GMAT Club Tests | Re M41-15

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

Statistics : Posted by Bunuel • on 06 May 2024, 00:14 • Replies 1 • Views 21

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