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Difference between revisions of "2000 AIME I Problems/Problem 1"

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| Powers of <math>2</math>: || <math>2</math> || <math>4</math> || <math>8</math> || <math>16</math> || <math>32</math> || <math>64</math> || <math>128</math> || <math>256</math> || <math>512</math> || <math>1\boxed{0}24</math>
 
| Powers of <math>2</math>: || <math>2</math> || <math>4</math> || <math>8</math> || <math>16</math> || <math>32</math> || <math>64</math> || <math>128</math> || <math>256</math> || <math>512</math> || <math>1\boxed{0}24</math>
 
|-
 
|-
−
| Powers of <math>5</math>: || <math>5</math> || <math>25</math> || <math>125</math> || <math>625</math> || <math>3125</math> || <math>15625</math> || <math>78125</math> || <math>39\boxed{0}265</math>
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| Powers of <math>5</math>: || <math>5</math> || <math>25</math> || <math>125</math> || <math>625</math> || <math>3125</math> || <math>15625</math> || <math>78125</math> || <math>39\boxed{0}625</math>
 
|}
 
|}
  

Revision as of 16:21, 31 January 2009

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 (Problems • Answer Key • Resources)
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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