1984 AIME Problems/Problem 2

Revision as of 20:10, 24 May 2021 by Ishudobby (talk | contribs) (Solution)

Problem

The integer $n$ is the smallest positive multiple of $15$ such that every digit of $n$ is either $8$ or $0$. Compute $\frac{n}{15}$.

Solution

Any multiple of 15 is a multiple of 5 and a multiple of 3.

Any multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units digit of $n$ must be 0.



The sum of the digits of any multiple of 3 must be divisible by 3. If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $.  For this number to be divisible by 3,$a$must be divisible by 3. We also know that$a>0$since$n$is positive. Thus$n$must have at least three copies of the digit 8.

The smallest number which meets these two requirements is 8880. Thus the answer is$ (Error compiling LaTeX. Unknown error_msg)\frac{8880}{15} = \boxed{592}$.

See also

1984 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions