Question

The $7$ -digit numbers $\underline{74A52B1}$ and $\underline{326AB4C}$ are each multiples of $3$ . Which of the following could be the value of $C$ ?

$1$

$2$

$3$

$5$

$8$

Answer: A

Concepts:divisibility digits

Difficulty rating: 1300

Solution:

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.

2014 AMC 8 Problem 21

Answer: A

Difficulty rating: 1300

Solution:

Problem 21 in Other Years