Difference between revisions of "2014 AIME I Problems/Problem 3"
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We note that <math>\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1</math>. | We note that <math>\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1</math>. | ||
| − | Hence, <math>\dfrac{n}{m}</math> is irreducible if <math>\dfrac{1000}{m}</math> is | + | Hence, <math>\dfrac{n}{m}</math> is irreducible if <math>\dfrac{1000}{m}</math> is irreducible, and <math>\dfrac{1000}{m}</math> is irreducible if <math>m</math> is not divisible by 2 or 5. Thus, the answer to the question is the number of integers between 999 and 501 inclusive that are not divisible by 2 or 5. |
We note there are 499 numbers between 501 and 999, and | We note there are 499 numbers between 501 and 999, and | ||
Revision as of 11:31, 27 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 
and is irreducible.
We note that 
.
Hence, is irreducible if 
is irreducible, and is irreducible if 
is not divisible by 2 or 5. Thus, the answer to the question is the number of integers between 999 and 501 inclusive that are not divisible by 2 or 5.
We note there are 499 numbers between 501 and 999, and
- 249 are divisible by 2
- 99 are divisible by 5
- 49 are divisible by 10
Using the Principle of Inclusion and Exclusion, we get that there are 
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 2 |
Followed by Problem 4 | |
| 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. 
