A Pre-RMO question! -18

$n$ is divisible by $15$ . In other words, $n$ is divisible by $3$ and $5$ . Let's deal with them separately.

Divisible by $5$ : $n$ must end with $0$ or $5$ . Since the digits must be $0$ or $4$ , the last digit must be $0$ .

Divisible by $3$ : The sum of the digits of $n$ must be divisible by $3$ . Because extra $0$ s don't affect the sum of the digits, and $4$ is not divisible by $3$ , we need at least three $4$ s.

• ends with $0$
• has at least three $4$ s

Thus, the smallest value of $n$ is $4440$ . $\dfrac{4440}{1110} = \boxed{4}$ .

Find $\cfrac{n}{1110}$ is actually a hint.

$\cfrac{n}{1110} = k \implies n = 1110*k$

The smallest integer $k = 4$ makes n contains 0 and 4 only

The number will have atleast 3 four to be divisible by 3 and must end in 0, 5 to be divisible by 5. Smallest number is $\boxed{4440}$ satisfting both and hence being divisible by 15