Let
N=2^{4}* 5^{7}.
What is the sum of all N's positive factors?
Generalize.
(In reply to
a solution by Charlie)
Without using directly the sigma function, it is possible to relay on the sum of geometric sequences. When the first term is 1 the terms of the sequence are the successive potences of the ratio.
1+a+a^2+ ... +a^n = [a^(n+1)1]/(a1)
For (a, b)=(2, 5)
S= (2^(m+1)1)*(5^(n+1)1)/4
For (m,n) = (4,7)
S= (2^51)*(5^81)/4 = 31*390624/4 = 3027336
For general number (a,b) and potences (m,n)
S=[(a^(m+1)1)*(b^(n+1)1]/[(a1)*(b1)]
which is true only if there is no pair (i, j) that a^m*b^n = a^i*b^j
Edited on March 9, 2018, 5:48 am

Posted by armando
on 20180309 03:42:06 