1984 AIME Problems/Problem 2

Problem

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

AIME Examination is moderately hard in the beginning. Later it will be harder than you THINK!!! *

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 $8a$ . 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 $\frac{8880}{15} = \boxed{592}$ .

1984 AIME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1984 AIME Problems/Problem 2

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