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2008 AMC 10A Problem 3

Problem 3 of 25EasierAlgebraNumber Theory

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?

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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.
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