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]

12 Nov 2009, 04:38

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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

Last edited by Bunuel on 30 Mar 2012, 01:30, edited 1 time in total.

Edited the question and added the OA

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Re: least common multiple -12^12 [#permalink]

12 Nov 2009, 05:12

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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: least common multiple -12^12 [#permalink]

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: least common multiple -12^12 [#permalink]

12 Nov 2009, 05:42

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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Re: least common multiple -12^12 [#permalink]

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: least common multiple -12^12 [#permalink]

12 Nov 2009, 09:43

sriharimurthy wrote:

You are spot on
The answer is indeed C

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Re: least common multiple -12^12 [#permalink]

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: least common multiple -12^12 [#permalink]

30 Mar 2012, 01:29

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essarr wrote:

sriharimurthy wrote:

jade3 wrote:

Thus, k will also be in the form of : (2^a)*(3^b)

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: least common multiple -12^12 [#permalink]

30 Mar 2012, 04:43

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essarr wrote:

sriharimurthy wrote:

jade3 wrote:

Thus, k will also be in the form of : (2^a)*(3^b)

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]

31 Mar 2012, 12:48

ahhhhh I see it now; thanks so much bunuel & karishma, that clarified it

Re: For how many values of k is 12^12 the least common multiple [#permalink] 31 Mar 2012, 12:48

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