1984 AIME Problems/Problem 2

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 1

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

Solution 2

Notice how $8 \cdot 10^k \equiv 8 \cdot (-5)^k \equiv 5 \pmod{15}$ for all integers $k \geq 2$ . Since we are restricted to only the digits $8,0$ , because $8\equiv -7 \pmod{15}$ we can't have an $8$ in the optimal smallest number. We can just 'add' fives to quickly get $15 \equiv 0 \pmod{15}$ to get our answer. Thus the answer is $80+800+8000=\boxed{8880}.$

1984 AIME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1984 AIME Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

See also