Difference between revisions of "1984 AIME Problems/Problem 2"
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== See also == | == See also == | ||
| − | * [[ | + | {{AIME box|year=1984|num-b=1|num-a=3}} |
| − | * [[ | + | * [[AIME Problems and Solutions]] |
| − | * [[ | + | * [[American Invitational Mathematics Examination]] |
| + | * [[Mathematics competition resources]] | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 13:20, 6 May 2007
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. Thus must have at least three copies of the digit 8.
The smallest number which meets these two requirements is 8880. Thus 
is our answer.
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 | ||
