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Re: least common multiple -12^12 [#permalink]
12 Nov 2009, 04:12

3

This post received KUDOS

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.

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.
_________________

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.
_________________

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} !!
_________________

Re: least common multiple -12^12 [#permalink]
12 Nov 2009, 08:43

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.

Re: least common multiple -12^12 [#permalink]
29 Mar 2012, 13:54

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?

Re: least common multiple -12^12 [#permalink]
30 Mar 2012, 00:29

1

This post received KUDOS

Expert's post

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 kcannot 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.

Re: least common multiple -12^12 [#permalink]
30 Mar 2012, 03:43

2

This post received KUDOS

Expert's post

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.