2006 AIME I Problems/Problem 3
Problem
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is of the original integer.
Solution
The number can be represented as , where is the leftmost digit, and is the rest of the number. We know that . Thus has to be 7 since can not have 7 as a factor, and the smallest can be and have a factor of is We find that is 25, so the number is 725.
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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