We find the remainder when each of the terms in the sum raised. $$(2^{40}+12^{41}+23^{42}+67^{43}+87^{44})\equiv 1+1+1+1+1 \equiv 5 \mod11$$So raising this to the $45!+46$ power, we get that $10 | 45!$, so the remainder when the exponent is divided by $10$ is $6$. We now derive $$5^6\equiv 15625 \equiv 1420*11+5 \equiv 5 \mod11$$QED.