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2 answers

Congratulations Sir! You are a genius.

2006-09-09 05:57:12 · answer #1 · answered by ash_m_79 6 · 0 0

let a,b e Z

a|b <=> there exists k e Z, b = k * a

suppose that's true ...

then b * n = (k * a) * n = (a * n) * k

so, then exists k e Z, (a*n) * k = b * n,

therefore (a*n) | (b*n)

2006-09-09 12:59:23 · answer #2 · answered by cybrdng 2 · 0 0

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