# Number system Problem

• Jul 16th 2012, 11:02 PM
timkuc
Number system Problem
If N= 144 How many sets of two values (a,b) are there for which the LCM (a,b) is 144 ??

What the general formula for this (with proof) ??:confused:
• Jul 16th 2012, 11:14 PM
richard1234
Re: Number system Problem
Note that $144 = 2^43^2$. Therefore one of a or b must be divisible by $2^4$, and one of a or b must be divisible by $3^2$. How many ordered pairs (a,b) satisfy these conditions?

This method definitely generalizes, but I doubt there is a general formula since it becomes one giant counting problem when N has more prime divisors.
• Jul 17th 2012, 05:08 AM
timkuc
Re: Number system Problem
Why a or b will be divisible by 2^4 or 3^2 ????

. Very confused :(
• Jul 17th 2012, 05:47 AM
emakarov
Re: Number system Problem
In order for LCM(a,b) to be $2^43^2$, a and b must have the form, respectively, $2^x3^u$ and $2^y3^v$ for some nonnegative integers x, y, u and v. Further, $\text{LCM}(2^x3^u,2^y3^v)=2^{\max(x,y)}3^{\max(u,v ))}$. Therefore, $\max(x,y)=4$ and $\max(u,v)=2$.
• Jul 17th 2012, 05:57 AM
timkuc
Re: Number system Problem
Yeah dude You are awesome :) thanks twice
• Jul 17th 2012, 06:09 PM
timkuc
Re: Number system Problem
@emakarov : We can't say intuitively that ia or b will be divisible by 2^4 or 3^2 ,right ? (can we ?)
• Jul 18th 2012, 03:39 AM
emakarov
Re: Number system Problem
Quote:

Originally Posted by timkuc
We can't say intuitively that ia or b will be divisible by 2^4 or 3^2 ,right ?

We can. If the maximum power of 2 that divides either a or b is, say, 3, then there is no reason to include 2^4 as a factor of LCM(a, b). In the notation of post #4, since 2^4 is the maximum power of 2 that divides 144, we have max(x, y) = 4, so x = 4 or y = 4 or both. This means that 2^4 divides a or b or both. Similarly for 3^2.
• Jul 19th 2012, 04:14 AM
timkuc
Re: Number system Problem
(Bow)