Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (fixed typo in 5^8)
(→‎Solution)
Line 5: Line 5:
 
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.
  
−
{| cellspacing="8" cellpadding="8"
+
<math>2^1 = 2</math> | <math>5^1 = 5</math>
−
|-
+
<math>2^2 = 4</math> | <math>5 ^ 2 =25</math>
−
| || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10
+
<math>2^3 = 8</math> | <math>5 ^3 = 125</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}625</math>
+
<math>2^8 = 256</math> | <math>5^8 = 390625</math>
−
|}
 
  
 
We see that <math>5^8</math> generates the first zero, so the answer is <math>\boxed{008}</math>.
 
We see that <math>5^8</math> generates the first zero, so the answer is <math>\boxed{008}</math>.

Revision as of 23:19, 7 August 2010

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.

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

                 .
                 .
                 .

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

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

See also

2000 AIME I (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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_1&oldid=35581"