2005 AMC 10A Problem 24

Attempt Problem 24 of the 2005 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10A solutions, or check the answer key.

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24.

For each positive integer m>1,m \gt 1, let P(m)P(m) denote the greatest prime factor of m.m. For how many positive integers nn is it true that both P(n)=nP(n) = \sqrt{n} and P(n+48)=n+48?P(n + 48) = \sqrt{n + 48}?

00

11

33

44

55

Answer: B
Concepts:primeperfect squaredifference of squares
Difficulty rating: 2120
Solution:

The condition P(n)=nP(n) = \sqrt{n} means nn is the square of a prime q,q, and likewise n+48=p2n + 48 = p^2 for a prime p.p. Then 48=p2q2=(pq)(p+q).48 = p^2 - q^2 = (p - q)(p + q). Checking the same-parity factorizations of 48,48, only pq=2, p+q=24p - q = 2,\ p + q = 24 yields primes, giving (p,q)=(13,11)(p, q) = (13, 11) and n=121.n = 121. So there is exactly one such n.n.

Thus, the correct answer is B.

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