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

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== Solution ==
 
== Solution ==
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{{solution}}
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If a factor of <math>10^{n}</math> has a <math>2</math> and a <math>5</math> in its [[prime factorization]], then that factor will end in a <math>0</math>. Therefore, we have left to consider the case when the two factors have the <math>2</math>s and the <math>5</math>s separated, so we need to find the first power of 2 or 5 that contains a 0.
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For <math>n = 1:</math> <cmath>2^1 = 2 , 5^1 = 5</cmath>
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<math>n = 2:</math> <cmath>2^2 = 4 , 5 ^ 2 =25</cmath>
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<math>n = 3:</math> <cmath>2^3 = 8 , 5 ^3 = 125</cmath>
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and so on, until,
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<math>n = 8:</math> <math>2^8 = 256</math> | <math>5^8 = 390625</math>
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We see that <math>5^8</math> contains the first zero, so <math>n = \boxed{8}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=I|before=First Question|num-a=2}}
 
{{AIME box|year=2000|n=I|before=First Question|num-a=2}}
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 16:44, 16 December 2020

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 (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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