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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 Spring '10
 Coluos
 Multiplication, #, main diagonal

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