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12^12 = (3^12)(2^24) phi(12^12) = phi(3^12) * phi(2^24) = 2^(24-1) * (2-1) *3^12 *(3-1) = 2^23 * 3^12 * 2 = 2^24 * 3^12

Well, I set it up: 12^12 = (3^12)(4^12) = (3^12)(2^24) sigma(12^12) = (1-2^25)/(1-2) * (1-3^13)/(1-3)

Let n=12^12. Find sigma(n). sigma(n) = the sum of the positive divisors of n.

Find eight different numbers n with phi(n) = 240. phi(n) = the number of positive integers less than or equal to n, that are relatively prime to n.