BTGmoderatorDC wrote:

Let n and k be positive integers with k ≤ n. From an n × n array of dots, a k × k array of dots is selected. The figure above shows two examples where the selected k × k array is enclosed in a square. How many pairs (n, k) are possible so that exactly 48 of the dots in the n × n array are NOT in the selected k × k array?

A. 1
B. 2
C. 3
D. 4
E. 5

OA C

Source: Official Guide

The number of dots in n*n array = n^2, and in those in k*k = k^2, so the dots in the n X n array are NOT in the selected k X k array = n^2 - k^2 = 48, where n > k.

n^2 − k^2 = 48;

(n−k)*(n+k) = 48, since the product 48 is even, at least one of n-k or n+k must be even. Moreover, both n-k or n+k must be even, else if one is even and the other is odd, then n and k will not be positive integers. For example: Say (n−k)*(n+k) = 48 = 1*48. Then n - k = 1 and n + k = 48 => n = 24.5, not a psotive integer

So, 48 can be written as a product of two even numbers as (2, 24); (4, 12); and (6, 8). There are three pairs.

The correct answer: C

Hope this helps!

-Jay
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