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Joined: 02 Sep 2009
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Question Stats:
47% (01:08) correct 53% (01:47) wrong based on 59 sessions
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Math Expert
Joined: 02 Sep 2009
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Re M2833
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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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Joined: 17 Jun 2014
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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 !



Math Expert
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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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Intern
Joined: 17 Jun 2014
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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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Re M2833
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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



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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



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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










