Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

4

This post received KUDOS

1

This post was BOOKMARKED

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

2

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!

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

3

This post received KUDOS

Expert's post

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?

gmatclubot

Re: For how many values of k is 12^12 the least common multiple
[#permalink]
23 Aug 2014, 09:21

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...

My swiss visa stamping is done and my passport was couriered through Blue dart. It took 5 days to get my passport stamped and couriered to my address. In...