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10 Feb 2012, 17:42
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
67% (01:33) correct
33%
(01:49)
wrong
based on 2378
sessions
History
Date
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If n is a positive integer and n^2 is divisible by 96, then the largest positive integer that must divide n is
(A) 6
(B) 12
(C) 24
(D) 36
(E) 48
(A) 6
(B) 12
(C) 24
(D) 36
(E) 48
This is how I am trying to solve. The prime factorization of 96 is (\(2^5\))(3), so \(n^2\) must share those
prime factors.
Lets go straight into answer choices
A --> Is out as 6^6 = 36 is not divisible by 96
B --> 12 = \(2^2\) * 3. Not divisible by 96 because 12 has \(2^2\) * 3 as its prime factors. We need \(2^5\) or more.
C --> 24 . Prime factors are \(2^3\) * 3. Therefore \(n^2\) = {\(2^3\) * 3}\(^2\) = \(2^6\) * \(3^2\). OK its true.
D --> 36 . Prime factors are \(2^4\) * \(3^4\). We need \(2^5\) or more.
E --> 96. Prime factors are \(2^5\) * 3. So n ^2 is divisible by 96 and also by 48. So this can also be true??
Any thoughts guys?
prime factors.
Lets go straight into answer choices
A --> Is out as 6^6 = 36 is not divisible by 96
B --> 12 = \(2^2\) * 3. Not divisible by 96 because 12 has \(2^2\) * 3 as its prime factors. We need \(2^5\) or more.
C --> 24 . Prime factors are \(2^3\) * 3. Therefore \(n^2\) = {\(2^3\) * 3}\(^2\) = \(2^6\) * \(3^2\). OK its true.
D --> 36 . Prime factors are \(2^4\) * \(3^4\). We need \(2^5\) or more.
E --> 96. Prime factors are \(2^5\) * 3. So n ^2 is divisible by 96 and also by 48. So this can also be true??
Any thoughts guys?
10 Feb 2012, 17:57
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enigma123
Given: \(n^2=96*k=2^5*3*k\), where \(k\) is some positive integer. \(n=\sqrt{2^5*3*k}=2^2*\sqrt{2*3*k}\) --> the least value of \(k\) for which \(n\) is an integer (hence the least value of \(n\)) is for \(k=2*3\) --> \(n=2^2*\sqrt{3*2*3*2}=24\), hence \(n\) in any case is divisible by 24 (24 is the largest positive integer that MUST divide n evenly).
Answer: C.
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Hope it helps.
20 Feb 2012, 01:50
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enigma123
Let me tell you why your method didn't work.
"The prime factorization of 96 is (\(2^5\))(3), so \(n^2\) must share those
prime factors."
You are right. n^2 must share those prime factors. But remember, it is a square so its prime factors have even powers. So n^2 must have 2^6*3^2. We know n^2 has one 3. It must have another 3 since it is a square. Similarly, we know n^2 has 2^5. It must have another 2 to make 2^6 so that the power of 2 is even.
n^2 = 2^6 * 3^2 (at least)
n = 2^3 * 3 = 24 (atleast)
Therefore, n is definitely divisible by 24 at least.
2^2*\(\sqrt{2*3*k}\)
