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Difficulty:
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Question Stats:
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50%
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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
A. 23
B. 24
C. 25
D. 26
E. 27
30 Mar 2012, 01:29
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essarr
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.
12 Nov 2009, 05:12
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jade3
\(6^6 = (2^6)*(3^6)\)
\(8^8 = 2^{24}\)
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.
