In each case show your work and justify your answers. (a) If 9k+5 and 2k+1 have the same last digit for some natural number k, find the last digit of k. (b) If S = ∑ 19 i=1 i!, then find the last digit of S 2 . (c) Denote the last digit of a natural number a by b. Show that the last digit of a n is the same as the last digit of b n where is n ≥ 1 is a natural number. (d) Inspired by part (c), show that the last digit of a 4k+1 for all natural numbers k is the same as the last digit of a. [Euler’s Theorem is not allowed.]