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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]  12 Nov 2009, 03:38
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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, 00:30, 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, 04:12
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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

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

$$8^8 = 2^{24}$$

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, 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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compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : word-problems-made-easy-87346.html

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Re: least common multiple -12^12 [#permalink]  12 Nov 2009, 04:42
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: least common multiple -12^12 [#permalink]  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 $$thanks^{10}$$ !!
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compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : word-problems-made-easy-87346.html

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Re: least common multiple -12^12 [#permalink]  12 Nov 2009, 08:43
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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
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Re: least common multiple -12^12 [#permalink]  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!
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Re: least common multiple -12^12 [#permalink]  30 Mar 2012, 00:29
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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: least common multiple -12^12 [#permalink]  30 Mar 2012, 03:43
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Expert's post
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]  31 Mar 2012, 11:48
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]  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?
Re: For how many values of k is 12^12 the least common multiple   [#permalink] 23 Aug 2014, 09:21
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