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:

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- Jul 16th 2012, 11:02 PMtimkucNumber 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 PMrichard1234Re: Number system Problem
Note that $\displaystyle 144 = 2^43^2$. Therefore one of a or b must be divisible by $\displaystyle 2^4$, and one of a or b must be divisible by $\displaystyle 3^2$. 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 AMtimkucRe: Number system Problem
Why a or b will be divisible by 2^4 or 3^2 ????

.__Very confused :(__ - Jul 17th 2012, 05:47 AMemakarovRe: Number system Problem
In order for LCM(a,b) to be $\displaystyle 2^43^2$, a and b must have the form, respectively, $\displaystyle 2^x3^u$ and $\displaystyle 2^y3^v$ for some nonnegative integers x, y, u and v. Further, $\displaystyle \text{LCM}(2^x3^u,2^y3^v)=2^{\max(x,y)}3^{\max(u,v ))}$. Therefore, $\displaystyle \max(x,y)=4$ and $\displaystyle \max(u,v)=2$.

- Jul 17th 2012, 05:57 AMtimkucRe: Number system Problem
Yeah dude You are awesome :) thanks twice

- Jul 17th 2012, 06:09 PMtimkucRe: 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 AMemakarovRe: Number system Problem
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 AMtimkucRe: Number system Problem
(Bow)