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 $10^na+b$ , where a is the leftmost digit, and b is the rest of the number. It satisfies $b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na$ . 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.

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2006 AIME I Problems/Problem 3

Problem

Solution

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