Difference between revisions of "2000 AIME I Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | {{ | + | If there is a 2 and a 5 in one of the factors, then that factor will have a 0 in it. Therefore, the worst case scenario is where the 2's and the 5's are separated. Therefore, we need to find which <math>2^n</math> or <math>5^n</math> produces a 0 first. |
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| + | Thinking back to our powers of 2, <math>2^{10}</math> is the first power of 2 with a 0 in it. | ||
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| + | For our powers of 5, after a little multiplication, we find that <math>5^8=390625</math> is the first power of 5 with a 0 in it. | ||
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| + | <math>\boxed{008}</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}} | ||
Revision as of 09:43, 13 November 2007
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 there is a 2 and a 5 in one of the factors, then that factor will have a 0 in it. Therefore, the worst case scenario is where the 2's and the 5's are separated. Therefore, we need to find which 
or 
produces a 0 first.
Thinking back to our powers of 2, 
is the first power of 2 with a 0 in it.
For our powers of 5, after a little multiplication, we find that 
is the first power of 5 with a 0 in it.
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 | ||
