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 
Since we must account for the two 's, the sum of the rest of the digits must be one more than a multiple of The only way to make this happen is with seven additional '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 | ||
