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Difficulty:
35%
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Question Stats:
68% (01:32) correct
32%
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wrong
based on 2585
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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?
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.
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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.
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General Discussion
enigma123
I think you misinterpreted the question: the correct answer must divide any n, which satisfy the condition that n^2 is divisible by 96. So, we should find the least value of n for which n^2 is a multiple of 96.
Follow the links in my previous post for similar questions to understand the concept better.
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Bunuel
Can you pls explain the last argument of your answer i.e.
hence \(n\) in any case is divisible by 24 (24 is the largest positive integer that MUST divide n evenly).
docabuzar
The question asks about the largest positive integer that must divide n, since the lowest value of n is 24, then the largest positive integer that must divide n is 24 (so even for the lowest value of n it's divisible by 24).
Hope it's clear.
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cavana
Joined: 20 Jan 2017
Last visit: 20 Jun 2018
Posts: 33
Location: United States (NY)
Schools: CBS '20 (A)
GMAT 1: 610 Q34 V41
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Products:
1) Since n is squared, it should have two identical sets of prime factors
2) Since n^2 is divisible by 96, all prime factors of 96 should be the prime numbers of n^2.
3) The prime factors of 96 are 2*2*2*2*2*3. A squared number that includes all these prime factors should have an additional factor 2 and an additional factor 3.
4) n^2 has to be at least 2^6*3^2, and n has to be 2^3*3=24
5) If n has to be at least 24, then the largest integer it MUST be divisible by is 24.
Posted from my mobile device
2) Since n^2 is divisible by 96, all prime factors of 96 should be the prime numbers of n^2.
3) The prime factors of 96 are 2*2*2*2*2*3. A squared number that includes all these prime factors should have an additional factor 2 and an additional factor 3.
4) n^2 has to be at least 2^6*3^2, and n has to be 2^3*3=24
5) If n has to be at least 24, then the largest integer it MUST be divisible by is 24.
Posted from my mobile device
Bunuel
Bunuel hello, i have two questions
how from this \(n=\sqrt{2^5*3*k}\)= how you got this i mean \(2^2\) outside of radical sign wheras only one 2 is left under radical sign, but we have five 2s
2^2*\(\sqrt{2*3*k}\)here- \(n=2^2*\sqrt{3*2*3*2}=24\) --- but 3*2*3*2 = 36
how can 3*2*3*2 be equal 24
dave13
1. \(n=\sqrt{2^5*3*k}=\sqrt{(2^4*2)*3*k}=\sqrt{((2^2)^2*2)*3*k}=2^2*\sqrt{2*3*k}\).
2. \(n=2^2*\sqrt{3*2*3*2}=4*\sqrt{36}=4*6=24\).
Hope it helps.
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cavana
Dear VeritasKarishma,
Would you please explain point 5 - I've followed everything from point 1 to point 4, but it lost me at point 5, which says "if n has to be at least 24, then the largest integer it must be divisible by is 24". The thing is, "n has to be at least 24" doesn't mean that n has to be 24 (nor does it say anywhere in the prompt." If n equaled 48, it still follows everything that the prompt asks. Let's try plugging n - 48 into the question:
"Say n is a positive integer 48, and n^2 is divisible by 96 (to be exact, n^2 = 96 * 24), then the largest positive integer that must divide 48 is 48" => This statement is perfectly fine and there's nothing wrong with n being 48. Does it say anywhere that n has to be 24, or does it only say n has to be at least 24?
I'm very confused by this question and I've looked through Bunuel's similar questions and I understand the process, I just didn't understand why it has to be the least value - what part in the sentence indicates that n has to be least value? Thank you very much. I appreciate your help.
shabuzen102
n^2 has 2^5*3 as a factor. Since it is a square, it MUST have 2^6 * 3^2 as factors and can have other factors too such as n^2 can be 2^6 * 3^2 *5^2 or 2^6 * 3^4 or 2^6 * 3^2 * 11^2 etc.
The smallest value of n is then 2^3 * 3 = 24. Of course, if n^2 can take many different values, n can take as many different corresponding values such as 2^3 * 3* 5 or 2^3 * 3^2 or 2^3 * 3* 11 and so on...
What will certainly be a factor of n? 2^3 * 3 (= 24). We don't know about other additional factors.
n can take various values including 24. e.g. 24, 48, 72, 96 ... and so on
Which number MUST divide each of these values of n? 24.
Can I say that 48 MUST divide each of these values of n? No. Because 48 cannot divide 24.
Can I say that 72 MUST divide each of these values of n? No. Because 72 cannot divide 24 and 48.
and so on...
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Bunuel
KarishmaB
Sould the Question NOT be amended to highlight n as the minimum positive integer..in order to bring clarity in understanding the Question stem..and more importantly, to leaad to an UNIQUE SOLUTION ?
I think it should read..
If n is the MINIMUM 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
lest..24 (in case of n^2 = 576...divisble by 96)
36 (in case of n^2 = 1296..diivisible by 96)
48 (in case of n^2 = 2304..diivisible by 96)..
all of these ie, 24 or 36 or 48 will be possible soltions to this Question..!!
Is my argument valid?
KarishmaB
Sould the Question NOT be amended to highlight n as the minimum positive integer..in order to bring clarity in understanding the Question stem..and more importantly, to leaad to an UNIQUE SOLUTION ?
I think it should read..
If n is the MINIMUM 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
lest..24 (in case of n^2 = 576...divisble by 96)
36 (in case of n^2 = 1296..diivisible by 96)
48 (in case of n^2 = 2304..diivisible by 96)..
all of these ie, 24 or 36 or 48 will be possible soltions to this Question..!!
Is my argument valid?
PRamesh2008
When we use "must divide n" in the question stem, it is clear that for every value of n, the positive integer should divide it. So n could take its minimum value or any greater value, the positive integer must divide it.
When we get that n = 24 * Some positive integer, we know that 24 "must divide n." Can we say the same for any other positive integer greater than 24? No.
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KarishmaB
Hi KarishmaB,
I have a question, why the question asking about the largest number which can divive n, but your answer is about the at leat number?
Yuki2202
I am finding the least value of n i.e. the actual value of n will be either the value that I find or a positive multiple of it.
So I found that n is at least 24. So n could be 24, 48, 72, 96 etc.
What is the greatest value by which n must be divisible? 24.
n must be divisible by 2, also by 3, also by 4 etc but the greatest value by which it must be divisible is 24, which is the least value of n.
Check factorization here: https://youtu.be/Kd-4cH4cqHw
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KarishmaB
I think I misunderstood the question. I thought that the question is asking about the largest number that is divisible by n.
Thank you so much for your explanation.
I think I misunderstood the question. I thought that the question is asking about the largest number that is divisible by n.
Thank you so much for your explanation.
enigma123
\(n^2 = 96*x = 2^5*3*x\)
But for a number to be perfect square, the exponents of all it's distinct prime factors should be divisible by 2
i.e. \(n^2_{minimum} = 2^5*3*(2*3) = 2^6*3^2\)
i.e. \(n_{minimum} = 2^3*3 = 24\)
Answer: Option C
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