Modular Arithmetic 3

Modular Arithmetic 3 - Problem 4 In multiplication tables...

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Problem # 4 In multiplication tables, the last row is always a reverse of the first row. Since N - 1 is always a coprime with N, then according to Problem #5, the last row must be a permutation of the first one. It's a very specific permutation. The number a m in column m staisfies a m = m·(N - 1) (mod N). But we have m·(N - 1) = -m = N - m (mod N) which exactly means that the last row is the reverse of the first. Problem # 5 In multiplication tables modulo N, rows corresponding to numbers coprime with N contain permutations of the first row. Let m be coprime to N. Let a and b be two different remainders of division by N. Then ma = mb (mod N) would imply m(a - b) = 0 (mod N) which, in turn, leads to (a - b) = 0 (mod N). Assuming b < a, we have a positive number less than N divisible by N. Contradiction. Hence a = b. Problem # 6 For prime (N + 1), multiplication tables offer multiple and simultaneous solutions to the rook problem: on an N×N board position N rooks so that none may capture another. To solve, select
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