Problem Solving (PS) | Re: P = 1^1 + 2^2 + 3^3 + 4^4 + 5^5 +............+48^{48}+49^{49}+50^{50}

We will split the series 1^1 + 2^2 + 3^3….50^50 in to two sets each consisting of even and odd numbers .

So we have ,

Set1 = 2^2+4^4 + 6^6…50^50

Set2 = 1^1+3^3+7^7….49^49

so the problem reduces to ( ( Set 1 + Set 2) / 8), our approach would be to find the remainders of the two sets when individually divided by 8 and then add them up to get the final remainder.

Considering Set 1 we observe

Except 2^2 every other even number is a power of 8 , so the remainder of the first set comes out