2000 AIME I Problems/Problem 1: Difference between revisions
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n = 8:<math>2^8 = 256</math> | <math>5^8 = 390625</math> | n = 8:<math>2^8 = 256</math> | <math>5^8 = 390625</math> | ||
We see that <math>5^8</math> | We see that <math>5^8</math> contains the first zero, so n = <math>\boxed{008}</math>. | ||
== See also == | == See also == | ||
Revision as of 16:31, 5 January 2018
Problem
Find the least positive integer 
such that no matter how 
is expressed as the product of any two positive integers, at least one of these two integers contains the digit 
.
Solution
If a factor of has a 
and a 
in its prime factorization, then that factor will end in a . Therefore, we have left to consider the case when the two factors have the s and the s separated, so we need to find the first power of 2 or 5 that contains a 0.
For n = 1:![\[2^1 = 2 , 5^1 = 5\]](http://latex.artofproblemsolving.com/8/0/0/800db9e5191bc4e4a5a3b45161e9fb126ee24ca8.png)
n = 2:![\[2^2 = 4 , 5 ^ 2 =25\]](http://latex.artofproblemsolving.com/3/5/b/35b9714a58da6c3e2555b8c22cf9c2d71234075f.png)
n = 3:![\[2^3 = 8 , 5 ^3 = 125\]](http://latex.artofproblemsolving.com/b/9/b/b9b6c237c6b9f7c3394ebc7c9a2f05b3d34e07fe.png)
and so on, until,
n = 8:
| 
We see that 
contains the first zero, so n = 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
