2014 AMC 8 Problem 8

Question

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?

0

1

2

3

4

Answer: D

Concepts:divisibility digits

Difficulty rating: 960

Small hint:

The total paid by

11

members must be divisible by

11

.

Big hint:

For a three-digit number, use the alternating-sum test for divisibility by

11

.

Solution:

Note that since

11

people paid the same amount, then the resulting sum is divisible by

11.

Therefore,

\underline{1} \underline{A} \underline{2}

must be divisible by

11.

Remember the divisibility rule for

11

: if we take the difference of the sums of alternating digits, and this difference is divisible by

11,

the whole number is divisible by

11.

For

\underline{1} \underline{A} \underline{2},

the aforementioned difference is

1 + 2 - A = 3 - A.

The only way for this to be divisible by

11

is if

A = 3.

Thus, D is the correct answer.