2007 AMC 10B Problems/Problem 24
Problem
Let 
denote the smallest positive integer that is divisible by both 
and 
and whose base-
representation consists of only 's and 
's, with at least one of each. What are the last four digits of 
Solution
For a number to be divisible by 
the last two digits have to be divisible by 
That means the last two digits of this integer must be
For a number to be divisible by the sum of all the digits must be divisible by 
The only way to make this happen is with 9 's. However, we also need one
The smallest integer that meets all these conditions is 
. The last four digits are 
See Also
| 2007 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
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| All AMC 10 Problems and Solutions | ||
