Difference between revisions of "1984 AIME Problems/Problem 2"
(→Solution) |
(→Solution) |
||
| Line 7: | Line 7: | ||
Any multiple of 5 ends in 0 or 5; since <math>n</math> only contains the digits 0 and 8, the units [[digit]] of <math>n</math> must be 0. | Any multiple of 5 ends in 0 or 5; since <math>n</math> only contains the digits 0 and 8, the units [[digit]] of <math>n</math> must be 0. | ||
| − | + | ||
Revision as of 19:21, 4 August 2015
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 
. For this number to be divisible by 3, must be divisible by 3. We also know that 
since is positive. Thus must have at least three copies of the digit 8.
The smallest number which meets these two requirements is 8880. Thus the answer is 
.
See also
| 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 | ||
| All AIME Problems and Solutions | ||
