Modular Power Train

Let us 'peel' the solution layer by layer.

Layer 1

Consider the problem of finding $i^2 \mod 4$ for a positive integer $i$ .

If $i=2m$ is even, then $i^2 \mod 4 = 4m^2 \mod 4 = 0$ , and

if $i=2m+1$ is odd, then $i^2 \mod 4 = 4(m^2+m)+1 \mod 4 = 1$

Layer 2

(ODD i) Now if $i$ is an odd number $i^{i^2} = i^{4k+1}$ for some $k$

and hence for odd $i$

$i^{i^2} \mod 10 = \left(i \mod 10\right)^{i^2} = \left(i \mod 10\right)^{4k}\left(i \mod 10\right)$

It is interesting to note that $1^4 \mod 10 = 3^4 \mod 10 = 7^4 \mod 10 = 9^4 \mod 10 = 1$ and $5^4 \mod 10 = 5$ and

further that $1^k \mod 4 = 1$ and $5^k \mod 4 = 5$

Hence, $\boxed{i^{i^2} \mod 10 = i \mod 10}$ for all odd numbers.

(EVEN i) If $i$ is an even number, $i^{i^2} = i^{4k}$ for some $k$ , which results in

$i^{i^2} \mod 10 = \left(i \mod 10\right)^{i^2} = \left(i \mod 10\right)^{4k}\left(i \mod 10\right)$

It is again interesting to see that for even $i$ , not divisible by 10, $2^4 \mod 10 \ 4^4 \mod 10 = 6^4 \mod 10 = 8^4 \mod 10 = 6$

Further, $6^k \mod 10 = 6$

Hence, for any even number which is not 0 modulo 10, $\boxed{i^{i^2} \mod 10 = 6}$

Layer 3

Now a table can be formed

Layer 4

Using the above table for evaluating the sum for various $n$ , it can be seen that $LHS = n \mod 10$ for $n=8,9,32,36,44$

Therefore, we have the solution to be $8+9+32+36+44=\boxed{129}$