abstract algebra?

Question

If a|c and b|c and gcd(a,b) , then ab|c .
[Hint: By Theorem 2.12, p. 75, since gcd(a,b)=1 , we can write 1 = ax+by for some integers x and y . End of Hint.]

Answer 1

If you want the proof, it is as follows. Assume gcd(a,b) = 1. Then there exist integers x and y such that 1 = ax + by. We multiply by c to get c = acx + bcy. Since a|c we have ab|bc, and since b|c we have ab|ac. Therefore ab divides acx + bcy, so ab|c.

Answer 2

When gcd(a,b)=1
and
a|c and b|c

We have ab | c.
If gcd(a,b)!=1 then this may not be true.

Eg.
2|100 and 5|100
since gcd(2,5) = 1
Hence 2*5 = 10|100

But on contrary
5|100
and 50|100
But 5*50 = 250 does not divide 100.
Since gcd(5,50) is not 1, it is 5.