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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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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Jun 2016, 08:51, edited 2 times in total.
Edited the question and added the OA
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Re: For how many values of k is 12^12 the least common multiple [#permalink]

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jade3 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



\(6^6 = (2^6)*(3^6)\)

\(8^8 = 2^{24}\)

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

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New post 12 Nov 2009, 05:15
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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Re: For how many values of k is 12^12 the least common multiple [#permalink]

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New post 12 Nov 2009, 05:42
Expert's post
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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Re: For how many values of k is 12^12 the least common multiple [#permalink]

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New post 12 Nov 2009, 05:53
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 \(thanks^{10}\) !! :wink:
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Re: For how many values of k is 12^12 the least common multiple [#permalink]

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sriharimurthy wrote:

\(6^6 = (2^6)*(3^6)\)

\(8^8 = 2^{24}\)

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.


You are spot on
The answer is indeed C
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Re: For how many values of k is 12^12 the least common multiple [#permalink]

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New post 29 Mar 2012, 14:54
sriharimurthy wrote:
jade3 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!
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Re: For how many values of k is 12^12 the least common multiple [#permalink]

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essarr wrote:
sriharimurthy wrote:
jade3 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.

Answer: C.

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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essarr wrote:
sriharimurthy wrote:
jade3 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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New post 31 Mar 2012, 12:48
ahhhhh I see it now; thanks so much bunuel & karishma, that clarified it
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Re: For how many values of k is 12^12 the least common multiple [#permalink]

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New post 23 Aug 2014, 10:21
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.

Answer: C.

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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New post 08 Dec 2015, 06:16
jade3 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


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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New post 14 Mar 2016, 01:31
Great Question
Here values of k are => 3^12 * 2^p for p=> [0,24]
so 25 values
hence C
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For how many values of k is 12^12 the least common multiple [#permalink]

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New post 30 Jun 2016, 11:09
jade3 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


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
Answer is D

Why am in overshooting by 1?
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Last edited by LogicGuru1 on 01 Jul 2016, 01:38, 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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New post 01 Jul 2016, 00:39
Expert's post
jade3 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
Answer is D


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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New post 01 Jul 2016, 01:50
VeritasPrepKarishma wrote:
jade3 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
Answer is D


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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For how many values of k is 12^12 the least common multiple   [#permalink] 01 Jul 2016, 01:50
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