For how many values of k is 12^12 the least common multiple of the positive integers 6^6, 8^8 and k?

A. 23
B. 24
C. 25
D. 26
E. 27

6^6 = (2^6)*(3^6)

8^8 = 2^{24}

Now we know that the least common multiple of the above two numbers and k is:

12^{12} = (2*2*3)^{12} = (2^{24})*(3^{12})

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

Now, b has to be equal to 12 since in order for (2^{24})*(3^{12}) to be a common multiple, at least one of the numbers must have the terms 2^{24} and 3^{12} as its factors. (not necessarily the same number).

We can see that 8^8 already takes care of the 2^{24} part.
Thus, k has to take care of the 3^{12} part of the LCM.

This means that the value k is (2^a)*(3^{12}) where a can be any value from 0 to 24 (both inclusive) without changing the value of the LCM.

Thus K can have 25 values. Choice (c).

Cheers.

8^8 = 2^(24)

Similarly for the other numbers.

Sorry for that confusion. Wasn't able to get a 2 digit power using the math function. If any one knows how to do it please do let me know.
Cheers.

Edited your post. Please check if I didn't mess it up accidentally. To get two digit power just put the power in {}, eg. 2^{24} and mark with [m] button.

Nope, you didn't mess it up.. Only made it better! Thanks Brunel!
Infact thanks^{10} !!

You are spot on
The answer is indeed C