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 14:20, 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 |