2006 AIME I Problems/Problem 3

Problem

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer.

Solution

Suppose the original number is $N = \overline{a_na_{n-1}\ldots a_1a_0},$ where the $a_i$ are digits and the first digit, $a_n,$ is nonzero. Then the number we create is $N_0 = \overline{a_{n-1}\ldots a_1a_0},$ so $N = 29N_0.$ But $N$ is $N_0$ with the digit $a_n$ added to the left, so $N = N_0 + a_n \cdot 10^n.$ Thus, $N_0 + a_n\cdot 10^n = 29N_0$ $a_n \cdot 10^n = 28N_0.$ The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number $10^n$ is never divisible by $7,$ so $a_n$ must be divisible by $7.$ But $a_n$ is a nonzero digit, so the only possibility is $a_n = 7.$ This gives $7 \cdot 10^n = 28N_0$ or $10^n = 4N_0.$ Now, we want to minimize both $n$ and $N_0,$ so we take $N_0 = 25$ and $n = 2.$ Then $N = 7 \cdot 10^2 + 25 = \boxed{725},$ and indeed, $725 = 29 \cdot 25.$ $\square$

2006 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2006 AIME I Problems/Problem 3

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