For 4.23(b), remember the method in 4.3(b), where you had $$31^{999} + 65^{100} \equiv (-1)^{999} + 1^{100} \equiv -1 + 1 = 0.$$ As a further warmup to 4.23(b), answer this question:  what is $(-\pi)^{2017} + (\pi)^{2017}$?

For 4.26, use 4.25(b), which shows that a number is equivalent to the alternating sum of its digits (right to left) modulo $11$.