# 2021 AMC 12B Problems/Problem 10

## Problem

Two distinct numbers are selected from the set $\{1,2,3,4,\dots,36,37\}$ so that the sum of the remaining $35$ numbers is the product of these two numbers. What is the difference of these two numbers?

$\textbf{(A) }5 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8\qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

## Solution

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

Therefore, $703-x-y=xy$

Rearranging, $xy+x+y=703$. We can factor this equation by SFFT to get

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

Looking at the possible divisors of $704 = 2^6\cdot11$, $22$ and $32$ are within the constraints of $0 < x \leq y \leq 37$ so we try those:

$(x+1)(y+1) = 22\cdot32$

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

$x = 21, y = 31$

Therefore, the difference $y-x=31-21=\boxed{\textbf{(E) }10}$.

~ SoySoy4444

~MathFun1000 ($\LaTeX$ help)

## Video Solution (Just 2 min!)

~Education, the Study of Everything

~ pi_is_3.14

~IceMatrix

## See Also

 2021 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.