# Divisibility (gcd)

• December 15th 2008, 09:48 PM
Sea
Divisibility (gcd)
Show that: $c\mid ab \Rightarrow c\mid (a,b).(b,c)$

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(a,b)=gcd(a,b)
(b,c)=gcd(b,c)
• December 15th 2008, 11:40 PM
o_O
Are you sure that's the correct statement?

For example, $6 \mid (8 \times 9)$ but $(8,9) \cdot (9,6) = (1)(3) = 3$ and $6 \not{\mid} \ 3$