Bunuel
Nope, you didn't mess it up.. Only made it better! Thanks Brunel!
Infact \(thanks^{10}\) !!
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| Author: | jade3 [ 12 Nov 2009, 04:38 ] |
| Post subject: | For how many values of k is 12^12 the least common multiple |
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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| Author: | Bunuel [ 30 Mar 2012, 01:29 ] |
| Post subject: | Re: For how many values of k is 12^12 the least common multiple |
essarr 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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| Author: | sriharimurthy [ 12 Nov 2009, 05:12 ] |
| Post subject: | Re: For how many values of k is 12^12 the least common multiple |
jade3 \(6^6 = (2^6)*(3^6)\) \(8^8 = 2^{24}\) 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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| Author: | sriharimurthy [ 12 Nov 2009, 05:15 ] |
| Post subject: | Re: For how many values of k is 12^12 the least common multiple |
Quote: 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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| Author: | Bunuel [ 12 Nov 2009, 05:42 ] |
| Post subject: | Re: For how many values of k is 12^12 the least common multiple |
sriharimurthy 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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| Author: | sriharimurthy [ 12 Nov 2009, 05:53 ] |
| Post subject: | Re: For how many values of k is 12^12 the least common multiple |
Bunuel Nope, you didn't mess it up.. Only made it better! Thanks Brunel! Infact \(thanks^{10}\) !!
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| Author: | essarr [ 29 Mar 2012, 14:54 ] |
| Post subject: | Re: For how many values of k is 12^12 the least common multiple |
sriharimurthy 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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| Author: | KarishmaB [ 30 Mar 2012, 04:43 ] |
| Post subject: | Re: For how many values of k is 12^12 the least common multiple |
essarr 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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| Author: | LogicGuru1 [ 30 Jun 2016, 11:09 ] |
| Post subject: | For how many values of k is 12^12 the least common multiple |
jade3 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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| Author: | KarishmaB [ 01 Jul 2016, 00:39 ] |
| Post subject: | Re: For how many values of k is 12^12 the least common multiple |
jade3 Quote: 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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| Author: | LogicGuru1 [ 01 Jul 2016, 01:50 ] |
| Post subject: | For how many values of k is 12^12 the least common multiple |
VeritasPrepKarishma 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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| Author: | bumpbot [ 11 Mar 2025, 01:06 ] |
| Post subject: | Re: For how many values of k is 12^12 the least common multiple |
Automated notice from GMAT Club BumpBot: A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions. This post was generated automatically. |
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