2014 AMC 8 Problem 21

Question

The

7

-digit numbers

\underline{74A52B1}

and

\underline{326AB4C}

are each multiples of

3

. Which of the following could be the value of

C

?

1

2

3

5

8

Use the divisibility-by-

3

test on both numbers.

Compare what the first number says about

A+B

with what the second number needs.

For

\underline{74A52B1}

to be divisible by

3

, the digit sum

19+A+B

must be a multiple of

3

. Hence

A+B

is

1

less than a multiple of

3

.

For

\underline{326AB4C}

to be divisible by

3

, the digit sum

15+A+B+C

must be a multiple of

3

. Since

15

is already divisible by

3

,

A+B+C

must be divisible by

3

.

Therefore

C

must be

1

more than a multiple of

3

. Among the answer choices, only

1

has that form.

Thus, A is the correct answer.

Problem 21 in other years