2000 AIME II Problem 4

Below is the professionally curated solution for Problem 4 of the 2000 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AIME II solutions, or check the answer key.

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Concepts:factor countingprime factorization

Difficulty rating: 2070

4.

What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?

Solution:

Write N=2amN = 2^a m with mm odd. The odd divisors of NN are exactly the divisors of m,m, so d(m)=6.d(m) = 6. Every even divisor is 2k2^k (for 1ka1 \le k \le a) times an odd divisor, so there are ad(m)=6aa \cdot d(m) = 6a of them, and 6a=126a = 12 gives a=2.a = 2.

So N=4mN = 4m where mm is the smallest odd number with exactly 66 divisors. The shapes are p5p^5 (smallest 35=2433^5 = 243) and p2qp^2 q (smallest 325=453^2 \cdot 5 = 45), so m=45m = 45 and N=445=180.N = 4 \cdot 45 = 180.

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