ boxed{sqrt{7}} ]# Method III:1. Extend Sides AB and CD, Let E be the Intersection: - In right triangle BCE: [ angle CBE = 60^circ, angle BEC = 30^circ, BE = 2 cdot BC = 4, CE = sqrt{3} cdot BC = 2 sqrt{3} ] - In triangle AEB: [ AE = AB + BE = 1 + 4 = 5 ] [ DE = sqrt{3} cdot 2 = 3 sqrt{3} ] [ angle AED = 30^circ ]2. Apply the Law of Cosines in ADE: [ AD^{2} = AE^{2} + DE^{2} - 2 cdot AE cdot DE cdot cos(30^circ) ] [ AD^{2} = 25 + 27 - 2 cdot 5 cdot 3 sqrt{3} cdot frac{sqrt{3}}{2} = 25 + 27 - 30 cdot frac{3}{2} = 25 + 27 - 45 = 7 ] [ AD = sqrt{7} ]3. Conclusion: [ boxed{sqrt{7}} ]

🧠:Okay, let's see. I need to find the total surface area of a prism that's circumscribed about a sphere, and the area of its base is given as S. Hmm, alright. Let me start by understanding what this means. A prism circumscribed about a sphere... So, the sphere is tangent to all the faces of the prism? That makes sense. That would mean the sphere is inscribed in the prism, right? So the sphere touches each face of the prism exactly once. First, let's recall some properties. If a sphere is inscribed in a prism, the prism must be a right prism because otherwise, the sphere wouldn't be tangent to all the lateral faces. So, the prism is right, meaning the sides are rectangles, and the top and bottom faces are congruent polygons. The sphere touches the top, bottom, and all the lateral faces. Now, the radius of the sphere must be related to the dimensions of the prism. Since the sphere is tangent to the top and bottom faces, the height of the prism (the distance between the two bases) must be equal to twice the radius of the sphere. Let me write that down: if the radius of the sphere is r, then the height h of the prism is 2r. But also, the sphere is tangent to the lateral faces. For a prism circumscribed about a sphere, the sphere must fit perfectly within the prism. That means that in the base of the prism, which is a polygon, the sphere must be tangent to all the sides of the base. Wait, but the sphere is three-dimensional. Hmm, so maybe the projection of the sphere onto the base is a circle inscribed in the base polygon. Therefore, the base polygon must have an incircle tangent to all its sides. A polygon that has an incircle is called a tangential polygon, and for a polygon to be tangential, it must be convex and all its sides must be tangent to a single circle. Moreover, in such polygons, the sum of the lengths of opposite sides are equal. For example, in a tangential quadrilateral, the sums of the two pairs of opposite sides are equal. But the problem just says "a prism," so the base could be any polygon, right? But since it's circumscribed about a sphere, the base must be a tangential polygon. So, the base is a tangential polygon with area S, and the prism is right, with height equal to 2r, where r is the radius of the inscribed sphere. The total surface area of the prism would be the sum of the lateral surface area and twice the area of the base. The formula is: Total Surface Area = 2S + Lateral Surface Area. Now, I need to find the lateral surface area. The lateral surface area of a right prism is the perimeter of the base multiplied by the height. So, Lateral Surface Area = Perimeter of Base * Height. But we know the height is 2r. However, we don't know the perimeter of the base directly. But perhaps we can relate the perimeter to the area S and the radius r? Since the base is a tangential polygon, its area can be expressed as S = r * p / 2, where p is the perimeter of the base. Wait, no, hold on. For a tangential polygon, the area is equal to the inradius multiplied by the semiperimeter. Wait, that's for a circle inscribed in a polygon. Wait, in a triangle, the area is r*(a+b+c)/2, where r is the inradius. Similarly, in a tangential polygon, the area is equal to the inradius multiplied by the semiperimeter. Wait, let me confirm that. Yes, for a tangential polygon, the area is indeed given by S = r * p / 2, where p is the perimeter. Wait, in a triangle, area is r*(a+b+c)/2, which would be r*(p)/2, where p is the perimeter. Wait, but actually, semiperimeter is (a+b+c)/2, so area is r * semiperimeter. So maybe for a tangential polygon, area S = r * semiperimeter. Hmm, now I'm confused. Let me check.For a triangle: S = r * s, where s is the semiperimeter. For a tangential quadrilateral: S = r * s, where s is the semiperimeter as well. So, in general, for any tangential polygon, the area is equal to the inradius multiplied by the semiperimeter. Therefore, S = r * (p / 2), where p is the perimeter. So S = (r * p)/2, which implies that p = 2S / r. Wait, so perimeter p = 2S / r. But in our case, the inradius of the base polygon would be equal to the radius of the sphere? Hmm, no. Wait, the sphere is inscribed in the prism, so the inradius of the base polygon is the same as the radius of the sphere. Because the sphere touches the lateral faces, so the distance from the center of the sphere to each lateral face is equal to r. But since the prism is right, the lateral faces are rectangles, and the inradius of the base polygon (which is tangent to all sides of the base) must be equal to the radius of the sphere. Therefore, the inradius of the base polygon is r. Therefore, the area of the base S = r * (p / 2), so p = 2S / r. Therefore, the perimeter of the base is p = 2S / r. Then, the lateral surface area is p * h = (2S / r) * 2r = 4S. Therefore, the total surface area is 2S + 4S = 6S. Wait, that seems straightforward. But let me double-check. So, if the base area is S, and it's a tangential polygon with inradius r, then S = r * semiperimeter. So semiperimeter is S / r, which means perimeter p = 2S / r. Then, the lateral surface area is perimeter times height. The height h is equal to 2r, since the sphere has radius r and the prism's height is twice the radius. So lateral surface area is (2S / r) * 2r = 4S. Then total surface area is 2S (for the two bases) plus 4S (lateral) equals 6S. Therefore, the total surface area is 6S. But let me think again. Is there any assumption here that might be incorrect? For example, the inradius of the base is equal to the radius of the sphere. Is that necessarily true? The sphere is inscribed in the prism, so the center of the sphere is equidistant from all the faces. The distance from the center to each lateral face is r, which would correspond to the inradius of the base. Similarly, the distance from the center to the top and bottom faces is also r, which gives the height of the prism as 2r. Yes, that makes sense. The inradius of the base (distance from center to the sides) is the same as the radius of the sphere. Therefore, the semiperimeter of the base is S / r, so perimeter is 2S / r. Therefore, lateral surface area is perimeter times height: (2S / r) * 2r = 4S. Thus total surface area is 2S + 4S = 6S. Wait, but this seems too straightforward. Let me verify with a simple case. Let's take a cube. If a cube is circumscribed about a sphere, then the sphere is the inscribed sphere of the cube. For a cube, the edge length is 2r, so the surface area is 6*(2r)^2 = 24r². But according to our formula, the base area S would be the area of one face, which is (2r)^2 = 4r². Then total surface area would be 6*4r² = 24r², which matches. So that works. Another example: a regular triangular prism. Suppose the base is an equilateral triangle with an incircle of radius r. The area of the base S = r * semiperimeter. The semiperimeter of an equilateral triangle is (3a)/2, where a is side length. So S = r * (3a)/2. But also, the area of an equilateral triangle is (√3/4)a². Therefore, setting these equal: (√3/4)a² = r*(3a)/2. Solving for a: (√3/4)a = (3/2)r ⇒ a = (3/2)r * (4/√3) = (6r)/√3 = 2√3 r. So the perimeter of the base is 3a = 6√3 r. Then the lateral surface area would be perimeter times height. The height h = 2r. So lateral surface area = 6√3 r * 2r = 12√3 r². The total surface area is 2*(√3/4)*(2√3 r)^2 + 12√3 r². Let's compute that. First, the base area S = √3/4 * (2√3 r)^2 = √3/4 * 4*3 r² = √3/4 *12 r² = 3√3 r². Then total surface area would be 2*3√3 r² + 12√3 r² = 6√3 r² + 12√3 r² = 18√3 r². Now, using our formula: total surface area should be 6S. Since S = 3√3 r², 6S = 18√3 r². Which matches. So that works as well. Therefore, the formula seems to hold. Thus, the total surface area of the prism circumscribed about a sphere with base area S is 6S. Final AnswerThe total surface area of the prism is boxed{6S}.

✅:To find the total surface area of a prism circumscribed about a sphere with the area of its base ( S ):1. Understanding the Prism and Sphere Relationship: - The prism is circumscribed about a sphere, meaning the sphere is tangent to all faces of the prism. - The prism is a right prism (height ( h ) is twice the radius ( r ) of the sphere, so ( h = 2r )). - The base of the prism is a tangential polygon (it has an incircle with radius ( r )).2. Properties of the Tangential Polygon: - The area ( S ) of a tangential polygon is given by ( S = r cdot frac{p}{2} ), where ( p ) is the perimeter of the base. - Solving for the perimeter ( p ), we get ( p = frac{2S}{r} ).3. Calculating the Lateral Surface Area: - The lateral surface area of a prism is the product of the perimeter of the base and the height. - Substituting the values, the lateral surface area is ( frac{2S}{r} cdot 2r = 4S ).4. Total Surface Area Calculation: - The total surface area of the prism is the sum of the lateral surface area and twice the base area. - Total surface area = ( 2S + 4S = 6S ).Thus, the total surface area of the prism is (boxed{6S}).

🔑:To find the total surface area of the prism described in the problem, we follow these steps:1. Define Variables: - Let ( S ) be the area of the base of the prism. - Let ( p ) be the semi-perimeter of the base of the prism. - Let ( r ) be the radius of the sphere around which the prism is circumscribed.2. Relationship between Base Area and Semi-Perimeter: - Given that the prism is circumscribed around a sphere, we utilize the property that the area of the base of the prism can be expressed in terms of the semi-perimeter and the radius of the sphere. - For a polygon circumscribed around a circle, the area ( S ) is given by: [ S = p cdot r ] - This equation relates the semi-perimeter ( p ), the radius ( r ), and the area of the base ( S ).3. Calculate the Area of the Lateral Surface: - The lateral surface area of the prism consists of rectangles, each with height equal to the diameter of the sphere (which is ( 2r )) and width equal to a side of the base polygon. - Therefore, the total lateral surface area can be represented as: [ text{Lateral Surface Area} = p cdot 2r ] However, for a prism with ( n ) sides, this becomes: [ text{Lateral Surface Area} = p cdot h ] where ( h = 2r ).4. Express Lateral Surface Area in Terms of ( S ): - From the above expression, we know: [ text{Lateral Surface Area} = 2p cdot r = 2 cdot S ]5. Total Surface Area Calculation: - The total surface area of the prism is the sum of the area of its bases (top and bottom) and the lateral surface area. - Since there are two bases, the area of the bases is: [ text{Area of Bases} = 2 cdot S ] - Combining all contributions, the total surface area is: [ text{Total Surface Area} = text{Area of Bases} + text{Lateral Surface Area} = 2S + 4S = 6S ] Conclusion:[boxed{6S}]

🧠:Okay, so I need to find a six-digit number N where all the digits are distinct, and when I multiply it by 2, 3, 4, 5, and 6, each of those products is a permutation of the original digits. Hmm, that sounds like a cyclic number problem or something related to permutations. Let me think step by step.First, let's understand the problem. N is a six-digit number with all different digits. Then 2N, 3N, 4N, 5N, 6N must each be permutations of those digits. So, each multiple just rearranges the digits but doesn't introduce any new digits or repeat any digits that aren't already in N. That means all the multiples must also be six-digit numbers, right? Because if multiplying by 6 made it a seven-digit number, that would have an extra digit which isn't in the original. So N must be such that even when multiplied by 6, it's still a six-digit number. Therefore, the maximum value N can be is such that 6N < 1,000,000. So N < 166,666.666... So N must be less than 166,667. But since N is a six-digit number, it must be between 100,000 and 166,666.Wait, but 166,666 * 6 = 999,996, which is a six-digit number. So N has to be between 100,000 and 166,666. Also, all digits in N must be distinct. So digits can't repeat. Let me note that down.Another thought: If N and its multiples are permutations, they all must have the same digits, just rearranged. Therefore, all these numbers must have the same digit sum. Since the digit sum is a permutation each time, the digit sum remains constant. The digit sum is important for divisibility rules, like divisibility by 3 or 9. So if N is divisible by 3, then all its multiples would also be divisible by 3. But since the digits are just permutations, they all have the same digit sum, so if N is divisible by 3, all multiples would be too, which is obvious. Not sure if that helps, but maybe.But perhaps more importantly, since N is six digits, and 6N is also six digits, N must start with a 1. Because if N starts with, say, 2, then 6N would start with at least 12, which is seven digits. Wait, no. Wait, if N is 200,000, then 6N is 1,200,000 which is seven digits. But our N must be less than 166,667 as before. So 166,666 * 6 is 999,996, which is six digits. Therefore, N must start with a 1. So the first digit is 1. That narrows it down. So N is in the range 100,000 to 166,666, with all distinct digits, starting with 1.Also, since N has six digits, all distinct, and starting with 1, the digits 1 is already used, so the other digits must be from 0-9 excluding 1. But wait, 0 can be included, but since it's a six-digit number, the first digit is 1, so 0 can be in the other positions. So digits allowed are 0,2,3,4,5,6,7,8,9, but each digit must be distinct. So N has digits 1 and five other distinct digits from 0,2-9.Another property: cyclic numbers. The most famous cyclic number is 142857. Let me recall, 142857 multiplied by 1 through 6 gives permutations:142857 × 1 = 142857142857 × 2 = 285714142857 × 3 = 428571142857 × 4 = 571428142857 × 5 = 714285142857 × 6 = 857142Wait, that's exactly what the problem is asking for! So maybe 142857 is the number. Let me check if all digits are distinct. 1,4,2,8,5,7 – yes, all distinct. So 142857 seems to fit. But let me confirm the problem's conditions. The problem states "a six-digit number N composed of distinct digits such that the numbers 2N, 3N, 4N, 5N, and 6N are permutations of its digits." So 142857 does satisfy this. So is that the answer?But wait, the problem might be expecting me to derive it rather than recalling it. Let me verify if this is indeed the case. Let me check 142857.First, check 2N: 142857 × 2 = 285714. The digits are 2,8,5,7,1,4. Original digits are 1,4,2,8,5,7. So yes, same digits.3N: 142857 × 3 = 428571. Digits 4,2,8,5,7,1. Same as original.4N: 142857 × 4 = 571428. Digits 5,7,1,4,2,8. Same.5N: 142857 ×5=714285. Digits 7,1,4,2,8,5. Same.6N: 142857 ×6=857142. Digits 8,5,7,1,4,2. Same. So all permutations. So yes, 142857 is the number. But how did I know that? Because it's a well-known cyclic number. But maybe the problem expects the process to arrive at it without prior knowledge.So, assuming I didn't know about 142857, how would I find it?Let me try to figure it out.First, since N is a six-digit number starting with 1, and when multiplied by 2-6, remains six digits. So N must be between 100,000 and 166,666. Also, the digits are distinct. So let's consider the structure.Since multiplying by 1 through 6 gives cyclic permutations, perhaps the number is related to 1/7. Because 1/7 = 0.142857142857..., and the repeating cycle is 142857. That's where the cyclic number comes from. So maybe the number is derived from the decimal expansion of 1/7. But maybe that's too much prior knowledge.Alternatively, let's try to find such a number through reasoning.Let me note that since N, 2N, ..., 6N are permutations, they must all have the same digits. Therefore, they must all have the same digit sum, same number of even digits, etc. Also, since multiplying by 2, 4, 5, 6, the last digit might have some constraints.For example, the last digit of N multiplied by 1-6 should be a permutation of the digits. Therefore, the last digit of N must be such that when multiplied by 2-6, it results in digits that are in the original set. Let's think about possible last digits.Possible last digits of N (since digits are distinct and N starts with 1):Possible last digits: 0,2,3,4,5,6,7,8,9 (can't be 1). But also, considering that 5N must end with a digit that's a permutation. If the last digit of N is 5, then 5*5=25, so 5N would end with 5, but since digits must be distinct, N can't have two 5s. Wait, N has distinct digits, so if N ends with 5, then 5 is already a digit in N, and 5N would end with 5 (since 5*5=25, ends with 5). But 5 is already in N, so that's okay. Wait, but if N has a 5, then 2N, 3N, etc., must also have digits that include 5 if it's a permutation. But 2N might not necessarily end with 5. Let's check.Alternatively, if the last digit is 0, then 0 is part of N. Then 2N, 3N, etc., would end with 0 as well only if the multiplication doesn't carry over. But multiplying 0 by 2-6 is still 0, so last digit would be 0. So all multiples would end with 0, but since the original number has a 0, that's okay. However, in this case, all permutations would have 0 in the last digit. But permutations can have 0 in any position except the first. However, if all multiples end with 0, then their first digits would be different. But wait, for example, if N ends with 0, then 2N ends with 0, but the first digit would be different. But if N is, say, 102345, multiplying by 2 would be 204690, which has digits 2,0,4,6,9,0 – but duplicate 0s, which is invalid. So N cannot have a 0 as the last digit because 2N would have two 0s. Wait, but N has distinct digits, so if N has a 0, it's only once. So multiplying N by 2 would result in a number that has a 0 in the last digit (if N ended with 0), but 0 was already present once. However, multiplying by 2 might produce another 0 if there's a carryover. Wait, no. If N ends with 0, then 2N ends with 0, 3N ends with 0, etc. But since N already has one 0, all multiples would also have at least one 0 (in the last digit). But since they are permutations, they must have exactly one 0. Therefore, multiplying N by any number 1-6 must not create another 0 elsewhere. But if the last digit is 0, then all multiples end with 0, so the 0 is fixed in the last digit. However, the permutations must rearrange the digits. But if all multiples have 0 in the last digit, then the original number must have 0 in the last digit, but the permutations would require 0 to be in different positions. Contradiction. Therefore, the last digit of N cannot be 0. Therefore, N cannot contain a 0. Because if N contains a 0 in any position, when multiplied by 2-6, the 0 could become another digit, but if the last digit is non-zero, then multiplying might carry over. Wait, this is getting complicated. Maybe it's safer to assume that N doesn't contain 0. Because if N contains 0, then when multiplied by some number, it might result in a 0 in a different position, but since all digits are permutations, the number of zeros must remain the same. So if N has a 0, all multiples must have exactly one 0. But multiplying N by 2, for example, could create a new 0 if a digit becomes 0 through multiplication. But since digits are distinct, original N has one 0. For example, suppose N has a 0 in the second digit: 102345. Multiplying by 2 gives 204690, which has two 0s. Invalid. Therefore, N cannot contain a 0. Therefore, all digits of N are from 1-9, with distinct digits. So digits are 1, and five others from 2-9, all distinct.So, N starts with 1, and has digits from 2-9, all distinct. Now, let's think about the last digit.Since multiplying by 2-6 will change the last digit, but all the resulting last digits must be permutations of N's digits. Therefore, the last digit of N must be such that when multiplied by 2-6, the last digits of those multiples are all in N's digits.For example, suppose the last digit is d. Then 2d mod 10, 3d mod 10, 4d mod 10, 5d mod 10, 6d mod 10 must all be digits in N. Since N has digits 1, and five others from 2-9.Also, since N is divisible by 1, and all permutations (which are multiples) must be divisible by different numbers. Not sure.Alternatively, let's consider possible last digits d and check if 2d, 3d, 4d, 5d, 6d modulo 10 are distinct and are among the digits of N.Wait, but the digits of N are 1 and five others. So the last digits of the multiples must be in the set {1, a, b, c, d, e}, where a,b,c,d,e are five distinct digits from 2-9.Wait, but the last digit of N is d, so d is one of 2-9. Let's consider possible d.Let me check each possible d from 2-9 and see if 2d, 3d, 4d, 5d, 6d mod 10 are all in the set {1, a, b, c, d, e} where a,b,c,d,e are distinct digits from 2-9, and d is among them.But this might take time. Let's try:d=2:2*2=4, 3*2=6, 4*2=8, 5*2=10→0, 6*2=12→2. So last digits would be 4,6,8,0,2. But 0 is not allowed (since N doesn't have 0), and 2 is already in N. But 0 is not part of N, so invalid.d=3:2*3=6, 3*3=9, 4*3=12→2, 5*3=15→5, 6*3=18→8. Last digits: 6,9,2,5,8. So these must be in N's digits. N has 1 and five others. If d=3, then digits needed are 1,3,6,9,2,5,8. That's seven digits, which is too many. N is six-digit. So invalid.d=4:2*4=8, 3*4=12→2, 4*4=16→6, 5*4=20→0, 6*4=24→4. Last digits:8,2,6,0,4. Again 0 is invalid. So reject.d=5:2*5=10→0, 3*5=15→5, 4*5=20→0, 5*5=25→5, 6*5=30→0. Last digits:0,5,0,5,0. Multiple zeros and repeats. Invalid.d=6:2*6=12→2, 3*6=18→8, 4*6=24→4, 5*6=30→0, 6*6=36→6. Last digits:2,8,4,0,6. Again 0 is invalid.d=7:2*7=14→4, 3*7=21→1, 4*7=28→8, 5*7=35→5, 6*7=42→2. Last digits:4,1,8,5,2. So digits needed: 1,7 (since d=7), and 4,1,8,5,2. Wait, digits would be 1,7,4,8,5,2. That's six digits. All distinct, no zeros. Possible! Let's check:Original digits: 1,7,4,8,5,2 (since N starts with 1 and ends with 7). Then the multiples would end with 4,1,8,5,2 respectively. So the digits in N are exactly these. So this could be a candidate. So maybe d=7.d=8:2*8=16→6, 3*8=24→4, 4*8=32→2, 5*8=40→0, 6*8=48→8. Last digits:6,4,2,0,8. 0 invalid.d=9:2*9=18→8, 3*9=27→7, 4*9=36→6, 5*9=45→5, 6*9=54→4. Last digits:8,7,6,5,4. So digits needed:1,9,8,7,6,5,4. Seven digits, invalid.So only d=7 gives six distinct digits without 0. So last digit must be 7. Therefore, N ends with 7, and the digits are 1,7,4,8,5,2. So N is 1 followed by some permutation of 4,8,5,2,7. Wait, but the last digit is 7, so the number is 1 _ _ _ _ 7, with digits 1,4,8,5,2,7.Now, let's see. If N is such a number, then 2N, 3N, 4N, 5N, 6N should be permutations of these digits. Let's see. For example, the known cyclic number 142857 ends with 7. Let me check if that's the case. Yes, 142857 ends with 7. And its multiples:2*142857=2857143*142857=4285714*142857=5714285*142857=7142856*142857=857142All of these are permutations. So the digits are indeed 1,4,2,8,5,7. So this fits. So N=142857.But how to arrive at this without prior knowledge? Let's think.We know N starts with 1, ends with 7, and has digits 1,2,4,5,7,8.Now, let's try to construct such a number. Since multiplying by 2-6 cycles the digits, perhaps the number is a cyclic permutation. So if we can find a number where multiplying by 2 shifts the digits cyclically. For example, 142857. Multiplying by 2 gives 285714, which is a cyclic shift. Similarly, multiplying by 3 gives another shift, etc. But how to find this number.Another approach: Since N is a cyclic number, it's related to the fraction 1/7. Let's calculate 1/7:1/7 = 0.142857142857...So the repeating decimal is 142857. Hence, 142857 is the cyclic number. Therefore, this number multiplied by 2-6 gives permutations.But even if I didn't remember that, perhaps through considering the properties of the multiples.Given that N is 1 followed by 4,2,8,5,7. Let's check if multiplying by 2 gives a permutation.142857 *2=285714. Yes, digits are 2,8,5,7,1,4. Permutation.Similarly, *3=428571, etc.But how would I construct this? Let's suppose that multiplying by 2 shifts the digits by two places. For example, if N is ABCDEF, then 2N is CDEFAB or something. But not sure. Alternatively, maybe the multiplication causes a cyclic permutation.Alternatively, consider that since the number is cyclic, each multiplication by 2 shifts the digits in a certain way. Let's try to see if we can reconstruct it.Let me denote the digits of N as d1 d2 d3 d4 d5 d6, where d1=1 and d6=7.We know that 2N must be a permutation. Let's attempt to compute 2N.N=1 a b c d 7Then 2N must be a six-digit number. Let's compute 2*N:But since N is between 100,000 and 166,666, 2N is between 200,000 and 333,332. So the first digit of 2N must be 2 or 3.But 2N is a permutation of the digits 1,2,4,5,7,8. So the first digit of 2N must be one of these digits. So possible first digits are 2,4,5,7,8. But 2N is between 200,000 and 333,332, so first digit can be 2 or 3. But 3 is not among the digits of N (since digits are 1,2,4,5,7,8). Therefore, 2N must start with 2.Therefore, 2N starts with 2. Similarly, 3N must start with a digit from the set. 3*N is between 300,000 and 500,000 (since N up to 166,666). So first digit of 3N is 3,4, or 5. But 3 is not in the digits of N, so 3N must start with 4 or 5. But digits are 1,2,4,5,7,8. So first digit is 4 or 5.Similarly, 4N would be between 400,000 and 666,664. First digit could be 4,5,6, but 6 is not in the digits. So first digit 4,5,7,8. But since 4N is a permutation, first digit must be one of the digits of N: 4,5,7,8.5N is between 500,000 and 833,330. First digit could be 5,7,8.6N is between 600,000 and 999,996. First digit could be 6,7,8,9. But 6 is not a digit in N, so first digit must be 7,8,9. But digits are only up to 8. So first digit of 6N is 8 or 9. But 9 is not in N, so 6N must start with 8.Therefore:- 2N starts with 2- 3N starts with 4 or 5- 4N starts with 5,7,8- 5N starts with 7 or 8- 6N starts with 8Given that, let's try to reconstruct N.Starting with N=1abc d7.Multiply by 2: 2N=2??? ??, which is a permutation. Also, since N ends with 7, 2*7=14, so the last digit of 2N is 4. Therefore, 2N ends with 4. But since 2N is a permutation of N's digits, which include 4. Therefore, the last digit of 2N is 4.Similarly, multiplying N by 3, the last digit is 1 (since 3*7=21). So 3N ends with 1.Multiplying N by 4, last digit is 8 (4*7=28).Multiplying N by 5, last digit is 5 (5*7=35).Multiplying N by 6, last digit is 2 (6*7=42).So:- 2N ends with 4- 3N ends with 1- 4N ends with 8- 5N ends with 5- 6N ends with 2So these are the last digits of the multiples. Since all are permutations, the digits 1,2,4,5,8 must be in N. Which they are (since we have digits 1,2,4,5,7,8).Now, let's use this to reconstruct the number.Start with N=1 a b c d 7We know:2N ends with 4, so 2*N mod 10=4. Since N ends with 7, 2*7=14, so yes, last digit 4.Similarly, 3N ends with 1: 3*7=21, so last digit 1.So the last digits are determined.Now, let's try to build N step by step.Let's consider the known number 142857. Let's see how its multiples behave.N=1428572N=2857143N=4285714N=5714285N=7142856N=857142Notice that each multiple is a cyclic permutation. So shifting the digits. For example, multiplying by 2 shifts the digits two places to the right. Similarly, multiplying by 3 shifts three places, etc. But this is a property of cyclic numbers.Given that, perhaps we can model N as a cyclic number where multiplying by 2 results in a shift of two digits. Let's try to see that.Suppose N is d1 d2 d3 d4 d5 d6Then 2N is d3 d4 d5 d6 d1 d2But for N=142857, 2N=285714. Which is d3 d4 d5 d6 d1 d2 = 4 2 8 5 7 1 → Wait, no. Wait, 2N=285714, which is digits 2,8,5,7,1,4. Original N is 1,4,2,8,5,7. So 2N is shifting left by two digits? Let's see:Original: 1 4 2 8 5 72N: 2 8 5 7 1 4So it's like moving each digit two positions to the left (with wrap-around). Similarly, multiplying by 3 shifts three positions:3N: 4 2 8 5 7 1 → Wait, 3N=428571, which is digits 4,2,8,5,7,1. Original N is 1,4,2,8,5,7. So shifting left by three positions:Original: 1 4 2 8 5 7Shift left by three: 8 5 7 1 4 2 → But 3N is 428571, which is different. Hmm, maybe it's shifting right?Original: 1 4 2 8 5 7Shift right by one: 7 1 4 2 8 5 (which is 714285, which is 5N). Shift right by two: 5 7 1 4 2 8 (which is 571428, which is 4N). Shift right by three: 8 5 7 1 4 2 (which is 857142, which is 6N). Shift right by four: 2 8 5 7 1 4 (which is 285714, which is 2N). Shift right by five: 4 2 8 5 7 1 (which is 428571, which is 3N). So multiplying by 2 is equivalent to shifting right by four positions, multiplying by 3 is shifting right by five positions, etc. This seems related to modular arithmetic.Alternatively, since the number is related to 1/7=0.(142857), the cyclic nature comes from the decimal expansion. Each multiple corresponds to a different starting point in the cycle.But maybe this is going too deep. Let's try to reconstruct the number.We know N starts with 1 and ends with 7, with digits 1,2,4,5,7,8.Let's attempt to write N as 1 a b c d 7.We need to find a,b,c,d such that when multiplied by 2-6, the products are permutations.Given that N=1 a b c d 7.Let's compute 2N:2*N = 2*(100000 + 10000a + 1000b + 100c +10d +7) = 200000 + 20000a + 2000b + 200c +20d +14 = 200014 + 20000a + 2000b +200c +20d.But since 2N is a permutation, it must start with 2. Therefore, the first digit is 2, so:200014 + ... must be between 200000 and 299999.Given that N is between 100,000 and 166,666, 2N is between 200,000 and 333,332. But since 2N must start with 2, N must be less than 150,000, because 2*150,000=300,000. Wait, but 166,666*2=333,332. But earlier, we saw that 2N must start with 2, which requires that 2N < 300,000. Therefore, N must be less than 150,000. So N is between 100,000 and 149,999.But our earlier assumption was up to 166,666, but since 2N must start with 2, N must be less than 150,000. So this tightens the range.Therefore, N is between 100,000 and 149,999, starts with 1, ends with 7, digits 1,2,4,5,7,8.So let's correct that: N is between 100,000 and 149,999, so the second digit 'a' must be between 0 and 4, but since digits are distinct and from 2,4,5,7,8 (excluding 1 and 7, which are already in N), wait, digits are 1,2,4,5,7,8. So 'a' can be 2,4,5,8. Because N is between 100,000 and 149,999, so the second digit is 0-4, but digits are 1,2,4,5,7,8. Therefore, 'a' can be 2 or 4.Because 0 is excluded, 1 is already first digit, 3 is not in the digits, so possible second digits are 2,4. So a=2 or 4.Case 1: a=2Then N=1 2 b c d 7. Digits used:1,2,7. Remaining digits to assign:4,5,8.So b,c,d must be some permutation of 4,5,8.Let's suppose N=12bcd7, with b,c,d from 4,5,8.Now, compute 2N. Let's take N=124857. Wait, let's test this.Wait, but if N=124857:2N=249714. Check if this is a permutation. Original digits:1,2,4,8,5,7. 2N digits:2,4,9,7,1,4. Wait, there's a 9 which is not in the original digits. Also, duplicate 4s. So invalid.Alternatively, maybe N=142857. Wait, even though a=4 here. Let's check.N=142857. Let's check 2N=285714. Digits:2,8,5,7,1,4. All in the original set. Good.But in this case, a=4. So let's consider case a=4.Case 2: a=4Then N=1 4 b c d 7. Digits used:1,4,7. Remaining digits:2,5,8.So b,c,d must be permutations of 2,5,8.Let's try N=142857. As before. b=2, c=8, d=5.Check 2N=285714. Valid permutation.3N=428571. Check digits:4,2,8,5,7,1. Valid.4N=571428. Digits:5,7,1,4,2,8. Valid.5N=714285. Digits:7,1,4,2,8,5. Valid.6N=857142. Digits:8,5,7,1,4,2. Valid.So this works. Therefore, N=142857.But how to arrive here without knowing it?Suppose we proceed systematically.Take N=14bcd7. Remaining digits:2,5,8. So b,c,d are 2,5,8 in some order.Let's try b=2:N=142cd7. c and d are 5 and 8.Options: c=5, d=8: N=142587Check 2N=285174. Digits:2,8,5,1,7,4. Original digits:1,4,2,5,8,7. So digits match. So 285174 is a permutation. Good.Now check 3N=142587*3=427761. Digits:4,2,7,7,6,1. 6 is not in original digits, and duplicate 7s. Invalid.So reject this.Next option: c=8, d=5: N=1428572N=285714. Permutation.3N=142857*3=428571. Permutation.4N=142857*4=571428. Permutation.5N=714285. Permutation.6N=857142. Permutation.This works. So N=142857.Therefore, through this systematic approach, we can arrive at the solution.

✅:The six-digit number ( N ) that satisfies the given conditions is ( boxed{142857} ).Step-by-Step Explanation:1. Identify Constraints: - ( N ) must be a six-digit number with distinct digits. - ( 2N, 3N, 4N, 5N, ) and ( 6N ) must be permutations of ( N )'s digits. - ( N ) must start with 1 (since ( 6N ) must also be six digits, limiting ( N ) to less than 166,667).2. Analyze Digit Permutations: - All multiples of ( N ) must have the same digits as ( N ), just rearranged. - The last digit of ( N ) must produce valid last digits for all multiples (2N to 6N), which are permutations of ( N )'s digits.3. Determine Possible Last Digits: - Testing possible last digits (2-9) reveals that only ending with 7 works, as the resulting last digits (4, 1, 8, 5, 2) are permutations of ( N )'s digits without introducing new digits or duplicates.4. Constructing the Number: - ( N ) starts with 1 and ends with 7, with remaining digits 2, 4, 5, 8. - Testing permutations of these digits while ensuring multiples are cyclic permutations leads to the known cyclic number 142857.5. Verification: - ( 2 times 142857 = 285714 ) - ( 3 times 142857 = 428571 ) - ( 4 times 142857 = 571428 ) - ( 5 times 142857 = 714285 ) - ( 6 times 142857 = 857142 ) - All results are confirmed permutations of the original digits (1, 4, 2, 8, 5, 7).Thus, the number ( boxed{142857} ) meets all criteria.

🔑:To find the six-digit number ( N ) consisting of different digits such that the numbers ( 2N, 3N, 4N, 5N, 6N ) are merely permutations of ( N ), we can proceed as follows:1. Identify a candidate ( N ) Consider the repeating decimal for ( frac{1}{7} ): [ frac{1}{7} = 0.overline{142857} ] Here, the repeating sequence is ( 142857 ). This provides a candidate for ( N ).2. Calculate ( 2N, 3N, 4N, 5N, 6N ) Let's verify that multiplying ( N = 142857 ) by 2, 3, 4, 5, and 6 results only in permutations of ( N ). [ 2N = 2 times 142857 = 285714 ] [ 3N = 3 times 142857 = 428571 ] [ 4N = 4 times 142857 = 571428 ] [ 5N = 5 times 142857 = 714285 ] [ 6N = 6 times 142857 = 857142 ]3. Verification Every result from the calculations should only be permutations of the original number ( 142857 ). - ( 285714 ) is a permutation containing digits ( 1, 2, 4, 5, 7, 8 ) - ( 428571 ) is a permutation containing digits ( 1, 2, 4, 5, 7, 8 ) - ( 571428 ) is a permutation containing digits ( 1, 2, 4, 5, 7, 8 ) - ( 714285 ) is a permutation containing digits ( 1, 2, 4, 5, 7, 8 ) - ( 857142 ) is a permutation containing digits ( 1, 2, 4, 5, 7, 8 )Hence, the six-digit number ( N ) that satisfies the given condition is ( 142857 ). Conclusion:[boxed{142857}]

🧠:Okay, let's try to tackle this problem. So, we have a sequence of integers defined by a recurrence relation, and some conditions. The goal is to prove that all terms of the sequence are perfect squares. Let me start by breaking down the problem step by step.First, let's look at the recurrence relation: a_{n+1} = 3a_n - 3a_{n-1} + a_{n-2} for n ≥ 2. Hmm, this is a linear recurrence relation with constant coefficients. Maybe I can find its characteristic equation to solve it explicitly. The characteristic equation for such a recurrence would be r^{3} - 3r^{2} + 3r - 1 = 0. Let me check that: if we assume a solution of the form r^n, then plugging into the recurrence gives r^{n+1} = 3r^n - 3r^{n-1} + r^{n-2}, which simplifies to r^{3} - 3r^{2} + 3r - 1 = 0. So, that's the characteristic equation.Now, let's factor this equation. Maybe it has a repeated root? Let me try to factor it. Let's see if r=1 is a root. Plugging r=1: 1 - 3 + 3 - 1 = 0. Yes, r=1 is a root. So we can factor out (r - 1). Let's perform polynomial division or use synthetic division.Dividing r^3 - 3r^2 + 3r - 1 by (r - 1):Using synthetic division:1 | 1 -3 3 -1 1 -2 1 1 -2 1 0So the quotient is r^2 - 2r + 1, which factors further into (r - 1)^2. Hence, the characteristic equation is (r - 1)^3 = 0. So, all roots are r = 1 with multiplicity 3.Therefore, the general solution of the recurrence relation is a_n = (A + Bn + Cn^2)(1)^n = A + Bn + Cn^2. So, the sequence is a quadratic function of n. That is, a_n is of the form An^2 + Bn + C. Wait, but the coefficients here are constants, right? So, A, B, C are constants determined by the initial conditions.But we need to find the initial conditions. The problem gives us condition 2: 2a_1 = a_0 + a_2 - 2. Let's express a_0, a_1, a_2 in terms of the quadratic formula. Since a_n = An^2 + Bn + C, then:a_0 = A(0)^2 + B(0) + C = Ca_1 = A(1)^2 + B(1) + C = A + B + Ca_2 = A(2)^2 + B(2) + C = 4A + 2B + CThen condition 2: 2a_1 = a_0 + a_2 - 2Substituting:2(A + B + C) = C + (4A + 2B + C) - 2Left side: 2A + 2B + 2CRight side: C + 4A + 2B + C - 2 = 4A + 2B + 2C - 2Set equal:2A + 2B + 2C = 4A + 2B + 2C - 2Subtract 2A + 2B + 2C from both sides:0 = 2A - 2So, 2A - 2 = 0 → 2A = 2 → A = 1So, the quadratic coefficient A is 1. Therefore, the general form is a_n = n^2 + Bn + C.So now, the sequence is a quadratic function a_n = n² + Bn + C. The coefficients B and C can be determined by the initial terms a₀, a₁, but since the recurrence is of order 3, we might need three initial terms. Wait, actually, the recurrence starts from n=2, so the terms a₀, a₁, a₂ are needed to compute a₃ and onwards. But since the general solution is quadratic, once we have a quadratic function, the recurrence relation should hold for all n. Let me verify that.Suppose a_n = n² + Bn + C. Let's check if the recurrence a_{n+1} = 3a_n - 3a_{n-1} + a_{n-2} holds.Compute a_{n+1} = (n+1)^2 + B(n+1) + C = n² + 2n +1 + Bn + B + CCompute 3a_n - 3a_{n-1} + a_{n-2}:3a_n = 3(n² + Bn + C) = 3n² + 3Bn + 3C-3a_{n-1} = -3[(n-1)^2 + B(n-1) + C] = -3(n² - 2n +1 + Bn - B + C) = -3n² +6n -3 -3Bn +3B -3C+a_{n-2} = (n-2)^2 + B(n-2) + C = n² -4n +4 + Bn -2B + CAdd them up:3n² + 3Bn + 3C -3n² +6n -3 -3Bn +3B -3C +n² -4n +4 + Bn -2B + CLet's combine like terms:n² terms: 3n² -3n² +n² = n²n terms: 3Bn +6n -3Bn -4n + Bn = (0Bn) + (6n -4n) + Bn = 2n + BnConstant terms: 3C -3 -3C +3B +4 -2B + C = (3C -3C + C) + (-3 +4) + (3B -2B) = C +1 + BSo overall, the RHS is n² + (B + 2)n + (C + B +1)But the LHS is a_{n+1} = n² + 2n +1 + Bn + B + C = n² + (B + 2)n + (1 + B + C)Comparing LHS and RHS:LHS: n² + (B + 2)n + (1 + B + C)RHS: n² + (B + 2)n + (C + B +1)They are the same. So the recurrence holds for any quadratic sequence. Therefore, the general solution is indeed quadratic, and the coefficients B and C can be determined by initial terms. But in our case, the problem states that for any natural number m, there exists a block of m consecutive terms that are all perfect squares. So the sequence is such that no matter how large m is, we can find m consecutive squares somewhere in the sequence. Our goal is to show that all terms are squares.Given that the sequence is quadratic, a_n = n² + Bn + C. Let me denote this quadratic as Q(n) = n² + Bn + C. The problem is to prove that Q(n) is a perfect square for all integers n ≥ 0.Given that for any m, there exists a k such that Q(k), Q(k+1), ..., Q(k + m -1) are all perfect squares. So, arbitrarily long sequences of consecutive squares in Q(n). Our aim is to show Q(n) is a square for all n.First, let's note that a quadratic polynomial that is a square for infinitely many n must be a square of a linear polynomial. Because if a quadratic Q(n) is a square for infinitely many n, then Q(n) - (pn + q)^2 has infinitely many roots, hence must be identically zero. Therefore, Q(n) must be a perfect square polynomial. However, in this problem, the condition is weaker: there are arbitrarily long runs of consecutive n where Q(n) is a square. But not necessarily that the entire polynomial is a square. Hmm, but maybe the existence of arbitrarily long runs forces the polynomial itself to be a square.Wait, if Q(n) is quadratic, and there are infinitely many n where Q(n) is a square, then under certain conditions, Q(n) must be a square. But here it's even stronger: there are arbitrarily long consecutive sequences. So perhaps this forces Q(n) to be a square.Alternatively, suppose that Q(n) is not a square polynomial. Then Q(n) can only be a square at finitely many n, unless it is a square polynomial. But in our case, we have infinitely many n where Q(n) is a square, and even more, arbitrarily long consecutive sequences.Wait, but how can a quadratic that isn't a square have infinitely many square values? For example, Pell's equation: x² - Dy² = 1 has infinitely many solutions, but here we have Q(n) = n² + Bn + C. If Q(n) is not a square polynomial, can it still take square values infinitely often?Yes, for example, take Q(n) = n² + 1. This is not a square polynomial, but it takes square values at n=0: 0 +1=1, n=2: 4 +1=5 (not square), n=3: 9 +1=10 (not square), etc. Wait, but n² +1 is 1,2,5,10,... which aren't squares except at n=0. So that's only one square.Another example: Q(n) = n² + 2n. Then Q(n) = n(n + 2). This can be a square if n and n +2 are both squares multiplied by some factors. But it's not clear. For example, n=2: 2*4=8, not square; n=7: 7*9=63, not square; n=0: 0, which is square. So again, only a few squares.Alternatively, Q(n) = (n + k)^2 + c. If c is non-zero, then this might not be a square infinitely often.But in our problem, we have that there are arbitrarily long sequences where Q(n) is a square. So, for example, if m is 1000, there exists a k such that Q(k), Q(k+1), ..., Q(k+999) are all squares. That seems extremely restrictive.So, how can a quadratic polynomial have 1000 consecutive integer inputs where it outputs squares? That seems only possible if Q(n) is a square polynomial, i.e., Q(n) = (an + b)^2. Because otherwise, the gaps between consecutive squares would eventually exceed 1, but since the polynomial is quadratic, the difference between consecutive values Q(n+1) - Q(n) = 2n + B +1. If Q(n) is a square, say s_n^2, then s_{n+1}^2 - s_n^2 = 2n + B +1. However, the difference between consecutive squares is (s_{n+1} - s_n)(s_{n+1} + s_n). So unless s_{n} is linear, this difference would not be linear in n. Wait, let's think.Suppose that Q(n) = (pn + q)^2. Then Q(n) is a square for all n. Then the difference Q(n+1) - Q(n) = (p(n+1) + q)^2 - (pn + q)^2 = [ p(n+1) + q + pn + q ][ p(n+1) + q - pn - q ] = (2pn + p + 2q)(p) = 2p^2n + p^2 + 2pq.On the other hand, our Q(n) is n² + Bn + C, so Q(n+1) - Q(n) = ( (n+1)^2 + B(n+1) + C ) - (n² + Bn + C ) = 2n +1 + B.So, if Q(n) = (pn + q)^2, then Q(n+1) - Q(n) = 2p^2n + p^2 + 2pq, which should equal 2n +1 + B.Therefore, equate coefficients:2p^2 = 2 ⇒ p^2 =1 ⇒ p=1 or -1.p^2 + 2pq =1 + B.Since p=1 or p=-1.If p=1:2*1^2 =2, which matches. Then p^2 + 2pq =1 + 2q =1 + B ⇒ 2q = B ⇒ q = B/2.If p=-1:2*(-1)^2 =2, which is also okay. Then p^2 + 2pq =1 + 2*(-1)q =1 - 2q =1 + B ⇒ -2q = B ⇒ q = -B/2.But since Q(n) = (pn + q)^2, if p=1, then Q(n) = (n + q)^2. If p=-1, Q(n)= (-n + q)^2 = (q -n)^2. However, since n starts from 0,1,2,..., (q -n)^2 would eventually become squares of negative numbers as n increases, but squares are the same as their positive counterparts. However, since q is a constant, for large n, (q -n)^2 = (n - q)^2, so it's similar to shifting n. But whether p=1 or p=-1, the quadratic would essentially be a square of a linear polynomial. Therefore, if Q(n) is a square polynomial, then it can be written as (n + c)^2 or (c -n)^2. But let's check if our quadratic can be written like that.Given Q(n) = n² + Bn + C. If Q(n) is a square, then Q(n) = (n + c)^2 = n² + 2cn + c². Therefore, comparing coefficients:2c = Bc² = CTherefore, C = (B/2)^2. So, in order for Q(n) to be a perfect square polynomial, we must have C = (B/2)^2. Then Q(n) = n² + Bn + (B/2)^2 = (n + B/2)^2.But in our problem, the quadratic is given with integer coefficients, right? Because the sequence a_n is a sequence of integers. So Q(n) must be integer for all n. If Q(n) = (n + B/2)^2, then B must be even, since B/2 has to be an integer for Q(n) to be an integer square. Let's denote B = 2d, so Q(n) = (n + d)^2, where d is an integer. Therefore, if Q(n) is a square polynomial, then it must be of the form (n + d)^2 with integer d, which would make a_n = (n + d)^2.So, if we can show that Q(n) must be such a square polynomial, then we are done. So the problem reduces to showing that the quadratic Q(n) = n² + Bn + C must be a perfect square, given that there are arbitrarily long sequences of consecutive n where Q(n) is a square.Alternatively, if Q(n) is not a perfect square polynomial, then it can have only finitely many squares. But in our case, the condition is stronger: there are arbitrarily long runs of consecutive squares. How can this happen?Suppose that Q(n) is not a perfect square polynomial. Then, the equation Q(n) = y² has only finitely many solutions (n, y) unless Q(n) is a square polynomial. Wait, is that true? For quadratics, it's known that they can have infinitely many square values. For example, Pell-type equations. Wait, but Pell equations are of the form x² - Dy² = 1, which are hyperbolas. But our case is Q(n) = y², which is a quadratic in n. Let's see.Suppose Q(n) = n² + Bn + C. Let's complete the square:Q(n) = n² + Bn + C = (n + B/2)^2 + (C - B²/4).So, Q(n) = (n + B/2)^2 + (C - B²/4). Therefore, Q(n) is a square if (n + B/2)^2 + (C - B²/4) is a square. Let me set k = n + B/2. Then Q(n) = k² + (C - B²/4). So, k² + D = y², where D = C - B²/4.Thus, y² - k² = D ⇒ (y - k)(y + k) = D.So, for each integer k, if D can be factored as two integers whose product is D, then we can set y - k = d and y + k = D/d, leading to y = (d + D/d)/2 and k = (D/d - d)/2. However, since k must be an integer (as n is integer and B/2 must be integer if Q(n) is to be integer when B is even). Wait, but in our problem, the sequence a_n is integer, so Q(n) must be integer for all n. Therefore, Q(n) = n² + Bn + C is integer for all n. Therefore, B and C must be integers. Therefore, D = C - B²/4 must be a rational number. However, since Q(n) is integer for all n, then if B is even, say B = 2d, then D = C - d², which is integer. If B is odd, then B/2 is a half-integer, so (n + B/2) is a half-integer, and (n + B/2)^2 is a quarter-integer. Then Q(n) = (n + B/2)^2 + (C - B²/4) must be integer. So, (n + B/2)^2 is a quarter-integer, so (C - B²/4) must also be a quarter-integer to make the sum integer. Therefore, C must be of the form (B² + 4k)/4 for some integer k. Therefore, if B is odd, then C is not an integer unless B² is congruent to 0 mod 4, but B is odd, so B² ≡1 mod4, so (B² +4k)/4 would be (1 +4k)/4, which is not integer. Therefore, if B is odd, C cannot be integer. Therefore, B must be even. Therefore, B is even, so let's set B=2d for some integer d. Then, D = C - (2d)^2 /4 = C - d². So D = C - d². Therefore, Q(n) = (n + d)^2 + (C - d²). But Q(n) must be integer, so since (n + d)^2 is integer, C - d² must be integer. Since C is integer and d is integer, this holds.Therefore, Q(n) = (n + d)^2 + e, where e = C - d² is integer.So, the equation Q(n) = y² becomes y² = (n + d)^2 + e. Therefore, y² - (n + d)^2 = e ⇒ (y - (n + d))(y + (n + d)) = e.So, e must be factored into two integers whose product is e. Let me denote the factors as s and t, so that s*t = e, and y - (n + d) = s, y + (n + d) = t. Then, solving for y and n:Adding the two equations: 2y = s + t ⇒ y = (s + t)/2Subtracting: 2(n + d) = t - s ⇒ n + d = (t - s)/2 ⇒ n = (t - s)/2 - dTherefore, for each factorization s*t = e, we have a solution n = (t - s)/2 - d, provided that s and t have the same parity (so that t - s is even), and n must be an integer. Therefore, the number of solutions (n, y) to Q(n) = y² depends on the number of factorizations of e into two integers s and t of the same parity.Now, if e=0, then Q(n) = (n + d)^2, so all terms are squares. If e≠0, then the number of solutions is finite if e is non-zero and has only finitely many factorizations. However, if e is a perfect square, then there could be infinitely many solutions, but in our case, Q(n) must have infinitely many solutions, in fact, arbitrarily long sequences of consecutive solutions.But if e≠0, then Q(n) = (n + d)^2 + e. Suppose e is positive. Then as n increases, Q(n) grows quadratically, while the difference between consecutive squares is roughly 2n. So, the gaps between consecutive squares near Q(n) are about 2y ≈ 2(n + d). But Q(n) itself is (n + d)^2 + e, so the next square after (n + d)^2 is (n + d +1)^2 = (n + d)^2 + 2(n + d) +1. The difference between Q(n) and (n + d)^2 is e, so if e is less than 2(n + d) +1, then Q(n) would lie between (n + d)^2 and (n + d +1)^2, hence not a square. Similarly, for negative e, Q(n) = (n + d)^2 + e. If e is negative, then Q(n) could be a square if (n + d)^2 + e = m^2, which requires that (n + d)^2 - m^2 = -e ⇒ (n + d - m)(n + d + m) = -e. Again, similar to the positive case.In any case, if e ≠0, the equation Q(n) = y² can have only finitely many solutions unless e=0, because for large n, the value Q(n) = (n + d)^2 + e will lie between two consecutive squares, which are too far apart. Therefore, unless e=0, the quadratic can't have infinitely many square values, let alone arbitrarily long sequences of consecutive squares.But in our problem, it's given that for any m, there exists a k such that Q(k), Q(k+1), ..., Q(k + m -1) are all squares. Therefore, if e ≠0, this would require that there are arbitrarily long sequences of consecutive integers n where (n + d)^2 + e is a square. But as n grows, the term e becomes negligible compared to (n + d)^2, so the difference between consecutive squares is about 2(n + d), so for (n + d +1)^2 - (n + d)^2 = 2(n + d) +1. Therefore, to have (n + d)^2 + e = m^2, the difference e must be equal to m^2 - (n + d)^2. But if we need consecutive n's to satisfy this, then for n, n+1, ..., n + m -1, each (n + i + d)^2 + e must be a square.But if e ≠0, for large enough n, the values (n + i + d)^2 + e would lie between (n + i + d)^2 and (n + i + d +1)^2, so they cannot be squares. Therefore, unless e=0, we cannot have arbitrarily long sequences of squares. Therefore, e must be zero. Therefore, Q(n) = (n + d)^2. Hence, all terms a_n are perfect squares.Therefore, the conclusion is that the quadratic must be a perfect square polynomial, hence all terms are squares.But let me check if this reasoning is solid. Suppose e ≠0. Then, can we have arbitrarily long sequences of n where (n + d)^2 + e is a square? For example, take e =1. Then we have (n + d)^2 +1 = m^2 ⇒ m^2 - (n + d)^2 =1 ⇒ (m - (n + d))(m + (n + d))=1. The only integer solutions are m - (n + d) =1 and m + (n + d)=1, which gives m =1, n + d=0, so n = -d. Or m - (n + d) = -1 and m + (n + d) = -1, which gives m = -1, n + d=0. Thus, the only solutions are n = -d. Hence, only one solution. Therefore, even for e=1, there is only a finite number of solutions. Similarly, for other non-zero e, there would be only finite solutions.Therefore, the only way to have infinitely many solutions, and in particular, arbitrarily long sequences of consecutive solutions, is if e=0, i.e., Q(n) is a perfect square polynomial. Therefore, the given conditions force e=0, so Q(n) = (n + d)^2, hence all terms a_n are perfect squares.Therefore, we can conclude that all terms of the sequence are perfect squares.But let's verify this with the initial conditions. Remember that we had from condition 2: 2a₁ = a₀ + a₂ -2. Since a_n = (n + d)^2, let's compute this:a₀ = (0 + d)^2 = d²a₁ = (1 + d)^2a₂ = (2 + d)^2Condition 2: 2*(1 + d)^2 = d² + (2 + d)^2 -2Compute RHS: d² + (4 + 4d + d²) -2 = 2d² +4d +4 -2 = 2d² +4d +2LHS: 2*(1 + 2d + d²) = 2 +4d +2d²Thus, LHS = 2d² +4d +2, RHS =2d² +4d +2. So equality holds for any d. Therefore, condition 2 is automatically satisfied for any d. Therefore, the quadratic sequence a_n = (n + d)^2 satisfies all the given conditions.Furthermore, the third condition is that for any m, there exists k such that a_k, ..., a_{k + m -1} are all squares. But since a_n is always a square, this condition is trivially satisfied by choosing k=0, for example. Therefore, the only sequences that satisfy all three conditions are quadratic sequences that are perfect squares for all n, which are of the form a_n = (n + d)^2 for some integer d.Therefore, all terms of the sequence are perfect squares.Final Answerboxed{All terms of the sequence are perfect squares.}

✅:Given the sequence of integers ({a_{0}, a_{1}, a_{2}, cdots}) that satisfies the following conditions:1. (a_{n+1} = 3a_{n} - 3a_{n-1} + a_{n-2}) for (n geq 2)2. (2a_{1} = a_{0} + a_{2} - 2)3. For any (m in mathbb{N}), there exists an integer (k) such that (a_{k}, a_{k+1}, cdots, a_{k+m-1}) are all perfect squares.We need to prove that all terms of the sequence are perfect squares.First, we analyze the recurrence relation (a_{n+1} = 3a_{n} - 3a_{n-1} + a_{n-2}). The characteristic equation is (r^3 - 3r^2 + 3r - 1 = 0), which factors as ((r - 1)^3 = 0). Thus, the general solution is a quadratic polynomial (a_n = An^2 + Bn + C).Using condition 2, we substitute (a_0 = C), (a_1 = A + B + C), and (a_2 = 4A + 2B + C) into the equation (2a_1 = a_0 + a_2 - 2). Solving this, we find (A = 1), so the sequence is (a_n = n^2 + Bn + C).Given condition 3, there exist arbitrarily long sequences of consecutive perfect squares. We consider the quadratic polynomial (Q(n) = n^2 + Bn + C). If (Q(n)) is not a perfect square polynomial, it can only take square values finitely many times, contradicting the existence of arbitrarily long sequences of squares. Therefore, (Q(n)) must be a perfect square polynomial.We complete the square for (Q(n)) and find that it must be of the form ((n + d)^2) for some integer (d). This satisfies all conditions, including the recurrence relation and the initial condition. Thus, all terms (a_n) are perfect squares.[boxed{All terms of the sequence are perfect squares.}]

🔑: Problem:Given the integer sequence left{a_{0}, a_{1}, a_{2}, cdots right}, which satisfies the conditions:1. a_{n+1} = 3a_{n} - 3a_{n-1} + a_{n-2}, n = 2,3, cdots2. 2a_{1} = a_{0} + a_{2} - 2,3. For any m in mathbb{N}, there exists k such that[a_{k}, a_{k+1}, cdots, a_{k+m-1}]are all perfect squares.Prove: All terms in the sequence left{a_{0}, a_{1}, a_{2}, cdots right} are perfect squares.1. From (1), for n geq 2, we have: [ a_{n+1} - a_{n} = 2left(a_{n} - a_{n-1}right) - left(a_{n-1} - a_{n-2}right). ]2. Define d_{n}=a_{n}-a_{n-1} for n=1,2, cdots. Then the equation can be rewritten as: [ d_{n+1} - d_{n} = d_{n} - d_{n-1}. ]3. This implies: [ d_{n+1} - d_{n} = d_{n} - d_{n-1} = cdots = d_{2} - d_{1}. ]4. From (2), we have: [ d_{2} - d_{1} = a_{2} - 2a_{1} + a_{0} = 2. ]5. Therefore, [ d_{n} = d_{1} + 2(n-1), quad n=1,2,3, cdots. ]6. Summing these differences, we can express (a_n) as: [ a_{n} = a_{0} + sum_{k=1}^{n} d_{k} = a_{0} + d_{1} n + n(n-1). ]7. This simplifies to: [ a_{n} = n^{2} + bn + c, quad n=0,1,2, cdots, ] where (b = d_{1} - 1 = a_{1} - a_{0} - 1) and (c = a_{0}).8. From (3), there exists a non-negative integer (t) such that both (a_{t}) and (a_{t+2}) are perfect squares. Thus, [ a_{t+2} - a_{t} notequiv 2 mod 4. ]9. From (1), we also have: [ a_{t+2} - a_{t} = 4t + 4 + 2b. ]10. Hence, (b) must be an even number. Let (b = 2lambda), then: [ a_{n} = (n + lambda)^{2} + c - lambda^{2}, quad n = 0,1,2, cdots. ]11. We need to show (c - lambda^{2} = 0). Suppose otherwise, i.e., (c - lambda^{2} neq 0). Then (c - lambda^{2}) would have only finitely many distinct values. Let the number of such values be (m).12. From (1), there exists (n_{0}) such that when (n geq n_{0}), the sequence (left{a_{n}right}) is strictly monotonic. Also, by the assumption (3), for (k geq n_{0}), we have: [ a_{k+i} = p_{i}^{2}, quad i=0,1,2, cdots, m. ] where (p_{i}) are distinct positive integers and (p_{0} < p_{1} < cdots < p_{m}).13. From (2), we find: [ c - lambda^{2} = p_{i}^{2} - (k + i + lambda)^{2} = (p_{i} - k - i - lambda)(p_{i} + k + i + lambda), quad i = 0,1,2, cdots, m. ]14. This implies (c - lambda^{2}) has at least (m+1) distinct values, which contradicts the earlier assumption that (c - lambda^{2}) can only have finitely many values.15. Therefore, (c - lambda^{2} = 0) must hold. Thus, [ a_{n} = left(n + frac{b}{2}right)^{2}, quad n = 0,1,2, cdots. ]16. This implies all terms in the sequence are perfect squares: [ a_{t} = left(t + frac{b}{2}right)^{2}. ] 17. Rewriting for simplicity, we get: [ a_{n} = left(n + frac{b}{2}right)^{2}, quad n = 0,1,2, cdots. ]18. Thus, every term in the sequence (left{ a_{0}, a_{1}, a_{2}, cdots right}) is a perfect square.Conclusion:[ boxed{All terms of the sequence are perfect squares.}]

🧠:Okay, let's see. So, the problem is about four characters—Kros, Yozhik, Nyusha, and Barash—who ate a total of 86 candies. Each of them ate at least 5 candies. The key points given are that Nyusha ate more than any of the others, and Kros and Yozhik together ate 53 candies. The question is how many candies Nyusha ate.First, let's break down the information. Total candies: 86. Each person ate at least 5. So, Kros + Yozhik + Nyusha + Barash = 86. Kros and Yozhik together ate 53, so that leaves Nyusha and Barash with 86 - 53 = 33 candies between them.We also know that Nyusha is the one who ate the most. So Nyusha's candies are more than Kros, Yozhik, and Barash individually. Since each ate at least 5, let's try to find possible numbers for Nyusha and Barash that add up to 33, with Nyusha's amount being the highest among all four.But wait, Kros and Yozhik have 53 together. Let's denote:K = Kros's candiesY = Yozhik's candiesN = Nyusha's candiesB = Barash's candiesSo, K + Y + N + B = 86Given that K + Y = 53, so N + B = 33Also, N > K, N > Y, N > B. And each of K, Y, N, B ≥ 5.So, we need to find N such that N is greater than K, Y, and B. But since K and Y are part of the 53 total between them, maybe we can find possible values for K and Y first, then see what N and B can be.But since K and Y could vary, but their total is fixed at 53. Let's think about what the minimum and maximum possible values for K and Y could be. Since each of them has to be at least 5, the minimum each could have is 5. So, for example, if K is 5, Y would be 48 (since 53 - 5 = 48). Similarly, if Y is 5, K is 48. But Nyusha has to be greater than both K and Y. So, in the case where one of K or Y is 48, Nyusha would have to be greater than 48. But Nyusha and Barash together have 33 candies. If Nyusha had to be greater than 48, but 48 + something is only 33? Wait, that doesn't add up. Wait, that seems impossible. Because if K and Y together have 53, and say K is 48, Y is 5, then Nyusha must be greater than 48. But Nyusha + Barash = 33. So if Nyusha is greater than 48, then 48 + something is 33? That's impossible because 48 is already more than 33. So that must mean that neither K nor Y can be as high as 48. Wait, this is a contradiction. So maybe my initial assumption is wrong here.Wait, hold on. If K + Y = 53, and each is at least 5, then the maximum possible value for either K or Y is 53 - 5 = 48. But if one of them is 48, then Nyusha has to be greater than 48. But Nyusha and Barash together have only 33. So 33 can't be split into Nyusha >48 and Barash ≥5. Because Nyusha would have to be at least 49, but 49 + 5 = 54, which is way more than 33. So that's impossible. Therefore, the maximum value that either K or Y can have is such that Nyusha can still be greater than them, given that Nyusha + Barash = 33.So let's formalize this. Let's suppose that K and Y are two numbers adding to 53, each at least 5. Let’s denote K ≥ Y without loss of generality. Then K can range from 26.5 (if they were equal) to 48 (if Y is 5). But since K and Y are integers (since you can't eat a fraction of a candy), K can be from 27 to 48. Wait, actually, if they have to be integers, then the maximum K can be is 53 - 5 = 48, as before. But if K is 48, then Nyusha has to be at least 49. But 49 is more than 33 (since Nyusha + Barash = 33). So that's impossible. Therefore, K cannot be 48. Similarly, K can't be 47. If K is 47, Y is 6, then Nyusha has to be greater than 47, which is again impossible. Similarly, K can't be 46, because Nyusha would have to be at least 47, but 47 + Barash's minimum 5 is 52, which is more than 33. So this seems like a problem. Therefore, K must be such that even the maximum possible K is less than or equal to N, but N + B = 33, so N is at most 33 - 5 = 28 (since B is at least 5). Therefore, N ≤28. Therefore, K and Y must both be less than or equal to 28. Wait, but K + Y =53. If both K and Y are ≤28, then their sum would be at most 28 +28=56, which is more than 53. Wait, no. Wait, if each of K and Y is ≤28, then the maximum sum would be 28 +28=56. But their actual sum is 53. So that's possible. So K and Y can be, for example, 28 and 25. But then N has to be greater than 28. But N + B =33, so N would need to be greater than 28, which would mean N ≥29. Then B would be 33 -29=4, but B must be at least 5. So that's impossible. Hmm, so this is a contradiction again.Wait, so maybe I need to approach this differently. Let's think about the maximum possible value of N. Since N has to be greater than both K and Y, and K + Y =53. Let’s denote that the maximum of K and Y is, let's say, M. Then N must be greater than M. Also, since K + Y =53, the maximum possible M is when one of them is as large as possible. For example, if K is 48 and Y is 5, then M=48. But as before, N must be >48, but N + B=33, so impossible. Therefore, M cannot be too large.Alternatively, perhaps the maximum M can be is such that N can be M +1 and B is at least 5. So N = M +1, and B =33 - N =33 - (M +1)=32 - M. Since B must be at least 5, 32 - M ≥5, so M ≤27. Therefore, the maximum possible M is 27, which would make N=28 and B=5. Let's check if that works. If M=27, then K and Y are 27 and 26, adding to 53. Then Nyusha is 28, which is more than both 27 and 26, and Barash is 5. That works. But wait, K and Y can also be other combinations. For example, if K=28 and Y=25, then M=28, but then N would have to be 29, B=4, which is invalid. So M must be at most 27. Therefore, the maximum value for the larger of K and Y is 27, leading to N=28, B=5. Let's verify this. If K=27, Y=26 (sum 53), N=28, B=5. Total is 27+26+28+5=86. Each has at least 5. Nyusha is the highest. That works. So in this case, Nyusha ate 28 candies.But are there other possibilities? Let's check.Suppose M=26. Then N=27, B=33 -27=6. Then K and Y can be, say, 26 and 27. Wait, but K + Y=53. If the maximum of K and Y is 26, then the other would be 53 -26=27. Wait, that contradicts M=26. Because if one is 27, then M=27. So actually, if M=26, the other would be 53 -26=27, which is higher than M=26. So that's not possible. Therefore, M must be at least 27. Wait, this is confusing. Let's think again.If K + Y=53, then the maximum of K and Y must be at least ceiling(53/2)=27 (since 53 is an odd number). So the larger of K and Y must be at least 27. Therefore, M ≥27. So the minimal M is 27, and the maximum M is, as before, constrained by N needing to be greater than M and B=33 -N ≥5.Therefore, for M=27, N=28, B=5. If M=28, then N=29, B=4, which is invalid. Therefore, only M=27 is possible. Therefore, the only possible value for N is 28.Wait, but let's check if K and Y can actually be 27 and 26. 27 +26=53. Yes. Then N=28, B=5. Total 27+26+28+5=86. Each ate at least 5. Nyusha is the highest. That works. Is there another way to split K and Y as 53 where both are less than 28? For example, K=25, Y=28. Then M=28, which would require N=29, B=4, invalid. So if Y=28, K=25, then Nyusha would need to be 29, but B=4. So invalid. Therefore, any split where one of K or Y is 28 or higher would require N to be 29 or higher, but then B would be less than 5. Hence, only splits where the maximum of K and Y is 27 are possible, leading to N=28 and B=5.Therefore, the answer must be 28.Wait, but let's check another possible split where K and Y are both 26.5. But since they have to be integers, you can't have half candies. So the closest would be 26 and 27, which we already considered. So the maximum of K and Y is 27, leading to N=28.Alternatively, if K and Y are 24 and 29. Wait, but 24 +29=53. Then M=29, so N would need to be 30, B=33 -30=3, which is less than 5. Invalid. Hence, M can't be 29. Therefore, M must be 27.Therefore, the only possible value for N is 28. Hence, Nyusha ate 28 candies.But wait, let's verify once again. Let's suppose K=25 and Y=28. Then M=28. Then N=29, B=33-29=4. But B=4 is less than 5. So invalid. So even if K=25 and Y=28, that doesn't work. Therefore, any combination where either K or Y is 28 or higher would result in B being less than 5. Therefore, the maximum possible M (the larger of K and Y) is 27. Therefore, K=27, Y=26. Then N=28, B=5. That works.Alternatively, K=26, Y=27. Same result. So regardless of which one is 26 or 27, the maximum M is 27. Therefore, Nyusha must have eaten 28 candies, and Barash 5. Total is 27+26+28+5=86. All constraints are satisfied.Therefore, the answer is 28.

✅:Solution:Given that Kros (K), Yozhik (Y), Nyusha (N), and Barash (B) ate a total of 86 candies, with each eating at least 5 candies, and the constraints:1. ( K + Y = 53 )2. ( N > K, N > Y, N > B )Step 1: Calculate the remaining candies for Nyusha and Barash: [ N + B = 86 - (K + Y) = 86 - 53 = 33 ]Step 2: Determine the maximum possible value for ( K ) or ( Y ). Since ( N ) must exceed both ( K ) and ( Y ), and ( N + B = 33 ), we analyze: - Let ( M = max(K, Y) ). Then ( N > M ), and ( B = 33 - N geq 5 ). - For ( B geq 5 ), ( N leq 33 - 5 = 28 ). Thus, ( M < 28 ). Step 3: Since ( K + Y = 53 ), the largest possible ( M ) (rounded down) is 27 (e.g., ( K = 27, Y = 26 )). Substituting: - ( N = 28 ) (the smallest integer greater than ( M = 27 )), - ( B = 33 - 28 = 5 ). Verification: - Total candies: ( 27 + 26 + 28 + 5 = 86 ), - Each ate at least 5 candies, - ( N = 28 ) is greater than ( K = 27 ), ( Y = 26 ), and ( B = 5 ). Final Answer: Nyusha ate (boxed{28}) candies.

🔑:To find out how many candies Niusha ate, let's start by setting up the conditions and analyzing them step by step.1. Given Data: - Four individuals named Krosh, Ezhik, Niusha, and Barash collectively ate 86 candies. - Each individual ate at least 5 candies. - Niusha ate more candies than any of the other three. - Krosh and Ezhik together ate 53 candies.2. Formulating Variables: Let ( K ) be the number of candies Krosh ate. Let ( E ) be the number of candies Ezhik ate. Let ( N ) be the number of candies Niusha ate. Let ( B ) be the number of candies Barash ate.3. Equations Setup: From the problem, we can set up the following equations: [ K + E + N + B = 86 quad text{(1)} ] [ K + E = 53 quad text{(2)} ]4. Manipulating Equations: Substitute the second equation into the first equation: [ 53 + N + B = 86 ] Simplify to get: [ N + B = 33 quad text{(3)} ]5. Minimum Values: Since each individual ate at least 5 candies: [ K geq 5, quad E geq 5, quad N geq 5, quad B geq 5 ]6. Finding Bounds: Niusha ate more candies than any of the others. Therefore: [ N > K, quad N > E, quad N > B ]7. Combining Information: With equations (2), we know ( K + E = 53 ). Thus: [ K + E = 53 quad implies K + E leq 53 ]8. Critical Values Analysis: If Niusha ate the most, we should ensure the maximum possible values are consistent: [ K leq 26 quad text{and} quad E leq 26 quad (text{since } K + E = 53) ]9. Minimum Candies by Krosh and Ezhik Analysis: To meet the sum requirement (Krosh and Ezhik together eating 53 candies), at least one of them must have eaten: [ text{At least } 27 text{ candies, otherwise the sum } K + E leq 52 text{ cannot be achieved.} ]10. Niusha’s Candy Consumption: Since we need ( N > K ) and ( N > E ), along with minimum values calculated, ( N geq 28 ). Since ( B geq 5 ) and we determined ( N + B = 33 ): [ text{If } N = 28, quad B = 33 - 28 = 5 ]11. Verification: All conditions must hold: [ K + E + N + B = 86 quad text{(checked)} ] With ( K + E = 53 ). Therefore, our calculations point to the exact sum, verifying correctness. Conclusion:Niusha ate ( boxed{28} ) candies.