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 nine 's. However, we also need one
The smallest integer that meets all these conditions is . The last four digits are
Video Solution by OmegaLearn
https://youtu.be/p5f1u44-pvQ?t=59
~ pi_is_3.14
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 |
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