For how many values of k is 12^12 the least common multiple

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.
\(a = 2*5\)
\(b = 2*5*7^2\)
\(c = 2^4*5^2\)

What 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,
\(a = 2^6*3^6\)
\(b = 2^{24}\)
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

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 \(thanks^{10}\) !!

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!

Lets use a quick example
What is the LCM of 2,4,9,12,
Factorise all the numbers one by one and write them in prime numbers raised to exponent form \((Prime1)^m\) X \((Prime2)^m\)...
\(2=2^1\)
\(4=2*2==>2^2\)
\(9=3*3==>3^2\)
\(12=4*3==>2*2*3==>2^2 * 3^1\)

NOW LCM OF THESE NUMBERS WILL TAKE THE HIGHEST POWER OF EACH PRIME FROM EACH NUMBER (ONE TIMES ONLY)
So LCM = \(2^2 * 3^3\)==> 4*9=36
Notice how \(2^1\) and \(3^1\) are not contributing towards the LCM at all.

Now apply the same logic to your question
You already know LCM is = \(12^{12}=(4*3)^{12}\)==> \((2^2*3)^{12}\) ==> \(2^{24}*3^{12}\)
Similarly \(6^6= (2*3)^6==>2^6*3^6\) So we know \(6^6\) is neither contributing 2's or 3's towards the LCM
\(8^8= (2^3)^8==> 2^{24}\) , So we know 8 is contributing all the \(2^{24}\)towards our LCM

Now we need a \(3^{12}\) to reach the LCM
Since K is the only remaining digit therefore K must contribute \(3^{12}\)
but it is also possible K can or cannot have \(2^m\) in it also and the values of \(2^m\) can vary from \(2^0 to 2^{24}\)

Remember for LCM we take the highest power, so \(2^{24}\) can be common in \(8^8\) as well as K

therefore total values of 2 in k = \(2^0 to 2^{24}\) (Total=25) and one compulsory value of \(3^{12}\) (total= 1)
Total=26 values
Answer is D

Why am in overshooting by 1?

Here is the problem in your solution.
When you say the possible values vary from 2^0 to 2^24 (that is 25 values) AND another value is 3^12, you are double counting 3^12.
Note that 2^0 = 1. So
2^0*3^12 = 3^12

Hence you have only 25 values.

Thanks Karishma ,
Just to clarify one more doubt,
4= 2*2 = \(2^2\) ==> The genreal form is \(2^q\)
Total possible factors of 4 = q+1 = 2+1 = 3 {1,2,4}
IS this the same thing that you mentioned :- I am counting 1 in \(3^{12}\) and also in \(2^0\) and i need to drop it one time.? RIGHT ??

In all such questions, does one need to ignore "1" in the final count ?

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