The divisors of 6 are 1,2,3 and 6.
The sum of the squares of these numbers is 1+4+9+36=50.
The sum of the squares of these numbers is 1+4+9+36=50.
Let sigma2(n) represent the sum of the squares of the divisors of n. Thus sigma2(6)=50.
Let SIGMA2 represent the summatory function of sigma2, that is SIGMA2(n)=∑sigma2(i) for i=1 to n.The first 6 values of SIGMA2 are: 1,6,16,37,63 and 113.
Find SIGMA2(1015) modulo 109.
I have to found SIGMA2(10^15)
for bruteforce I cant't solve in my life time
hint
1^2 + 2^2 + 3^2 + 4^2+ 5^2 + .... N^2 = n(n+1)(2n+1)/6
and what you can do is find answer.~!!!!!!
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