1984 AIME Problems/Problem 2
Problem
The integer is the smallest positive multiple of such that every digit of is either or . Compute .
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 only contains the digits 0 and 8, the units digit of must be 0.
The sum of the digits of any multiple of 3 must be divisible by 3. If has digits equal to 8, the sum of the digits of is aa>0nn$must have at least three copies of the digit 8.
The smallest number which meets these two requirements is 8880. Thus the answer is$ (Error compiling LaTeX. Unknown error_msg)\frac{8880}{15} = \boxed{592}$.
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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