MHB How Many Ordered Triples (a, b, c) Satisfy Given LCM Conditions?

[juantheron]

If $\bf{L.C.M}$ of $(a,b)$ is $432,$ and $\bf{L.C.M}$ of $(b,c)$ is $72,$ and $\bf{L.C.M}$ of $(c,a)$ is $432.$

Then the number of ordered pairs $(a,b,c)$ is

My Trail Solution:: First we will factorise in prime factor form.

$\bf{L.C.M}$ of $(a,b)$ is $ = 432 = 2^4 \times 3^3$

Similarly $\bf{L.C.M}$ of $(b,c)$ is $ = 72 = 2^3 \times 3^2$

Similarly $\bf{L.C.M}$ of $(c,a)$ is $ = 432 = 2^4 \times 3^3$

Now Let we assume that $a = 2^l\cdot 3^m$

and Similarly $b = 2^p\cdot 3^q$

and Similarly $c = 2^x\cdot 3^y$,

Now How can i solve after that

Help Required.

Thanks.

 

[kaliprasad]

If $\bf{L.C.M}$ of $(a,b)$ is $432,$ and $\bf{L.C.M}$ of $(b,c)$ is $72,$ and $\bf{L.C.M}$ of $(c,a)$ is $432.$

Then the number of ordered pairs $(a,b,c)$ is

My Trail Solution:: First we will factorise in prime factor form.

$\bf{L.C.M}$ of $(a,b)$ is $ = 432 = 2^4 \times 3^3$

Similarly $\bf{L.C.M}$ of $(b,c)$ is $ = 72 = 2^3 \times 3^2$

Similarly $\bf{L.C.M}$ of $(c,a)$ is $ = 432 = 2^4 \times 3^3$

Now Let we assume that $a = 2^l\cdot 3^m$

and Similarly $b = 2^p\cdot 3^q$

and Similarly $c = 2^x\cdot 3^y$,

Now How can i solve after that

Help Required.

Thanks.

This has more than one solution and we can approach as

First find power of 2

Higest from LCM of (b,c) = 3 so so b cannot have power > 3 and c cannot power > 3

So power or 2 in a = 4

Now in b and c can be (3,0), (3,1), (3,2), (3.3), (0,3),(1,3),(2,3) as these combinations given power 3

Similarly you can find the power of 3

In a = 3 an in bc = (2,0),(2,1)(2,2),(1,2),(0,2)

So a = 432 and for bc there are 35 sets

for example one is (2^3*3^2,1)

 

[I like Serena]

If $\bf{L.C.M}$ of $(a,b)$ is $432,$ and $\bf{L.C.M}$ of $(b,c)$ is $72,$ and $\bf{L.C.M}$ of $(c,a)$ is $432.$

Then the number of ordered pairs $(a,b,c)$ is

My Trail Solution:: First we will factorise in prime factor form.

$\bf{L.C.M}$ of $(a,b)$ is $ = 432 = 2^4 \times 3^3$

Similarly $\bf{L.C.M}$ of $(b,c)$ is $ = 72 = 2^3 \times 3^2$

Similarly $\bf{L.C.M}$ of $(c,a)$ is $ = 432 = 2^4 \times 3^3$

Now Let we assume that $a = 2^l\cdot 3^m$

and Similarly $b = 2^p\cdot 3^q$

and Similarly $c = 2^x\cdot 3^y$,

Now How can i solve after that

Help Required.

Thanks.

Enumerate.
Smartly.

What are the possible combinations of powers of 2 and 3 in b respectively c due to $\text{LCM}(b,c)=2^3 \times 3^2$?
In each case, what are the possible powers of 2 and 3 in a?