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If x is a positive integers and x^3 is divisible by 48 then which of t

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GMATinsight
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If x is a positive integers and x^3 is divisible by 48 then which of t [#permalink] New post   11 Jul 2017, 02:00
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If x is a positive integers and x^3 is divisible by 48 then which of the following is the least number of factors that x may have?

A) 1
B) 3
C) 6
D) 12
E) greater than 12

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C
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shashankism
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Re: If x is a positive integers and x^3 is divisible by 48 then which of t [#permalink] New post   12 Jul 2017, 07:01
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GMATinsight wrote:
If x is a positive integers and x^3 is divisible by 48 then which of the following is the least number of factors that x may have?

A) 1
B) 3
C) 6
D) 12
E) greater than 12

https://www.GMATinsight.com


x^3 is divisible by 48
48 = 2*2*2*2*3 = 2^4 * 3

So lets start pairing the factor in group of 3. (2*2*2)*2*3
So x must be having one factor 2 for (2*2*2) , one 2 for 2 and one 3 for 3.
So minimum value of x = 2^2 * 3^1

Least number of factors of x = (2+1)*(1+1) = 6

Answer C
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If x is a positive integers and x^3 is divisible by 48 then which of t [#permalink] New post   12 Jul 2017, 19:17
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GMATinsight wrote:
If x is a positive integers and x^3 is divisible by 48 then which of the following is the least number of factors that x may have?

A) 1
B) 3
C) 6
D) 12
E) greater than 12

https://www.GMATinsight.com


x^3 is divisible by 48

i.e. \(x^3 = 2^4 * 3 * a\)

but x^3 must be a perfect cube as well

Perfect Cube: A number with all its prime factors having their powers multiples of 3

i.e. Smallest possible value of \(x^3 = 2^4 * 3 * (2^2 * 3^2)\)

i.e. Smallest possible value of \(x^3 = 2^6 * 3^3\)

i.e. Smallest possible value of \(x = 2^2 * 3^1\)

i.e. Smallest number of factors \(x = (2+1)*(1+1) = 3*2 = 6\)

Answer: option C
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Re: If x is a positive integers and x^3 is divisible by 48 then which of t [#permalink] New post   11 Jul 2017, 02:59
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Ans :C
Min value of x^3 = 2^6*3^3
x=12 and factors of x=3*2=6

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Re: If x is a positive integers and x^3 is divisible by 48 then which of t [#permalink] New post   12 Jul 2017, 06:26
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Adityagmatclub wrote:
x^3 = 48p
x^3 = 2^3 * (2*3)P
as x and p are integers, p=(2^2*3^2)q, where q is an integers.

so minimum value of p is 36, no of factors of 36 is 6.

Ans 6 option c



Hi
Can you please explain your solution a little more as I am not able to understand it?

I solved it this way:

x^3= 2^4 * 3^1 *k

k is some positive integer that when multiplied with 48 gives x^3.

Then I took a cube root on both sides.
x= 2*sq.root 2 *cube root 3* cube root k

But I didn't get the answer..
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Re: If x is a positive integers and x^3 is divisible by 48 then which of t [#permalink] New post   11 Jul 2017, 02:20
x^3 = 48p
x^3 = 2^3 * (2*3)P
as x and p are integers, p=(2^2*3^2)q, where q is an integers.

so minimum value of p is 36, no of factors of 36 is 6.

Ans 6 option c
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Re: If x is a positive integers and x^3 is divisible by 48 then which of t [#permalink] New post   12 Jul 2017, 08:19
shashankism wrote:
GMATinsight wrote:
If x is a positive integers and x^3 is divisible by 48 then which of the following is the least number of factors that x may have?

A) 1
B) 3
C) 6
D) 12
E) greater than 12

https://www.GMATinsight.com


x^3 is divisible by 48
48 = 2*2*2*2*3 = 2^4 * 3

So lets start pairing the factor in group of 3. (2*2*2)*2*3
So x must be having one factor 2 for (2*2*2) , one 2 for 2 and one 3 for 3.
So minimum value of x = 2^2 * 3^1

Least number of factors of x = (2+1)*(1+1) = 6

Answer C



yeah, mine answer was wrong.
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Re: If x is a positive integers and x^3 is divisible by 48 then which of t [#permalink] New post   12 Jul 2017, 17:17
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GMATinsight wrote:
If x is a positive integers and x^3 is divisible by 48 then which of the following is the least number of factors that x may have?

A) 1
B) 3
C) 6
D) 12
E) greater than 12


Since x^3/48 = integer, we know that x^3 is a perfect cube divisible by 48, which equals 2^4 x 3^1. Since x^3 is a perfect cube, x^3 must have unique prime factors that are in quantities of multiples of 3.

Thus, the smallest value of x^3 is 2^6 x 3^3, so x = 2^2 x 3^1.

To determine the number of factors of x, we can add one to each exponent attached to each unique prime and then multiply. Thus, x has (2 + 1)(1 + 1) = (3)(2) = 6 factors.

Answer: C
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Re: If x is a positive integers and x^3 is divisible by 48 then which of t [#permalink] New post   27 Oct 2023, 17:34
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