2014 AMC 8

Complete problem set with solutions and individual problem pages

Problem 8 Easy

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 $\underline{1} \underline{A} \underline{2}. What is the missing digit A of this 3-digit number?

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    3

  • E.

    4

Answer:D

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}

 

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}

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