2006 AIME I Problems/Problem 3
Revision as of 21:24, 4 July 2006 by Joml88 (talk | contribs) (2006 AIME I Problem 3 moved to 2006 AIME I Problems/Problem 3)
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 a is the leftmost digit, and b is the rest of the number. It satisfies
. a has to be 7 since 10^n can not have 7 as a factor, and the smallest 10^n can be and have a factor of 2^2 is 10^2=100. We find that b is 25, so the number is 725.