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

Question

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

Bunuel · Accepted Answer

CrackverbalGMAT · Answer

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!

CrackVerbal Academics Team

Bunuel · Answer

n^3 is divisible by 24 implies  n^3 = 24k = 2^3*3k . Therefore, 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.

Alternatively, given  n^3 = 24k = 2^3*3k , we get  n=2*\sqrt {3k} . The smallest k making n an integer is 3^2, yielding n=2*3=6. Hence, the largest factor of n must be 6.

Answer: C.

ScottTargetTestPrep · Answer

Since a perfect cube has unique prime factors in multiples of 3 and since 24 = 2^3 x 3^1, we see that the smallest value of n^3 is 2^3 x 3^3, and thus the smallest value of n is 6. Since the largest factor of 6 is 6, the largest number that must be a factor of n is 6.

Answer: C

KarishmaB · Answer

Since n is an integer, each prime factor in  n^3  will have an exponent which is a multiple of 3.
For example, if  n = 2*3, n^3 = 2^3 * 3^3 . 
If  n = 3^2*5 , then  n^3 = 3^6*5^3  etc.

Given,  n^3  is divisible by 24 so  n^3  is a multiple of 24. 
 n^3 = 24k = 2^3 * 3 * k  
This means, n MUST have 2 and 3 as factors. It could have other primes too as factors but we know that it certainly DOES have a 2 and a 3.

Hence the numbers that MUST be a factor of n are 1, 2, 3 and 6. The largest number that MUST be a factor of n is 6.

Answer (C)

As an aside, note that  k  MUST be at least  3^2 . It could have other factors too with exponents as multiples of 3 (e.g.  k  could be  3^2  or  3^2*5^3  or  3^2*7^6*5^3  etc)

houston1980 · Answer

Can someone enlighten me? Is it not possible for n= 0? 0 is also an integer.

Bunuel · Answer

Yes, n can indeed be 0, and in such a case, it would be divisible by any integer. However, the question is asking for the largest number that MUST be a factor of n. While any number COULD be a factor of n, 6 is the largest number that MUST always be a factor of n, meaning 6 will be a factor of n for all possible values of n.

Bhuvan1212 · Answer

I don't understand this. Since there is a concept that goes, if you need to find the maximum number, we need to minimize the rest of the set. I under stand that we are finding the largest factor of n but what are we minimizing here? n^3? As in lowest term of n^3 which can be divided by 24?

Please help me understand this in simple terms.

Please help me understand this in simple terms.

Krunaal · Answer

Firstly, above highlighted is more of a logic applicable to certain type of problems depending if that operation is apt / valid for the question asked. In this question I don't see any reason to apply it.

n^3  being divisibly by  24  i.e.  2^3 * 3  implies that  n  has a  2  and a  3  =>  n = 2*3*k , where k is an integer greater than zero.

Since, the question asks the largest number that  must  be a factor of  n , and we don't know what  k  can be, we can definitely say that  6 has to be the largest factor  since  n = 6*k

benbitz · Answer

why does this mean that 6 is the largest number that HAS to factor into n? in theory, if both are divisible by 24, why can't 8 factor into both?

Bunuel · Answer

The question asks which of the following MUST divide n, not which COULD divide n. That's why we check for the SMALLEST possible value of n: if something divides the smallest possible n, then it will for sure divide all other valid values of n. However, the reverse is not true, if something divides larger possible values of n, it does not necessarily divide the smallest one.

The smallest possible value of n is 6, so the largest number that divides it is 6. Therefore, while ANY number could divide different valid n, 6 is the one that MUST divide any valid n.

Please also check similar questions given above for better understanding.

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

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