# 2014 AMC 8 Problems/Problem 8

## Problem

Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$. What is the missing digit $A$ of this $3$-digit number? $\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4$

## Solution 1

Since all the eleven members paid the same amount, that means that the total must be divisible by $11$. We can do some trial-and-error to get $A=3$, so our answer is $\boxed{\textbf{(D)}~3}$ ~SparklyFlowers

## Solution 2

We know that a number is divisible by $11$ if the odd digits added together minus the even digits added together (or vice versa) is a multiple of $11$. Thus, we have $1+2-A$ = a multiple of $11$. The only multiple that works here is $0$, as $11 \cdot 0 = 0$. Thus, $A = \boxed{\textbf{(D)}~3}$ ~fn106068

## Video Solution (CREATIVE THINKING)

~Education, the Study of Everything

## Video Solution

https://youtu.be/mHWTWk-xt0o ~savannahsolver

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