2000 AIME I Problems/Problem 1

Revision as of 16:21, 31 January 2009 by Skyhog (talk | contribs) (fixed typo in 5^8)

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.

1 2 3 4 5 6 7 8 9 10
Powers of $2$: $2$ $4$ $8$ $16$ $32$ $64$ $128$ $256$ $512$ $1\boxed{0}24$
Powers of $5$: $5$ $25$ $125$ $625$ $3125$ $15625$ $78125$ $39\boxed{0}625$

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