lcm of k,k+1,k+2 where k≡3(mod 4)

• Feb 11th 2010, 09:05 PM
kingwinner
lcm of k,k+1,k+2 where k≡3(mod 4)
Find the least common multiple of k,k+1, and k+2 where k≡3(mod 4).

Some ideas:
k≡3(mod 4) => k=3+4z for some z E Z
So k is odd, k+1 is even, and k+2 is odd.
Also, this question asks about the lcm of THREE numbers, so I don't think the the Euclidean algorithm (which only gives the gcd or lcm of TWO numbers) will help us in this question.

So how can we solve this?
Thanks for any help!

[also under discussion in math links forum]
• Feb 11th 2010, 09:16 PM
Drexel28
Quote:

Originally Posted by kingwinner
Find the least common multiple of k,k+1, and k+2 where k≡3(mod 4).

Some ideas:
k≡3(mod 4) => k=3+4z for some z E Z
So k is odd, k+1 is even, and k+2 is odd.
Also, this question asks about the lcm of THREE numbers, so I don't think the the Euclidean algorithm (which only gives the gcd or lcm of TWO numbers) will help us in this question.

So how can we solve this?
Thanks for any help!

Hint: $\text{lcm}\left(a,b,c\right)=\text{lcm}\left(\text {lcm}(a,b),c\right)$
• Feb 11th 2010, 11:07 PM
kingwinner
Quote:

Originally Posted by Drexel28
Hint: $\text{lcm}\left(a,b,c\right)=\text{lcm}\left(\text {lcm}(a,b),c\right)$

Is this always true? How can we prove it?
• Feb 12th 2010, 05:41 AM
dooping
Since k is odd, k+1 is even, and k+2 is odd
then
GCD(k,k+1,k+2)=1 => LCM(k,k+1,k+2)=k(k+1)(k+2)
• Feb 12th 2010, 01:46 PM
kingwinner
Quote:

Originally Posted by dooping
Since k is odd, k+1 is even, and k+2 is odd
then
GCD(k,k+1,k+2)=1 => LCM(k,k+1,k+2)=k(k+1)(k+2)

How do we actually formally prove these two claims?