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Re: If n is an integer and n3 is divisible by 24, what is the la [#permalink]
SAHITHI97 wrote:
why 6 but not 12?


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Re: If n is an integer and n3 is divisible by 24, what is the la [#permalink]
please explain this once again i didnt understand why 6
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If n is an integer and n3 is divisible by 24, what is the la [#permalink]
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sandy 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

-----ASIDE---------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
-----ONTO THE QUESTION!---------------------

If n³ is divisible by 24, then we can say that n³= 24k for some positive integer k

Since 24 = (2)(2)(2)(3), we can write: n³= (2)(2)(2)(3)k for some positive integer k

So, what does this tell us about n?
Since there's an 8 hiding in the prime factorization of n³, we can conclude that 8 is a divisor of n³, which means 2 must be a divisor of n
It also tells us that 3 must be a divisor of n
These are the only two guaranteed conclusions we can draw about the divisors of n

If 2 and 3 must be divisors of n, then we can be certain that and is divisible by 6

Answer: C
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Re: If n is an integer and n3 is divisible by 24, what is the la [#permalink]
GreenlightTestPrep wrote:
sandy 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

-----ASIDE---------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
-----ONTO THE QUESTION!---------------------

If n³ is divisible by 24, then we can say that n³= 24k for some positive integer k

Since 24 = (2)(2)(2)(3), we can write: n³= (2)(2)(2)(3)k for some positive integer k

So, what does this tell us about n?
Since there's an 8 hiding in the prime factorization of n³, we can conclude that 8 is a divisor of n³, which means 2 must be a divisor of n
It also tells us that 3 must be a divisor of n
These are the only two guaranteed conclusions we can draw about the divisors of n

If 2 and 3 must be divisors of n, then we can be certain that and is divisible by 6

Answer: C



ohhhh i got it now thank you sir
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Re: If n is an integer and n3 is divisible by 24, what is the la [#permalink]
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I have follow the process below-
n^3/24 so we can say that n^3 must have at least 3 two's and 1 three. As n^3= 2^3 so n=2^1 or 2 and again as n^3= 3 or 3^1, thus n will be 3^3. so the n could be 2X3X3X3= 54. The factors of 54 are- 1, 2, 3, 6, 9, 18, 27 etc. So from the answer choices 6 is the largest one. Is it a correct way?
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Re: If n is an integer and n3 is divisible by 24, what is the la [#permalink]
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SAHITHI97 wrote:
why 6 but not 12?



N should contain atleast one 2 and one 3.so N Can values [6,12,18,...].if we take 6 or 12 then both options 6 and 12 are correct but if we take N as 18 only 6 is a factor of N but 12 is not.So 6 is the answer.

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Re: If n is an integer and n3 is divisible by 24, what is the la [#permalink]
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Re: If n is an integer and n3 is divisible by 24, what is the la [#permalink]
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