we have to subtract....some cases which give 1 and that number itself
8 Answers
Hint: try to write 30! as 2a3b5c7d11e13f17g19h23i29j
The important thing is dont get scared seeing this expression :D
g=h=i=j=1 (trivially?)
a = 15+7+3+1 = 26
b = 10+3+1 = 14
c = 6+1 = 7
d = 4
e = 2
now find the sum of all factors of this number?
and then sum of all divisors =(1+2+22.....+226)(1+3+32...+312)(......)(1+29)
no karna..
i meant that you have to divide by (2-1), (3-1), (5-1) and so on [3]
226.314.57.74.112.132.17.19.23.29
sum of all divisors = (20+..+226)(30+..+314)...(230+23)(290+29)
nishant sir, karna's answer seems perfectly fine except for the power of 3 in #6
we need to divide by (2-1)(3-1)... only when we apply the GP formula for sum.. He has written the previous step