2008 AMC 10A Problem 3

Below is the professionally curated solution for Problem 3 of the 2008 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10A solutions, or check the answer key.

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Concepts:sum of factorsfunction

Difficulty rating: 940

3.

For the positive integer n,n, let n\langle n \rangle denote the sum of all the positive divisors of nn with the exception of nn itself. For example, 4=1+2=3\langle 4 \rangle = 1 + 2 = 3 and 12=1+2+3+4+6=16.\langle 12 \rangle = 1 + 2 + 3 + 4 + 6 = 16. What is 6?\langle\langle\langle 6 \rangle\rangle\rangle?

66

1212

2424

3232

3636

Solution:

The positive divisors of 66 other than 66 are 1,2,1, 2, and 3,3, so 6=1+2+3=6.\langle 6 \rangle = 1 + 2 + 3 = 6.

Since applying the operation to 66 again returns 6,6, we get 6=6.\langle\langle\langle 6 \rangle\rangle\rangle = 6.

(A number equal to the sum of its proper divisors is called a perfect number, and 66 is the smallest.)

Thus, the correct answer is A.

Problem 3 in Other Years