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For how many values of k is 12^12 the least common multiple
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Updated on: 30 Jun 2016, 08:51
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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
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Originally posted by jade3 on 12 Nov 2009, 04:38.
Last edited by Bunuel on 30 Jun 2016, 08:51, edited 2 times in total.
Edited the question and added the OA




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Re: For how many values of k is 12^12 the least common multiple
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30 Mar 2012, 01:29
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: \(2^0*3^{12}=3^{12}\); \(2^1*3^{12}\); \(2^2*3^{12}\); ... \(2^{24}*3^{12}=12^{12}=LCM\). So, \(k\) can take total of 25 values. Answer: C. Hope it helps.
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Re: For how many values of k is 12^12 the least common multiple
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12 Nov 2009, 05:12
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. Thus K can have 25 values. Choice (c). Cheers.
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Re: For how many values of k is 12^12 the least common multiple
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12 Nov 2009, 05:15
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.
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Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/wordproblemsmadeeasy87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/workwordproblemsmadeeasy87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distancespeedtimewordproblemsmadeeasy87481.html



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Re: For how many values of k is 12^12 the least common multiple
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12 Nov 2009, 05:42
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.
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Re: For how many values of k is 12^12 the least common multiple
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12 Nov 2009, 05:53
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 \(thanks^{10}\) !!
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Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilationoftipsandtrickstodealwithremainders86714.html#p651942
Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/wordproblemsmadeeasy87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/workwordproblemsmadeeasy87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distancespeedtimewordproblemsmadeeasy87481.html



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Re: For how many values of k is 12^12 the least common multiple
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29 Mar 2012, 14:54
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? any help is appreciated, Thanks!



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Re: For how many values of k is 12^12 the least common multiple
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30 Mar 2012, 04:43
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. So k can take 25 values only
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Re: For how many values of k is 12^12 the least common multiple
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31 Mar 2012, 12:48
ahhhhh I see it now; thanks so much bunuel & karishma, that clarified it



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



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Re: For how many values of k is 12^12 the least common multiple
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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. (I think this is a 700 level question)



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Re: For how many values of k is 12^12 the least common multiple
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14 Mar 2016, 01:31
Great Question Here values of k are => 3^12 * 2^p for p=> [0,24] so 25 values hence C
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For how many values of k is 12^12 the least common multiple
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Updated on: 01 Jul 2016, 01:38
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 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?
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Originally posted by LogicGuru1 on 30 Jun 2016, 11:09.
Last edited by LogicGuru1 on 01 Jul 2016, 01:38, edited 1 time in total.



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Re: For how many values of k is 12^12 the least common multiple
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01 Jul 2016, 00:39
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 Quote: (Total=25) and one compulsory value of 3^12
Total=26 values Answer is D
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.
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For how many values of k is 12^12 the least common multiple
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01 Jul 2016, 01:50
VeritasPrepKarishma wrote: 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 Quote: (Total=25) and one compulsory value of 3^12
Total=26 values Answer is D
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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Re: For how many values of k is 12^12 the least common multiple
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29 Nov 2016, 18:30
K can take on any of the following values:
(3^12) (3^12)(2) (3^12)(2^2) (3^12)(2^3) (3^12)(2^4) (3^12)(2^5) (3^12)(2^6) (3^12)(2^7) (3^12)(2^8) (3^12)(2^9) (3^12)(2^10) (3^12)(2^11) (3^12)(2^12) (3^12)(2^13) . . . (3^12)(2^24)
Thus, there are 25 values that K can take on.
C.



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