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

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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, in other words whether <math>2^n</math> or <math>5^n</math> produces a 0 first.
 
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, in other words whether <math>2^n</math> or <math>5^n</math> produces a 0 first.
  
<math>2^1 = 2</math> | <math>5^1 = 5</math>
+
<cmath>2^1 = 2 | 5^1 = 5</cmath>
<math>2^2 = 4</math> | <math>5 ^ 2 =25</math>
+
<cmath>2^2 = 4 | 5 ^ 2 =25</cmath>
<math>2^3 = 8</math> | <math>5 ^3 = 125</math>
+
<cmath>2^3 = 8 | 5 ^3 = 125</cmath>
  
 
and so on, until,  
 
and so on, until,  

Revision as of 18:52, 1 March 2015

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

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