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Re: least common multiple -12^12 [#permalink]
12 Nov 2009, 04:12

5

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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

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

\(8^8 = 2^{24}\)

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.

Re: least common multiple -12^12 [#permalink]
12 Nov 2009, 04:15

Quote:

\(8^8 = 2^24\)

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. _________________

Re: least common multiple -12^12 [#permalink]
12 Nov 2009, 04:42

Expert's post

sriharimurthy wrote:

Quote:

\(8^8 = 2^24\)

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. _________________

Re: least common multiple -12^12 [#permalink]
12 Nov 2009, 04:53

Bunuel wrote:

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

Re: least common multiple -12^12 [#permalink]
12 Nov 2009, 08:43

1

This post was BOOKMARKED

sriharimurthy wrote:

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

\(8^8 = 2^{24}\)

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.

Re: least common multiple -12^12 [#permalink]
29 Mar 2012, 13:54

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?

Re: least common multiple -12^12 [#permalink]
30 Mar 2012, 00:29

1

This post received KUDOS

Expert's post

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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!

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: \(2^0*3^{12}=3^{12}\); \(2^1*3^{12}\); \(2^2*3^{12}\); ... \(2^{24}*3^{12}=12^{12}=LCM\).

Re: least common multiple -12^12 [#permalink]
30 Mar 2012, 03:43

4

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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

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]
23 Aug 2014, 09:21

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]
15 Sep 2015, 22:54

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