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# For how many values of k is 12^12 the least common multiple

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For how many values of k is 12^12 the least common multiple  [#permalink]

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Updated on: 30 Jun 2016, 07:51
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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

Originally posted by jade3 on 12 Nov 2009, 03:38.
Last edited by Bunuel on 30 Jun 2016, 07:51, edited 2 times in total.
Edited the question and added the OA
Math Expert
Joined: 02 Sep 2009
Posts: 52296
Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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30 Mar 2012, 00:29
5
6
essarr wrote:
sriharimurthy 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.

Hope it helps.
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Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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12 Nov 2009, 04:12
9
1
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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##### General Discussion
Manager
Joined: 29 Oct 2009
Posts: 196
GMAT 1: 750 Q50 V42
Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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12 Nov 2009, 04:15
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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http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

Math Expert
Joined: 02 Sep 2009
Posts: 52296
Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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12 Nov 2009, 04:42
sriharimurthy wrote:
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.

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.
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Joined: 29 Oct 2009
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Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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12 Nov 2009, 04:53
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 $$thanks^{10}$$ !!
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

Intern
Joined: 22 Jan 2012
Posts: 20
Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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29 Mar 2012, 13:54
sriharimurthy 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!
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8795
Location: Pune, India
Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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30 Mar 2012, 03:43
6
1
essarr wrote:
sriharimurthy 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.
$$a = 2*5$$
$$b = 2*5*7^2$$
$$c = 2^4*5^2$$

What 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,
$$a = 2^6*3^6$$
$$b = 2^{24}$$
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]

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31 Mar 2012, 11:48
ahhhhh I see it now; thanks so much bunuel & karishma, that clarified it
Manager
Joined: 15 Aug 2013
Posts: 247
Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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23 Aug 2014, 09:21
Bunuel wrote:
essarr 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

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.

Hope it helps.

Hi Bunuel,

I can see why K needs to have 3^12, but can't K have other values with the base 2? Meaning, why does the range only go from 2^0 to 2^24, why can't it be 2^-5 etc?
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Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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08 Dec 2015, 05:16
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

There are 3 numbers: 6^6 (in prime factors that is 2^6 * 3^6), 8^8 (that is 2^24) and k. LCM of these three numbers is given as: 12^12 (that is 3^12 * 2^24 )

First we can ignore k and find the LCM of the given two numbers (2^6 * 3^6) and (2^24) That is => 3^6 * 2^24 (Note that LCM of any two -or more- numbers is the product of all distinct prime factors with the greatest powers.)

So if 3^6 * 2^24 (LCM of the given two numbers) and k has a LCM of 3^12 * 2^24 then k must have the factor 3^12 (this is a necessity because other number is limited with 2^6 ) On the other hand -besides 3^12- k can take prime 2 to the power of 0 to 24 (2^0 to 2^24)

Therefore k can be any of the following: (3^12 and 2^0) or (3^12 and 2^1) or (3^12 and 2^2), ....., (3^12 and 2^24) that is 25 in total.

(I think this is a 700 level question)
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Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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14 Mar 2016, 00:31
Director
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For how many values of k is 12^12 the least common multiple  [#permalink]

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Updated on: 01 Jul 2016, 00:38
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

Lets use a quick example
What is the LCM of 2,4,9,12,
Factorise all the numbers one by one and write them in prime numbers raised to exponent form $$(Prime1)^m$$ X $$(Prime2)^m$$...
$$2=2^1$$
$$4=2*2==>2^2$$
$$9=3*3==>3^2$$
$$12=4*3==>2*2*3==>2^2 * 3^1$$

NOW LCM OF THESE NUMBERS WILL TAKE THE HIGHEST POWER OF EACH PRIME FROM EACH NUMBER (ONE TIMES ONLY)
So LCM = $$2^2 * 3^3$$==> 4*9=36
Notice how $$2^1$$ and $$3^1$$ are not contributing towards the LCM at all.

Now apply the same logic to your question
You already know LCM is = $$12^{12}=(4*3)^{12}$$==> $$(2^2*3)^{12}$$ ==> $$2^{24}*3^{12}$$
Similarly $$6^6= (2*3)^6==>2^6*3^6$$ So we know $$6^6$$ is neither contributing 2's or 3's towards the LCM
$$8^8= (2^3)^8==> 2^{24}$$ , So we know 8 is contributing all the $$2^{24}$$towards our LCM

Now we need a $$3^{12}$$ to reach the LCM
Since K is the only remaining digit therefore K must contribute $$3^{12}$$
but it is also possible K can or cannot have $$2^m$$ in it also and the values of $$2^m$$ can vary from $$2^0 to 2^{24}$$

Remember for LCM we take the highest power, so $$2^{24}$$ can be common in $$8^8$$ as well as K

therefore total values of 2 in k = $$2^0 to 2^{24}$$ (Total=25) and one compulsory value of $$3^{12}$$ (total= 1)
Total=26 values

Why am in overshooting by 1?
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Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
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Originally posted by LogicGuru1 on 30 Jun 2016, 10:09.
Last edited by LogicGuru1 on 01 Jul 2016, 00:38, edited 1 time in total.
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Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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30 Jun 2016, 23:39
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

Quote:

(Total=25) and one compulsory value of 3^12

Total=26 values

Here is the problem in your solution.
When you say the possible values vary from 2^0 to 2^24 (that is 25 values) AND another value is 3^12, you are double counting 3^12.
Note that 2^0 = 1. So
2^0*3^12 = 3^12

Hence you have only 25 values.
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For how many values of k is 12^12 the least common multiple  [#permalink]

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01 Jul 2016, 00:50
VeritasPrepKarishma 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

Quote:

(Total=25) and one compulsory value of 3^12

Total=26 values

Here is the problem in your solution.
When you say the possible values vary from 2^0 to 2^24 (that is 25 values) AND another value is 3^12, you are double counting 3^12.
Note that 2^0 = 1. So
2^0*3^12 = 3^12

Hence you have only 25 values.

Thanks Karishma ,
Just to clarify one more doubt,
4= 2*2 = $$2^2$$ ==> The genreal form is $$2^q$$
Total possible factors of 4 = q+1 = 2+1 = 3 {1,2,4}
IS this the same thing that you mentioned :- I am counting 1 in $$3^{12}$$ and also in $$2^0$$ and i need to drop it one time.? RIGHT ??

In all such questions, does one need to ignore "1" in the final count ?
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Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
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Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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29 Nov 2016, 17:30
K can take on any of the following values:

(3^12)
(3^12)(2)
(3^12)(2^2)
(3^12)(2^3)
(3^12)(2^4)
(3^12)(2^5)
(3^12)(2^6)
(3^12)(2^7)
(3^12)(2^8)
(3^12)(2^9)
(3^12)(2^10)
(3^12)(2^11)
(3^12)(2^12)
(3^12)(2^13)
.
.
.
(3^12)(2^24)

Thus, there are 25 values that K can take on.

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Re: For how many values of k is 12^12 the least common multiple  [#permalink]

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