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 < | + | 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>. For this number to be divisible by 3, </math>a<math> must be divisible by 3. We also know that </math>a>0<math> since </math>n<math> is positive. 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 the answer is <math>\frac{8880}{15} = \boxed{592} | + | The smallest number which meets these two requirements is 8880. Thus the answer is </math>\frac{8880}{15} = \boxed{592}$. |
== See also == | == See also == | ||
Revision as of 20:10, 24 May 2021
Problem
The integer 
is the smallest positive multiple of 
such that every digit of is either 
or 
. Compute 
.
Solution
Any multiple of 15 is a multiple of 5 and a multiple of 3.
Any multiple of 5 ends in 0 or 5; since only contains the digits 0 and 8, the units digit of must be 0.
The sum of the digits of any multiple of 3 must be divisible by 3. If has 
digits equal to 8, the sum of the digits of is 
a
a>0
n
n$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 | ||
