Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that we have two integers $a$ and $b$. Now say that $G = \gcd(a,b)$ and $L = \mathrm{lcm}(a,b)$. Now the value of $G$ and $L$ is given and another integer $c$'s value is given. How can we find $\gcd(a+c,b+c)$ and $\mathrm{lcm}(a+c,b+c)$ from $G$, $L$ and $c$?

What if we have $n$ arbitrary numbers. I know the GCD and LCM of those numbers but not the actual values of those numbers. Now I want to add $c$ with all of those numbers, what will be the new GCD, LCM of those numbers?

share|improve this question
Do your variables $c$ and $C$ refer to the same number? In that case please consistently use the same letter. – joriki Oct 22 '12 at 19:42
2  
The gcd(a+c,b+c) is a divisor of $a-b$. Therefore you have a finite number of possibilities as $c$ varies. – Pantelis Damianou Oct 22 '12 at 20:00
I believe there is a theorem (a special case of this) about Mersenne numbers which states that $\gcd ({2^n} - 1,{2^m} - 1) = {2^{\gcd (n,m)}} - 1$. – glebovg Oct 22 '12 at 20:25

1 Answer

Of course $\gcd (ka,kb) = |k|\gcd (a,b)$ for some integer $k \ne 0$ is easy to prove, but I do not think is a generalization you seek because this would mean the the gcd of any two numbers is somehow generated by $\gcd (a,b)$, where $a$ and $b$ are arbitrary. That does not seem likely. Unless $a$ and $b$ are coprime, which is trivial. The same goes for lcm.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.