2006 AIME I Problems/Problem 3
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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 , so the number is .
- It is quite obvious that , since the desired number can't be single or double digit, and cannot exceed . From , proceed as above.
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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