# Difference between revisions of "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

### Solution 1

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, $725 = 29 \cdot 25.$ $\square$

## Solution 1

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, $725 = 29 \cdot 25.$ $\square$

## See also

 2006 AIME I (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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