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suppose you have a pair of number a and b, whose largest common factor is 1.
And you multiply a by natural numbers then divide the result by b and get remainders. It must be b times until you get the same remainder when each the non negative numbers smaller than b appears once.
e.g
7 5
7*1=7 7|5=1...2
7*2=14 14|5=3...4
7*3=21 21|5=4...1
7*4=28 28|5=5...3
7*5=35 ...0
Apparently this theorem shall be proven by contradiction
But does anyone remember what this theorem is called?
X'(y-Xβ)=0
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Hi George Y,
It must be 40 years since I did this so what follows is 'covered in mathematical rust'.
Cannot recall that it is a named theorem at all. I had a quick trawl on Wiki but found nothing. But it may be there somewhere. Best guess is Gauss discovered/proved it first.
To prove:
(i) You won't get a zero remainder.
Assume you do. => axn (for some integer n) = b x m (for some other integer m)
=> a and b have a prime factor other than 1 =><=
(ii) Assume two integers give the same remainder
=> axn = axm for different integers n and m
=> a(n-m) = 0
=><= (i)
(i) and (ii) together => every remainder will occur once and once only.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Indeed bob it should be proved this way.
I already asked JaneFairFox for help, for she mentioned this theorem before.
X'(y-Xβ)=0
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