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If n is an integer and n^3 is divisible by 24, what is the largest num

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If n is an integer and n^3 is divisible by 24, what is the largest num [#permalink]

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New post Updated on: 18 Sep 2014, 01:04
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If n is an integer and n^3 is divisible by 24, what is the largest number that must be a factor of n?

(A) 1
(B) 2
(C) 6
(D) 8
(E) 12

Originally posted by anik89 on 18 Sep 2014, 00:56.
Last edited by Bunuel on 18 Sep 2014, 01:04, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If n is an integer and n^3 is divisible by 24, what is the largest num [#permalink]

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New post 18 Sep 2014, 01:14
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1
anik89 wrote:
If n is an integer and n^3 is divisible by 24, what is the largest number that must be a factor of n?

(A) 1
(B) 2
(C) 6
(D) 8
(E) 12


n^3 is divisible by 24 --> n^3 = 24k = 2^3*3k --> 2 and 3 must be prime factors of n (else how would they appear in n^3? Exponentiation does not produce primes). Could n be 2*3 = 6? Yes: n^3 = 216, which is in fact divisible by 24. So, the least value of n is 6.

Answer: C.

Or: \(n^3 = 24k = 2^3*3k\) --> \(n=2*\sqrt[3]{3k}\). The least value of k for which n comes out to be an integer is 3^2, in this case \(n=2*3=6\). So, the largest number that must be a factor of n is 6.

Answer: C.
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Re: If n is an integer and n^3 is divisible by 24, what is the largest num [#permalink]

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New post 18 Sep 2014, 01:15
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3
Bunuel wrote:
anik89 wrote:
If n is an integer and n^3 is divisible by 24, what is the largest number that must be a factor of n?

(A) 1
(B) 2
(C) 6
(D) 8
(E) 12


n^3 is divisible by 24 --> n^3 = 24k = 2^3*3k --> 2 and 3 must be prime factors of n (else how would they appear in n^3? Exponentiation does not produce primes). Could n be 2*3 = 6? Yes: n^3 = 216, which is in fact divisible by 24. So, the least value of n is 6.

Answer: C.

Or: \(n^3 = 24k = 2^3*3k\) --> \(n=2*\sqrt[3]{3k}\). The least value of k for which n comes out to be an integer is 3^2, in this case \(n=2*3=6\). So, the largest number that must be a factor of n is 6.

Answer: C.


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Hope it helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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If n is an integer and n^3 is divisible by 24, what is the largest num [#permalink]

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New post 31 Dec 2016, 05:30
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Hi Anik89,

For a number to be perfectly divisible by another number, the number in the denominator needs to cancel out completely with whatever number is present in the numerator. For e.g. 8/2 = 4, here 2 gets cancelled out leaving a denominator of 1.

Factorizing 24 we get 2^3 * 3. We need to cancel out the three 2's and one 3 with the numerator, so the numerator needs to have a minimum of three 2's and one 3. The numerator here is n^3, so the minimum possible value of n such that n^3 is perfectly divisible by 24 is n = 2 * 3. Notice that if n = 2 * 3, n^3 gives us 2^3 * 3^3 which cancels out the denominator of 24 (2^3 * 3).

Since the minimum possible value of n is 6, the number that will ALWAYS divide n out of the five answer choice is 6.

The best way to solve questions similar to this is to always find out the minimum value of the variable in the numerator which cancels out the denominator completely.

Hope this helps!

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Re: If n is an integer and n^3 is divisible by 24, what is the largest num [#permalink]

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New post 31 Dec 2016, 06:45
Bunuel wrote:
anik89 wrote:
If n is an integer and n^3 is divisible by 24, what is the largest number that must be a factor of n?

(A) 1
(B) 2
(C) 6
(D) 8
(E) 12


n^3 is divisible by 24 --> n^3 = 24k = 2^3*3k --> 2 and 3 must be prime factors of n (else how would they appear in n^3? Exponentiation does not produce primes). Could n be 2*3 = 6? Yes: n^3 = 216, which is in fact divisible by 24. So, the least value of n is 6.

Answer: C.

Or: \(n^3 = 24k = 2^3*3k\) --> \(n=2*\sqrt[3]{3k}\). The least value of k for which n comes out to be an integer is 3^2, in this case \(n=2*3=6\). So, the largest number that must be a factor of n is 6.

Answer: C.



Bunuel

I see that in your first approach, which is what I used as well, you took the least value for which n^3 is divisible by 24. Rightly so n must be 6. However I also see 12 as a choice. The question asks what is the largest number that must be a factor of n. Now if we let n=12, 12^3 is also divisible by by 24. So greatest factor of 12 is 12. Now 12 is greater than 6. And so why not the answer could be 12? As I am thinking about this more, may be the language of question is causing me trouble. Can you help me please ?

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Re: If n is an integer and n^3 is divisible by 24, what is the largest num [#permalink]

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New post 28 Mar 2017, 09:58
Bunuel

Shouldn't it be the smallest number rather than the largest number because we don't know if 2 and 3 are the ONLY prime factors of n?

I am really confused with the term "largest number".

Thanks
Re: If n is an integer and n^3 is divisible by 24, what is the largest num   [#permalink] 28 Mar 2017, 09:58
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