2014 AIME I Problems/Problem 3

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 1

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 $n+m =1000$ and $\dfrac{n}{m}$ is irreducible.

We note that $\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1$. Hence, $\dfrac{n}{m}$ is irreducible if $\dfrac{1000}{m}$ is irreducible, and $\dfrac{1000}{m}$ is irreducible if $m$ 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 numbers are divisible by 2
  • 99 numbers are divisible by 5
  • 49 numbers are divisible by 10

Using the Principle of Inclusion and Exclusion, we get that there are $499-249-99+49=200$ numbers between $501$ and $999$ are not divisible by either $2$ or $5$, so our answer is $\boxed{200}$.

Euler's Totient Function can also be used to arrive at 400 numbers relatively prime to 1000, meaning 200 possible fractions satisfying the necessary conditions.

Solution 2

If the initial manipulation is not obvious, instead ,consider the euclidean algorithm. Instead of using $\frac{n}{m}$ as the fraction to use the euclidean algorithm on, we can rewrite this as $\frac{500-x}{500+x}$ \[gcd(500+x,500-x)=gcd((500+x)+(500-x),500-x)=gcd(1000,500-x)\]. Thus, we want $gcd(1000,500-x)=1$. You can either proceed as solution 1, or consider that no even numbers work, limiting us to $250$ choices of numbers and restricting $x$ to be odd. If $x$ is odd, $500-x$ is odd, so the only possible common factors $1000$ and $500-x$ can share are multiples of $5$. Thus, we want to avoid these. There are $50$ multiples of $5$ less than $500$, so the answer is $250-50=\boxed{200}$.

Solution 3

Say $r=\frac{d}{1000-d}$; then $1<=d<=499$. If this fraction is reducible, then the modulus of some number for $d$ is the same as the modulus for $1000-d$. Since $1000=2^3*5^3$, that modulus can only be $2$ or $5$. This implies that if $d|2$ or $d|5$, the fraction is reducible. There are 249 cases where $d|2$, 99 where $d|5$, and 49 where $d|(2*5=10)$, so by PIE, the number of fails is 299, so our answer is $\boxed{200}$.

Solution 4

We know that the numerator of the fraction cannot be even, because the denominator would also be even. We also know that the numerator cannot be a multiple of 5 because the denominator would also be a multiple of 5. Proceed by listing out all the other possible fractions and we realize that the numerator and denominator are always relatively prime. We have 499 fractions to start with, and 250 with odd numerators. Subtract 50 to account for the multiples of 5, and we get $\boxed{200}$.

Solution 5

Let the numerator and denominator $x,y$ with $\gcd(x,y)$ and $x+y = 1000.$ Now if $\gcd(1000,x) = 1$ then $\gcd(x,1000-x) = \gcd(x,y) = 1.$ Therefore any pair that works satisfies $\gcd(x,1000)= 1.$ There are $\phi(1000) = 400$ pairs $x,y$ such that $x+y= 1000$ and $\gcd(x,y) = 1.$ However, exactly half of them work because of the condition $x<y.$ Therefore the desired answer is $\boxed{200}.$

Solution 6 (sort of cheese)

We notice that there are a total of $400$ fractions that are in simplest form and the numerator and denominator add up to $1000$ because the numerator and denominator have to be relatively prime so there are $\Phi(1000)=400$. Half of these are greater than $1$ so the answer is $400/2=200$

- bedwarspro

See also

2014 AIME I (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS