Difference between revisions of "2000 AIME I Problems/Problem 1"
(→Solution) |
(→Solution) |
||
| Line 8: | Line 8: | ||
<math>2^2 = 4</math> | <math>5 ^ 2 =25</math> | <math>2^2 = 4</math> | <math>5 ^ 2 =25</math> | ||
<math>2^3 = 8</math> | <math>5 ^3 = 125</math> | <math>2^3 = 8</math> | <math>5 ^3 = 125</math> | ||
| − | + | and so on, until, | |
| − | + | ||
| − | |||
<math>2^8 = 256</math> | <math>5^8 = 390625</math> | <math>2^8 = 256</math> | <math>5^8 = 390625</math> | ||
Revision as of 23:21, 7 August 2010
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, in other words whether 
or 
produces a 0 first.
| 
| 
| 
and so on, until,
| 
We see that 
generates the first zero, so the answer is 
.
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 | ||
