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

Find the last digit of the expression above.

The answer is 0.

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

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 $\boxed{0}$ .