Just need your brain

To find the second last digit we first need to find $\large{11^5+13^6+15^7} \pmod {100}$ . The last two digits of a power of $11$ repeat every $10$ powers, i.e.

$11^1 \equiv 11 \pmod {100}$

$11^2 \equiv 21 \pmod {100}$

$11^3 \equiv 31 \pmod {100}$

$11^4 \equiv 41 \pmod {100}$

and so on until

$11^{10} \equiv 01 \pmod {100}$ , and then we're back to

$11^{11} \equiv 11 \pmod {100}$ .

Therefore, we know already that the last two digits of $11^5$ are $51$ .

We can carry out the selfsame procedure with $13$ : this is a bit more messy. However, by calculating every power before $6$ of $13$ , and then calculating $13^6 \pmod {100}$ , this can be done, though quite slowly. We will finally find that the last two digits of $13^6$ are $09$ .

Thus far, we have that $11^5+13^6+15^7 \pmod {100} \equiv 60+15^5 \pmod {100}$ . We can calculate that $15^7 \equiv 75 \pmod {100}$ . Our final answer is therefore $60+75 \equiv 35 \pmod {100}$ , with the second last digit being $3$ .

@Brian Charlesworth @Sharky Kesa Do you perhaps know a quicker way? If the powers become quite large, my method will take a very long time.

can you tell me the short way??? like 8^80's unit digit?