# 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.

## Solutions

### 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 2

Let $N$ be the required number, and $N'$ be $N$ with the first digit deleted. Now, we know that $N<1000$ (because this is an AIME problem). Thus, $N$ has $1,$ $2$ or $3$ digits. Checking the other cases, we see that it must have $3$ digits. Let $N=\overline{abc}$, so $N=100a+10b+c$. Thus, $N'=\overline{bc}=10b+c$. By the constraints of the problem, we see that $N=29N'$, so $$100a+10b+c=29(10b+c).$$ Now, we subtract and divide to get $$100a=28(10b+c)$$ $$25a=70b+7c.$$ Clearly, $c$ must be a multiple of $5$ because both $25a$ and $70b$ are multiples of $5$. Thus, $c=5$. Now, we plug that into the equation: $$25a=70b+7(5)$$ $$25a=70b+35$$ $$5a=14b+7.$$ By the same line of reasoning as earlier, $a=7$. We again plug that into the equation to get $$35=14b+7$$ $$b=2.$$ Now, since $a=7$, $b=2$, and $c=5$, our number $N=100a+10b+c=\boxed{725}$.

Here's another way to finish using this solution. From the above, you have $$100a = 28(10b + c).$$ Divide by $4$, and you get $$25a = 7(10b + c).$$ This means that $25a$ has to be divisible by $7$, and hence $a = 7.$ Now, solve for $25 = 10b + c$, which gives you $a = 7, b = 2, c = 5$, giving you the number $\boxed{725}$

## Solution 3 (Quick)

Note that if we let the last digit be $c$ we must have $9c \equiv c \pmod{10}.$ Thus we either have $c=0$ which we can quickly check to be impossible (since the number after digit removal could be 10,20,30) or $c=5.$ Testing 5, 15, and 25 as the numbers after removal we find that our answer is clearly $29 \cdot 25 = 729.$

~Dhillonr25

(Note that the quick checking of six numbers was possible thanks to AIME problems having answers less than 1000).

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