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Solutions Math 310, Homework #5 Problem HW5.1. Write out the addition and multiplication tables for Z/5Z and Z/6Z. Solution. For Z/5Z, we have: + [0]5 [1]5 [2]5 [3]5 [4]5 [0]5 [0]5 [1]5 [2]5 [3]5 [4]5 [1]5 [1]5 [2]5 [3]5 [4]5 [0]5 [2]5 [2]5 [3]5 [4]5 [0]5 [1]5 [3]5 [3]5 [4]5 [0]5 [1]5 [2]5 [4]5 [4]5 [0]5 [1]5 [2]5 [3]5 · [0]5 [1]5 [2]5 [3]5 [4]5 [0]5 [0]5 [0]5 [0]5 [0]5 [0]5 [1]5 [0]5 [1]5 [2]5 [3]5 [4]5 [2]5 [0]5 [2]5 [4]5 [1]5 [3]5 [3]5 [0]5 [3]5 [1]5 [4]5 [2]5 [4]5 [0]5 [4]5 [3]5 [2]5 [1]5 and and for Z/6Z, we have: + [0]6 [1]6 [2]6 [3]6 [4]6 [5]6 [0]6 [0]6 [1]6 [2]6 [3]6 [4]6 [5]6 [1]6 [1]6 [2]6 [3]6 [4]6 [5]6 [0]6 [2]6 [2]6 [3]6 [4]6 [5]6 [0]6 [1]6 [3]6 [3]6 [4]6 [5]6 [0]6 [1]6 [2]6 [4]6 [4]6 [5]6 [0]6 [1]6 [2]6 [3]6 [5]6 [5]6 [0]6 [1]6 [2]6 [3]6 [4]6 · [0]6 [1]6 [2]6 [3]6 [4]6 [5]6 [0]6 [0]6 [0]6 [0]6 [0]6 [0]6 [0]6 [1]6 [0]6 [1]6 [2]6 [3]6 [4]6 [5]6 [2]6 [0]6 [2]6 [4]6 [0]6 [2]6 [4]6 [3]6 [0]6 [3]6 [0]6 [3]6 [0]6 [3]6 [4]6 [0]6 [4]6 [2]6 [0]6 [4]6 [2]6 [5]6 [0]6 [5]6 [4]6 [3]6 [2]6 [1]6 and Theorem HW5.2. Let m ≥ 2 be a composite integer (that is, not prime). Prove that there are elements [x]m and [y]m of Z/mZ with [x]m , [y]m 6= [0]m such that [x]m · [y]m = [0]m . Proof. Let m ≥ 2 be a composite integer. Then by definition there are integers x, y with 0 < x, y < m such that xy = m. As 0 < x, y < m, then m - x, y so that [x]m , [y]m 6= 0 in Z/mZ. However, [x]m · [y]m = [xy]m = [m]m = [0]m . Page 1 of 1