essarr wrote:

sriharimurthy wrote:

jade3 wrote:

Thus, k will also be in the form of : (2^a)*(3^b)

Hi, I'm trying to understand this question.. the explanation seems good, but I still can't seem to get a grasp of it..

why can we say that K is also in the form of

(2^a)*(3^b) ??

also, how do we consider the

6^6 term in this explanation?

any help is appreciated,

Thanks!

Here is my explanation:

LCM (Least Common Multiple) of 3 numbers a, b and c would be a multiple of each of these 3 numbers. So for every prime factor in these numbers, LCM would have the highest power available in any number e.g.

a = 2*5b = 2*5*7^2c = 2^4*5^2What is the LCM of these 3 numbers? It is

2^4*5^2*7^2 Every prime factor will be included and the power of every prime factor will be the highest available in any number.

So if,

a = 2^6*3^6b = 2^{24}k = ?

LCM

= 2^{24}*3^{12} What values can k take?

First of all, LCM has

3^{12}. From where did it get

3^{12}? a and b have a maximum

3^6. This means k must have

3^{12}.

Also, LCM has

2^{24} which is available in b. So k needn't have

2^{24}. It can have 2 to any power as long as it is less than or equal to 24.

k can be

2^{0}*3^{12} or

2^{1}*3^{12} or

2^{2}*3^{12} ...

2^{24}*3^{12}The power of 2 in k cannot exceed 24 because then, the LCM would have the higher power.

What about some other prime factor? Can k be

2^{4}*3^{12}*5? No, because then the LCM would have 5 too.

So k can take 25 values only

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