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Difference between revisions of "2014 AIME I Problems/Problem 3"

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== Solution ==
 
== Solution ==
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We have that the set of these rational numbers is from <math>\dfrac{1}{999}</math> to <math>\dfrac{499}{501}</math> where each each element <math>\dfrac{n}{m}</math> has <math>n+m =1000</math> but we also need <math>\dfrac{n}{m}</math> to be irreducible  
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We have that the set of these rational numbers is from <math>\dfrac{1}{999}</math> to <math>\dfrac{499}{501}</math> where each each element <math>\dfrac{n}{m}</math> has <math>n+m =1000</math> but we also need <math>\dfrac{n}{m}</math> to be irreducible.
  
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we note that <math>\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1</math>
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We note that <math>\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1</math>
 
hence <math>\dfrac{n}{m}</math> is irreducible iff <math>\dfrac{1000}{m}</math> isn't, which is equivalent to m not being divisible by 2 or 5.
 
hence <math>\dfrac{n}{m}</math> is irreducible iff <math>\dfrac{1000}{m}</math> isn't, which is equivalent to m not being divisible by 2 or 5.
 
so the question is equivalent to "how many numbers between 501 and 999 are not divisible by neither 2 or 5?"
 
so the question is equivalent to "how many numbers between 501 and 999 are not divisible by neither 2 or 5?"
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Using Principle of Inclusion Exclusion (PIE):
 
Using Principle of Inclusion Exclusion (PIE):
 
we get:
 
we get:
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<math>499-249-99+49=200</math> numbers between 501 and 999 arent divisible by neither 2 or 5 so our answer is <math>200</math>
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<math>499-249-99+49=200</math> numbers between 501 and 999 are not divisible by neither 2 or 5 so our answer is <math>200</math>

Revision as of 16:36, 14 March 2014

Problem 3

Find the number of rational numbers $r,$ $0<r<1,$ such that when $r$ 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 $\dfrac{1}{999}$ to $\dfrac{499}{501}$ where each each element $\dfrac{n}{m}$ has $n+m =1000$ but we also need $\dfrac{n}{m}$ to be irreducible.

We note that $\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1$ hence $\dfrac{n}{m}$ is irreducible iff $\dfrac{1000}{m}$ isn't, which is equivalent to m not being divisible by 2 or 5. so the question is equivalent to "how many numbers between 501 and 999 are not divisible by neither 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: $499-249-99+49=200$ numbers between 501 and 999 are not divisible by neither 2 or 5 so our answer is $200$

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