has this got a solution with inequalities or is it just here by accident? anyway, since we only need the remainder mod 100, and $(100k+x)^8=x^8 \bmod 100$, we only want the remainder mod 100 of $1^8+...+100^8$. (All multiples of 10 can be removed as well.) To know the remainder mod 100 it is sufficient to know it mod 4 and mod 25. (via Chinese Remainder Theorem) mod 4 we see it is obviously 50=2 mod 25 we have that $f(x)=x^8$ goes as follows: 0 -> 0 1 -> 1 2 -> 6 3 -> 11 4 -> 11 5 -> 0 6 -> 16 7 -> 1 8 -> 16 9 -> 21 10 -> 0 11 -> 6 12 -> 21 13 -> 21 14 -> 6 15 -> 0 16 -> 21 17 -> 16 18 -> 1 19 -> 16 20 -> 0 21 -> 11 22 -> 11 23 -> 6 24 -> 1 25 -> 0 which repeats from here since $(25k+x)^8=x^8 mod 25$, but I bet there are much more elegant ways to show that it ends in 30...

I did... but next to proving the pattern 0-12 is symmetric to 13-25 I think there must be more tricks to avoid the case analysis... seems a quite ugly solution like that

since $gcd(8,\phi(25))=4$, we can replace the sum of the 8-th powers mod 25 by the sum of the 4-th powers mod 25. it is clear that we can leave all terms divisible by 5 and that $x^4\equiv 1\; mod\; 5$ if $gcd(x,5)=1$. since $(5k+x)^4\equiv x^4-5kx\; mod\; 25$, we see that each of the numbers $1, 6, 11, 16, 21$ occurs the same number of times, thus $\frac{20}{5}=4$ times, thus the sum of all 8-th powers mod 25 is $4(1+6+11+16+21)=20$. taking this 4 times(for the numbers $1,...,25$,$26,...,50$,$51,...,75$,$76,...,100$), we get 5 as the remainder mod 25, thus the last two digits are 30. not much nicer, but without case analysis. Peter