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- May 25th 2011, 05:48 PM #1

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- May 25th 2011, 07:28 PM #2

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- May 31st 2011, 04:30 PM #3

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## This was not quite clear to me, so I found a way that might be easier

* c|b notation reads c divides b

Let (b,a) = c and (a,r) = d

c|b and c|a, then c|b-aq

r = b - aq, so c|r

c (less than or equal to) d

d|a and d|r, then d|aq + r

b = aq + r, so d|b

d (less than or equal to) c

Therefore d=c