# 2000 AIME I Problems/Problem 1

## Problem

Find the least positive integer $n$ such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit $0$.

## Solution

If a factor of $10^{n}$ has a $2$ and a $5$ in its prime factorization, then that factor will end in a $0$. Therefore, we have left to consider the case when the two factors have the $2$s and the $5$s separated, so we need to find the first power of 2 or 5 that contains a 0.

For $n = 1:$ $\[2^1 = 2 , 5^1 = 5\]$ $n = 2:$ $\[2^2 = 4 , 5 ^ 2 =25\]$ $n = 3:$ $\[2^3 = 8 , 5 ^3 = 125\]$

and so on, until, $n = 8:$ $2^8 = 256$ | $5^8 = 390625$

We see that $5^8$ contains the first zero, so $n = \boxed{8}$.

## See also

 2000 AIME I (Problems • Answer Key • Resources) Preceded byFirst Question Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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