2000 AIME I Problems/Problem 1

Revision as of 00:15, 25 March 2011 by Danielguo94 (talk | contribs) (Solution)

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, in other words whether $2^n$ or $5^n$ produces a 0 first.

$2^1 = 2$ | $5^1 = 5$ $2^2 = 4$ | $5 ^ 2 =25$ $2^3 = 8$ | $5 ^3 = 125$

and so on, until,

$2^8 = 256$ | $5^8 = 390625$

We see that $5^8$ generates the first zero, so the answer is $\boxed{008}$.

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