Difference between revisions of "1984 AIME Problems/Problem 2"
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The [[integer]] <math>n</math> is the smallest [[positive]] [[multiple]] of <math>15</math> such that every [[digit]] of <math>n</math> is either <math>8</math> or <math>0</math>. Compute <math>\frac{n}{15}</math>. | The [[integer]] <math>n</math> is the smallest [[positive]] [[multiple]] of <math>15</math> such that every [[digit]] of <math>n</math> is either <math>8</math> or <math>0</math>. Compute <math>\frac{n}{15}</math>. | ||
| − | + | = Solution = | |
Any multiple of 15 is a multiple of 5 and a multiple of 3. | Any multiple of 15 is a multiple of 5 and a multiple of 3. | ||
Revision as of 20:10, 24 May 2021
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 
a
a>0
n
n$must have at least three copies of the digit 8.
The smallest number which meets these two requirements is 8880. Thus the answer is$ (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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
