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

A number is divisible by $11$ if the difference between the sum of the digits in the odd-numbered slots (e.g. the ones slot, the hundreds slot, etc.) and the sum in the even-numbered slots (e.g. the tens slot, the thousands slot) is a multiple of $11$ . So $1 + 2 - A$ is equivalent to $0\pmod{11}$ . Clearly $1+2-A$ cannot be equal to $11$ or any multiple of $11$ greater than that. Also, if the expression $1+2-A$ is to be equal to a negative multiple of $A$ , $A$ must be 14 or greater, which violates the condition that A is a digit. So $3 - A = 0 \longrightarrow A = \boxed{\textbf{(D)}~3}$ .

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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 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

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