2014 AMC 8 Problems/Problem 8

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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, $A = \qquad\textbf{(D) }3$

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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

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