Difference between revisions of "2006 AIME A Problems/Problem 3"
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 14:55, 25 September 2007
Problem
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/29 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 II (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 | ||
| All AIME Problems and Solutions | ||
