# FidgetBoss 4000's 2019 Mock AMC 12B Problems/Problem 1

## Problem

If $x$ and $y$ are positive integers such that their product is $64,$ what is the sum of all distinct values for $xy+x+y?$ $\textbf{(A) }391\qquad\textbf{(B) }392\qquad\textbf{(C) }393\qquad\textbf{(D) }394\qquad\textbf{(E) }395\qquad$

## Solution 1

Use Simon's Favorite Factoring Trick to deduce $xy+x+y=(x+1)(y+1)-1$. We know $x$ and $y$ are positive integers and $xy=64$. So, we have the following possibilities for the pair $(x, y): (1, 64), (2, 32), (4, 16), (8, 8)$. Note that we do not have to account for flipping $x$ and $y$ because the question asks for only the $\emph{distinct}$ values for $xy+x+y$. If $(x, y)=(1, 64)$, then $(x+1)(y+1)-1=2\cdot 65-1=129$. If $(x, y)=(2, 32)$, then $3\cdot 33-1=98$. If $(x, y)=(4, 16)$, then $5\cdot 17-1=84$. If $(x, y)=(8, 8)$, then $9\cdot 9-1=80$. The sum of these distinct values is $129+98+84+80=227+164=\boxed{391}\textbf{(A)}$.

The problems on this page are copyrighted by FidgetBoss 4000's Mock American Mathematics Competitions. 