内容目录
Self powers
Problem 48
The series, 11 + 22 + 33 + ... + 1010 = 10405071317.
Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.
Go to the thread for problem 48 in the forum.
ans=0
i=1
F=10**10
for i in range(1,1001):
p=1
for j in range(i): p=p*i%F
ans=(ans+p)%F
print i
print ans