1999 CEMC Pascal Problems/Problem 15

Problem

For how many different values of $k$ is the $4$ -digit number $7k52$ divisible by $12$ ?

$\text{ (A) }\ 0 \qquad\text{ (B) }\ 1 \qquad\text{ (C) }\ 2 \qquad\text{ (D) }\ 3 \qquad\text{ (E) }\ 4$

Solution

We can use the fact that a number that is divisible by $12$ must be divisible by both $3$ and $4$ , since $12 = 3 \times 4$ .

The divisibility rule for $4$ states that that the last two digits must be divisible by $4$ . Since $52 / 4 = 13$ , which is an integer, we know that the number must be divisible by $4$ , no matter what the value of $k$ is.

The divisibility rule of $3$ states that the sum of the digits of the number must be divisible by $3$ .

Summing up the digits of the number, we get

$7 + k + 5 + 2 = k + 14$

This is divisible by $3$ when $k = 1$ , so it is divisible by $3$ when $k = 4$ and $k = 7$ . $k$ is a digit, so it must be an integer between $1$ and $9$ (inclusive)

This gives $\boxed {\textbf {(D) } 3}$ possible values of $k$ .

~anabel.disher

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1999 CEMC Pascal Problems/Problem 15

Problem

Solution