# Least common multiple - Greatest common divisor

• Oct 24th 2010, 07:08 AM
teps
Least common multiple - Greatest common divisor
So I have 2 positive integers a and b, a<b.
And their LCM minus GCD is 143.
Find a and b.
• Oct 24th 2010, 08:34 AM
HappyJoe
The problem is just to find any pair (a,b) for which LCM(a,b) - GCD(a,b) = 143?

You can take a=1, b=144. Their GCD is 1, while their LCM is 144.
• Oct 24th 2010, 01:00 PM
teps
No, the problem is to find all pairs, forgot to mention that sry. :)
• Oct 24th 2010, 01:40 PM
brennan
do you know the answer - because i think there are infinate answers - or have i been away from amths for so long im making stupid errors?
• Oct 24th 2010, 02:20 PM
Also sprach Zarathustra
LCM(a,b) - GCD(a,b) = 143

LCM(a,b)=k*GCD(a,b), k is some integer.

LCM(a,b) - GCD(a,b) = GCD(a,b) {k-GCD(a,b)}=143

hmmm...
• Oct 25th 2010, 05:29 AM
teps
Quote:

Originally Posted by Also sprach Zarathustra
LCM(a,b) - GCD(a,b) = 143

LCM(a,b)=k*GCD(a,b), k is some integer.

LCM(a,b) - GCD(a,b) = GCD(a,b) {k-GCD(a,b)}=143

hmmm...

You mean LCM(a,b) - GCD(a,b) = GCD(a,b) (k-1)=143=13*11, thank you from here on i make 4 cases, and get that:
(a,b)={(22,77);(11,154);(39,52);(13,156);(9,16);(1 ,144);(143,246)}
• Oct 25th 2010, 05:45 AM
Also sprach Zarathustra
Quote:

Originally Posted by teps
You mean LCM(a,b) - GCD(a,b) = GCD(a,b) (k-1)=143=13*11, thank you from here on i make 4 cases, and get that:
(a,b)={(22,77);(11,154);(39,52);(13,156);(9,16);(1 ,144);(143,246)}

Typo, thanks...

Good-job btw.