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M28-33

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M28-33  [#permalink]

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New post 16 Sep 2014, 00:30
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A
B
C
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Question Stats:

50% (01:19) correct 50% (01:29) wrong based on 206 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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Re M28-33  [#permalink]

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New post 16 Sep 2014, 00:30
3
3
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: M28-33  [#permalink]

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New post 01 Oct 2014, 19:46
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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M28-33  [#permalink]

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New post 02 Oct 2014, 01:41
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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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M28-33  [#permalink]

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New post 02 Oct 2014, 01:44
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 M28-33  [#permalink]

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New post 22 Jul 2016, 07:40
I think this is a high-quality question and I agree with explanation.
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Re: M28-33  [#permalink]

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New post 08 Feb 2017, 21:02
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: M28-33  [#permalink]

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New post 09 Feb 2017, 00:37
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: M28-33  [#permalink]

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New post 07 Jan 2018, 14:27
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: M28-33  [#permalink]

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New post 07 Jan 2018, 19:26
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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M28-33  [#permalink]

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New post 15 Apr 2018, 08:31
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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: M28-33  [#permalink]

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New post 15 Apr 2018, 08:39
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 re-read the solution carefully and read the discussion.

Hope it helps.
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M28-33  [#permalink]

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New post 22 Nov 2018, 08:41
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: M28-33  [#permalink]

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New post 23 Nov 2018, 21:00
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: M28-33  [#permalink]

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New post 24 Nov 2018, 00:24
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: M28-33  [#permalink]

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New post 25 Dec 2018, 06:19
Dears ScottTargetTestPrep VeritasKarishma
Why we couldn't take 1 as a value for X?
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Re: M28-33  [#permalink]

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New post 26 Dec 2018, 12:07
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Hi HisHo,

Because if x = 1, then the lcm of 1, 4^3 and 6^5 will not be 6^6.

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Re: M28-33 &nbs [#permalink] 26 Dec 2018, 12:07
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