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 500,000 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
6
This post received KUDOS
2
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
2
This post received KUDOS
Expert's post
4
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
5
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
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
For how many values of k is 12^12 the least common multiple [#permalink]
08 Dec 2015, 05:16
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.
(I think this is a 700 level question)
gmatclubot
For how many values of k is 12^12 the least common multiple
[#permalink]
08 Dec 2015, 05:16
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...