2005 AIME I Problem 3

Below is the professionally curated solution for Problem 3 of the 2005 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AIME I solutions, or check the answer key.

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Concepts:factor countingprimecombinations

Difficulty rating: 2070

3.

How many positive integers have exactly three proper divisors, each of which is less than 50?50? (A proper divisor of a positive integer nn is a positive integer divisor of nn other than nn itself.)

Solution:

An integer with exactly three proper divisors has exactly four divisors in total, so it is either n=pqn = pq with pp and qq distinct primes (proper divisors 1,p,q1, p, q) or n=p3n = p^3 with pp prime (proper divisors 1,p,p21, p, p^2).

In the first case we need pp and qq both less than 50.50. There are 1515 primes below 50,50, giving (152)=105\binom{15}{2} = 105 such numbers. In the second case we need p2<50,p^2 \lt 50, which holds for p=2,3,5,7,p = 2, 3, 5, 7, giving 44 more.

The total is 105+4=109.105 + 4 = 109.

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