Difference between revisions of "2014 AIME I Problems/Problem 3"
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<cmath>499-249-99+49=200</cmath> numbers between <math>501</math> and <math>999</math> are not divisible by either <math>2</math> or <math>5</math> so our answer is <math>\boxed{200}</math> | <cmath>499-249-99+49=200</cmath> numbers between <math>501</math> and <math>999</math> are not divisible by either <math>2</math> or <math>5</math> so our answer is <math>\boxed{200}</math> | ||
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| + | == See also == | ||
| + | {{AIME box|year=2014|n=I|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Revision as of 18:38, 14 March 2014
Problem 3
Find the number of rational numbers 
such that when 
is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.
Solution
We have that the set of these rational numbers is from 
to 
where each each element 
has 
but we also need to be irreducible.
We note that 
hence is irreducible if 
isn't, which is equivalent to m not being divisible by 2 or 5.
Therefore, the question is equivalent to "how many numbers between 501 and 999 are not divisible by either 2 or 5?"
We note there are 499 numbers between 501 and 999
- 249 are even (divisible by 2)
- 99 are divisible by 5
- 49 are divisible by 10 (both 2 and 5)
Using Principle of Inclusion Exclusion (PIE):
we get:
![\[499-249-99+49=200\]](http://latex.artofproblemsolving.com/a/6/9/a69b5c0017cc9b37399527494d68e2e84e8bd853.png)
numbers between 
and 
are not divisible by either 
or 
so our answer is 
See also
| 2014 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
