2021 AMC 12B Problems/Problem 10

The sum of the first $37$ integers is given by $n(n+1)/2$ , so $37(37+1)/2=703$ .

Therefore, $703-x-y=xy$

Rearranging, $xy+x+y=703$

$(x+1)(y+1)=704$

Looking at the possible divisors of $704 = 2^6*11$ , $22$ and $32$ are within the constraints of $0 < x <= y <= 37$ so we try those:

$(x+1)(y+1) = 22 * 32$

$x+1=22, y+1 = 32$

$x = 21, y = 31$

Therefore, the difference $y-x=31-21=10$ , choice E).

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