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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.

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. _________________

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. _________________

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 \(thanks^{10}\) !! _________________

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 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?

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!

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

We are given that \(12^{12}=2^{24}*3^{12}\) is the least common multiple of the following three numbers:

\(6^6=2^6*3^6\); \(8^8 = 2^{24}\); and \(k\);

First notice that \(k\) cannot have any other primes other than 2 or/and 3, because LCM contains only these primes.

Now, since the power of 3 in LCM is higher than the powers of 3 in either the first number or in the second, than \(k\) must have \(3^{12}\) as its multiple (else how \(3^{12}\) would appear in LCM?).

Next, \(k\) can have 2 as its prime in ANY power ranging from 0 to 24, inclusive (it cannot have higher power of 2 since LCM limits the power of 2 to 24).

For example \(k\) can be: \(2^0*3^{12}=3^{12}\); \(2^1*3^{12}\); \(2^2*3^{12}\); ... \(2^{24}*3^{12}=12^{12}=LCM\).

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

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.

Re: For how many values of k is 12^12 the least common multiple [#permalink]

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23 Aug 2014, 10:21

Bunuel wrote:

essarr wrote:

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

We are given that \(12^{12}=2^{24}*3^{12}\) is the least common multiple of the following three numbers:

\(6^6=2^6*3^6\); \(8^8 = 2^{24}\); and \(k\);

First notice that \(k\) cannot have any other primes other than 2 or/and 3, because LCM contains only these primes.

Now, since the power of 3 in LCM is higher than the powers of 3 in either the first number or in the second, than \(k\) must have \(3^{12}\) as its multiple (else how \(3^{12}\) would appear in LCM?).

Next, \(k\) can have 2 as its prime in ANY power ranging from 0 to 24, inclusive (it cannot have higher power of 2 since LCM limits the power of 2 to 24).

For example \(k\) can be: \(2^0*3^{12}=3^{12}\); \(2^1*3^{12}\); \(2^2*3^{12}\); ... \(2^{24}*3^{12}=12^{12}=LCM\).

So, \(k\) can take total of 25 values.

Answer: C.

Hope it helps.

Hi Bunuel,

I can see why K needs to have 3^12, but can't K have other values with the base 2? Meaning, why does the range only go from 2^0 to 2^24, why can't it be 2^-5 etc?

Re: For how many values of k is 12^12 the least common multiple [#permalink]

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15 Sep 2015, 23:54

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For how many values of k is 12^12 the least common multiple [#permalink]

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08 Dec 2015, 06:16

jade3 wrote:

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

There are 3 numbers: 6^6 (in prime factors that is 2^6 * 3^6), 8^8 (that is 2^24) and k. LCM of these three numbers is given as: 12^12 (that is 3^12 * 2^24 )

First we can ignore k and find the LCM of the given two numbers (2^6 * 3^6) and (2^24) That is => 3^6 * 2^24 (Note that LCM of any two -or more- numbers is the product of all distinct prime factors with the greatest powers.)

So if 3^6 * 2^24 (LCM of the given two numbers) and k has a LCM of 3^12 * 2^24 then k must have the factor 3^12 (this is a necessity because other number is limited with 2^6 ) On the other hand -besides 3^12- k can take prime 2 to the power of 0 to 24 (2^0 to 2^24)

Therefore k can be any of the following: (3^12 and 2^0) or (3^12 and 2^1) or (3^12 and 2^2), ....., (3^12 and 2^24) that is 25 in total.

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