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For how many values of k is 12^12 the least common multiple [#permalink]
12 Nov 2009, 04:38
Question Stats:
28% (01:02) correct
71% (08:45) wrong based on 7 sessions
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
Last edited by Bunuel on 30 Mar 2012, 01:30, edited 1 time in total.
Edited the question and added the OA
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Re: least common multiple -12^12 [#permalink]
12 Nov 2009, 05:12
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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. Thus K can have 25 values. Choice (c). Cheers.
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Re: least common multiple -12^12 [#permalink]
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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Re: least common multiple -12^12 [#permalink]
12 Nov 2009, 05:42
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Re: least common multiple -12^12 [#permalink]
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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Re: least common multiple -12^12 [#permalink]
12 Nov 2009, 09:43
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.
Thus K can have 25 values. Choice (c).
Cheers.
You are spot on The answer is indeed C
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Re: least common multiple -12^12 [#permalink]
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: least common multiple -12^12 [#permalink]
30 Mar 2012, 01:29
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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. So, k can take total of 25 values. Answer: C. Hope it helps.
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Re: least common multiple -12^12 [#permalink]
30 Mar 2012, 04:43
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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! 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*5b = 2*5*7^2c = 2^4*5^2What 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*3^6b = 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 [#permalink]
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
[#permalink]
31 Mar 2012, 12:48
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