Difference between revisions of "1984 AIME Problems/Problem 2"
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The sum of the digits of any multiple of 3 must be [[divisible]] by 3. If <math>n</math> has <math>a</math> digits equal to 8, the sum of the digits of <math>n</math> is <math>8a</math>. For this number to be divisible by 3, <math>a</math> must be divisible by 3. Thus <math>n</math> must have at least three copies of the digit 8. | The sum of the digits of any multiple of 3 must be [[divisible]] by 3. If <math>n</math> has <math>a</math> digits equal to 8, the sum of the digits of <math>n</math> is <math>8a</math>. For this number to be divisible by 3, <math>a</math> must be divisible by 3. Thus <math>n</math> must have at least three copies of the digit 8. | ||
− | The smallest number which meets these two requirements is 8880. Thus <math>\frac{8880}{15} = 592</math> | + | The smallest number which meets these two requirements is 8880. Thus the answer is <math>\frac{8880}{15} = \boxed{592}</math>. |
== See also == | == See also == |
Revision as of 17:09, 21 March 2009
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 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 |