60 Followers Problem

$\displaystyle \sum _{ n=1 }^{ 60 }{ { \left( 60n \right) }^{ \left( 60n \right) } }$

In the above summation $60n$ is always a multiple of $60$ or more precisely $10$ .

Therefore, all the terms are in form of $\overline{a0}$ where $a$ is a multiple of $6$ from $6$ to $360$ .

So $\overline{a0}^{\overline{a0}}$ is always going to end in a $0$ . Therefore, last digit of the summation is $\boxed{0}$ .

Given that the base is the multiple of 10, regardless of how much you exponent it, the last digit will always be zero. You can also pull the $60^{60}$ out of the equation and it will be

$\sum_{n=1}^{60} (60n)^{(60n)} = 60^{60}\times(\sum_{n=1}^{60}n^{60}(60n)^{60(n-1)})$

Regardless of the second sum, $60^{60}$ is a multiple of 10 and the final digit will always be 0