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 $n = \boxed{8}$ .

2000 AIME I (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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2000 AIME I Problems/Problem 1

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