2014 AIME I Problem 3

Below is the professionally curated solution for Problem 3 of the 2014 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AIME I solutions, or check the answer key.

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Concepts:Euler’s Totient Functiongreatest common divisorfraction

Difficulty rating: 2110

3.

Find the number of rational numbers r,r, 0<r<1,0 \lt r \lt 1, such that when rr is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.1000.

Solution:

Write r=abr = \frac{a}{b} in lowest terms with a+b=1000;a + b = 1000; since 0<r<1,0 \lt r \lt 1, we need 1a499.1 \le a \le 499. Because gcd(a,b)=gcd(a,1000a)=gcd(a,1000),\gcd(a, b) = \gcd(a, 1000 - a) = \gcd(a, 1000), the fraction is in lowest terms exactly when aa is coprime to 1000.1000.

There are φ(1000)=10001245=400\varphi(1000) = 1000 \cdot \frac{1}{2} \cdot \frac{4}{5} = 400 integers in [1,999][1, 999] coprime to 1000,1000, and they pair up as a1000aa \leftrightarrow 1000 - a (note a=500a = 500 is not coprime to 10001000), so exactly 200200 of them are less than 500.500. The answer is 200.200.

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