2000 AIME I Problems/Problem 1

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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 (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 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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