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Question Stats:
49% (01:49) correct 51% (02:17) wrong based on 135 sessions
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If the least common multiple of a positive integer \(x\), \(4^3\) and \(6^5\) is \(6^6\). Then \(x\) can take how many values? A. \(1\) B. \(6\) C. \(7\) D. \(30\) E. \(36\)
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16 Sep 2014, 01:30
Official Solution:If the least common multiple of a positive integer \(x\), \(4^3\) and \(6^5\) is \(6^6\). Then \(x\) can take how many values? A. \(1\) B. \(6\) C. \(7\) D. \(30\) E. \(36\) We are given that \(6^6=2^{6}*3^{6}\) is the least common multiple of the following three numbers: \(x\); \(4^3=2^6\); \(6^5 = 2^{5}*3^5\); First notice that \(x\) 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 second number or in the third, than \(x\) must have \(3^{6}\) as its multiple (else how \(3^{6}\) would appear in LCM?). Next, \(x\) can have 2 as its prime in ANY power ranging from 0 to 6, inclusive (it cannot have higher power of 2 since LCM limits the power of 2 to 6). Thus, \(x\) could take total of 7 values. Answer: C
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Re: M2833
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01 Oct 2014, 20:46
Hi Bunuel I do not understand why 7 but not 8 ? according to my count, it is supposed to be 3^6, 2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6 ????? I do not get it ! Can you explain it for me ! many thanks !



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langtuprovn2007 wrote: Hi Bunuel I do not understand why 7 but not 8 ? according to my count, it is supposed to be 3^6, 2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6 ????? I do not get it ! Can you explain it for me ! many thanks ! x could be: \(3^6\); \(2*3^6\); \(2^2*3^6\); \(2^3*3^6\); \(2^4*3^6\); \(2^5*3^6\); \(2^6*3^6\). 7 values!
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Woa !, Why did not I think about that ! Tks Bunuel my head was just mixed up when solving this problem !!!



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22 Jul 2016, 08:40
I think this is a highquality question and I agree with explanation.



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Re: M2833
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08 Feb 2017, 22:02
I did this as follows
4^3 = 2^6 6^5 = 2^5 & 3^5
6^6 = 2^6 3^6
So, 2^6 is fully deductible from 6^6 3^5 is fully deductible from 6^6
So the leftover values are 2^5, 3 , x
So should the answer be 6 or 7?
I chose 6



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Re: M2833
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09 Feb 2017, 01:37
bloomington wrote: I did this as follows
4^3 = 2^6 6^5 = 2^5 & 3^5
6^6 = 2^6 3^6
So, 2^6 is fully deductible from 6^6 3^5 is fully deductible from 6^6
So the leftover values are 2^5, 3 , x
So should the answer be 6 or 7?
I chose 6 x could be: \(3^6\); \(2*3^6\); \(2^2*3^6\); \(2^3*3^6\); \(2^4*3^6\); \(2^5*3^6\); \(2^6*3^6\). 7 values!
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Re: M2833
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07 Jan 2018, 15:27
to be the least common multipler doesnt mean that 6^6 is the greatest number? why can x be 2^2*6^6?
thank you



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Re: M2833
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07 Jan 2018, 20:26
arthurhernandes wrote: to be the least common multipler doesnt mean that 6^6 is the greatest number? why can x be 2^2*6^6?
thank you x cannot be 2^2*6^6. Where does it say that it can? x could be: \(3^6\); \(2*3^6\); \(2^2*3^6\); \(2^3*3^6\); \(2^4*3^6\); \(2^5*3^6\); \(2^6*3^6\). 7 values!
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Hi Bunuel , I don't understand even the beginning of the solution. We are given \(6^{6}\) = \(4^{4}\) and \(6^{5}\). I understand we can change \(4^{4}\) to \(2^{6}\), but how did you change \(6^{5}\) into \(3^{6}\)?
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Re: M2833
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15 Apr 2018, 09:39
msurls wrote: Hi Bunuel , I don't understand even the beginning of the solution. We are given \(6^{6}\) = \(4^{4}\) and \(6^{5}\). I understand we can change \(4^{4}\) to \(2^{6}\), but how did you change \(6^{5}\) into \(3^{6}\)? Nothing of that kin in the question. \(4^4 = (2^2)^4=2^8\), not 2^6, which is 4^3. Please reread the solution carefully and read the discussion. Hope it helps.
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I don't see any reason why this question must be marked as hard(95%)! Maybe medium (at most 65%) would be a better indicator :/



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Re: M2833
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23 Nov 2018, 22:00
I have doubt about the solution because the values could be : 2^0*3^6(in this case the HCF of all 3 numbers is 1) 2^1*3^6 2^2*3^6 2^3*3^6 2^4*3^6 2^5*3^6 2^6*3^6
2^0*3^5, 2^1*3^5, ...., 2^6*3^5
2^0*3^4, 2^1*3^4, ..... ,2^6*3^4
2^0*3^3, 2^1*3^3, ..... ,2^6*3^3
2^0*3^2, 2^1*3^2, ..... ,2^6*3^2
2^0*3^1, 2^1*3^1, ..... ,2^6*3^1
2^0*3^0(in this case the HCF of all 3 numbers is 1), 2^1*3^0, ..... ,2^6*3^0
Hence the values are 7*7 = 49



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Re: M2833
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24 Nov 2018, 01:24
Godsent wrote: I have doubt about the solution because the values could be : 2^0*3^6(in this case the HCF of all 3 numbers is 1) 2^1*3^6 2^2*3^6 2^3*3^6 2^4*3^6 2^5*3^6 2^6*3^6
2^0*3^5, 2^1*3^5, ...., 2^6*3^5
2^0*3^4, 2^1*3^4, ..... ,2^6*3^4
2^0*3^3, 2^1*3^3, ..... ,2^6*3^3
2^0*3^2, 2^1*3^2, ..... ,2^6*3^2
2^0*3^1, 2^1*3^1, ..... ,2^6*3^1
2^0*3^0(in this case the HCF of all 3 numbers is 1), 2^1*3^0, ..... ,2^6*3^0
Hence the values are 7*7 = 49 x could be: \(3^6\); \(2*3^6\); \(2^2*3^6\); \(2^3*3^6\); \(2^4*3^6\); \(2^5*3^6\); \(2^6*3^6\). 7 values!
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Re: M2833
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25 Dec 2018, 07:19
Dears ScottTargetTestPrep VeritasKarishmaWhy we couldn't take 1 as a value for X?



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Re: M2833
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26 Dec 2018, 13:07
Hi HisHo, Because if x = 1, then the lcm of 1, 4^3 and 6^5 will not be 6^6. Posted from my mobile device
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Re M2833
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20 Feb 2019, 19:01
I think this is a poorquality question and I agree with explanation. I agree with the explanation. But I didn't understand the question in the test. And I took me several minutes to understand the question even having read the explanation before. I think that the question should be rewritten.
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20 Feb 2019, 22:07
patto wrote: I think this is a poorquality question and I agree with explanation. I agree with the explanation. But I didn't understand the question in the test. And I took me several minutes to understand the question even having read the explanation before. I think that the question should be rewritten. The fact that you did not understood the question does not mean that it's a poor quality question. The question is grammatically and mathematically correct.
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If the least common multiple of a positive integer \(x\), \(4^3\) and \(6^5\) is \(6^6\). Then \(x\) can take how many values? We determine the LCM by counting the highest number of powers of a given prime that is repeated across the set * multiplied by nonrepeated powers If the LCM is \(6^6\) then only the primes 2 and 3 are present and the highest power of 3 must be contained within \(x\) itself since: \(4^3 = 2^6\) does not contain \(3^6\) and \(6^5 = (3^5 * 2^5)\) does not contain \(3^6\) Thus x must contain 3^6 1 solution is \(3^6\) (as determined already)..but what else can x contain? X can contain contain primes found within the LCM, so it can contain multiples of 2 up to the upper limit, which is given by the LCM of \(6^6 (2^6)\) Solution 2 = \(3^6 * 2^1\) Solution 3 = \(3^6 * 2^2\) Solution 4 = \(3^6 * 2^3\) Solution 5 = \(3^6 * 2^4\) Solution 6 = \(3^6 * 2^5\) Solution 7 = \(3^6 * 2^6\)
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