MathDB

6

Part of 2008 Korea Junior Math Olympiad

Problems(1)

8 \(n^3f_1(n) - 2nf_9(n) + n^2f_3(n)) where, f_s(n) = d_1^s+d_2^s+..+d_k^s

Source: KJMO 2008 p6

5/2/2019
If d1,d2,...,dkd_1,d_2,...,d_k are all distinct positive divisors of nn, we defi ne fs(n)=d1s+d2s+..+dksf_s(n) = d_1^s+d_2^s+..+d_k^s. For example, we have f1(3)=1+3=4,f2(4)=1+22+42=21f_1(3) = 1 + 3 = 4, f_2(4) = 1 + 2^2 + 4^2 = 21. Prove that for all positive integers nn, n3f1(n)−2nf9(n)+n2f3(n)n^3f_1(n) - 2nf_9(n) + n^2f_3(n) is divisible by 88.
number theoryDivisorsSum of powersSumdivisible