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 $\textbf{(D) }3$ . ~SparklyFlowers

Solution 2

We know that a number is divisible by 11 if the odd numbers added together minus the even numbers added together(or vice versa) is a multiple of $11$ . So, we have $1+2-A$ = a multiple of $ll$ . The only multiple that works here is 0, as 11x0 = 0. Thus, $A = \textbf{(D) }3$ ~fn106068

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AJHSME/AMC 8 Problems and Solutions

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

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2014 AMC 8 Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

See Also