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# M28-33

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Math Expert
Joined: 02 Sep 2009
Posts: 50009

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16 Sep 2014, 01:30
00:00

Difficulty:

95% (hard)

Question Stats:

48% (01:09) correct 52% (01:48) wrong based on 61 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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Math Expert
Joined: 02 Sep 2009
Posts: 50009

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16 Sep 2014, 01:30
2
1
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.

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Joined: 17 Jun 2014
Posts: 18

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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
Joined: 02 Sep 2009
Posts: 50009

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02 Oct 2014, 02:41
2
2
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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Joined: 17 Jun 2014
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02 Oct 2014, 02:44
Woa !, Why did not I think about that ! Tks Bunuel my head was just mixed up when solving this problem !!!
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Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
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22 Jul 2016, 08:40
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 22 May 2015
Posts: 2
Location: Australia
Schools: IMD Jan'18
GMAT 1: 620 Q44 V22
GPA: 3.11

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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
Math Expert
Joined: 02 Sep 2009
Posts: 50009

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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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Joined: 05 Mar 2015
Posts: 1

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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
Math Expert
Joined: 02 Sep 2009
Posts: 50009

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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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Status: Studying Quant
Joined: 04 Sep 2017
Posts: 103
GPA: 3.6
WE: Sales (Computer Software)

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15 Apr 2018, 09:31
1
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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Math Expert
Joined: 02 Sep 2009
Posts: 50009

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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 re-read the solution carefully and read the discussion.

Hope it helps.
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Re: M28-33 &nbs [#permalink] 15 Apr 2018, 09:39
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# M28-33

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