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06 May 2024, 00:14
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If \(S = x + (x + 1) + ... + (x + 80)\), where \(x\) is a positive integer, what is the minimum value of \(x\) such that S is the square of an integer?
A. 4
B. 9
C. 24
D. 41
E. 81
A. 4
B. 9
C. 24
D. 41
E. 81
06 May 2024, 00:14
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Official Solution:
If \(S = x + (x + 1) + ... + (x + 80)\), where \(x\) is a positive integer, what is the minimum value of \(x\) such that S is the square of an integer?
A. 4
B. 9
C. 24
D. 41
E. 81
\(S = x + (x + 1) + ... + (x + 80)\), represents the sum of 81 consecutive integers from \(x\) to \(x+80\), inclusive. Thus, \(S = \frac{{\text{first} + \text{last}}}{2} * \text{the number of terms} = \frac{{x + (x + 80)}}{2} * 81 = 81(x+40)\). Since 81 is a perfect square, \(9^2\), for the entire expression to be a perfect square \((x+40)\) must also be a perfect square. Given that \(x\) is a positive integer, the minimum value of \(x\) is 9.
Answer: B
If \(S = x + (x + 1) + ... + (x + 80)\), where \(x\) is a positive integer, what is the minimum value of \(x\) such that S is the square of an integer?
A. 4
B. 9
C. 24
D. 41
E. 81
\(S = x + (x + 1) + ... + (x + 80)\), represents the sum of 81 consecutive integers from \(x\) to \(x+80\), inclusive. Thus, \(S = \frac{{\text{first} + \text{last}}}{2} * \text{the number of terms} = \frac{{x + (x + 80)}}{2} * 81 = 81(x+40)\). Since 81 is a perfect square, \(9^2\), for the entire expression to be a perfect square \((x+40)\) must also be a perfect square. Given that \(x\) is a positive integer, the minimum value of \(x\) is 9.
Answer: B
27 Sep 2024, 21:24
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If we don't write the equation as a product of 2 numbers, it's almost impossible to get the answer in time. My natural instinct was to write it as 81x+3240 (=80*81/2).
