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:

Printable View

- 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 . Therefore one of a or b must be divisible by , and one of a or b must be divisible by . 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 , a and b must have the form, respectively, and for some nonnegative integers x, y, u and v. Further, . Therefore, and .

- 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)