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x is divisible by 144. If ³√x is an integer, then which of the followi

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Bunuel
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x is divisible by 144. If ³√x is an integer, then which of the followi [#permalink] New post   01 Oct 2019, 00:01
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x is divisible by 144. If \(\sqrt[3]{x}\) is an integer, then which of the following is \(\sqrt[3]{x}\) definitely divisible by?

I. 4
II. 8
III. 12

A. I only
B. II only
C. III only
D. I and II only
E. I and III only
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E
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Bunuel
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Re: x is divisible by 144. If ³√x is an integer, then which of the followi [#permalink] New post   14 Mar 2021, 01:55
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Bunuel wrote:
x is divisible by 144. If \(\sqrt[3]{x}\) is an integer, then which of the following is \(\sqrt[3]{x}\) definitely divisible by?

I. 4
II. 8
III. 12

A. I only
B. II only
C. III only
D. I and II only
E. I and III only


\(x=144k=2^4*3^2k\)

\(\sqrt[3]{x}=\sqrt[3]{144k}=\sqrt[3]{2^4*3^2k}=2\sqrt[3]{2*3^2k}=integer\)

As \(2\sqrt[3]{2*3^2k}\) is an integer, then the minimum value of \(k\) is \(2^2*3\) (to complete the powers of 2 and 3 under the cubic root). So, the minimum value of \(\sqrt[3]{x}\) is \(\sqrt[3]{x}=2\sqrt[3]{2*3^2k}=2\sqrt[3]{2*3^2*2^2*3}=12\).

12 is divisible by 4 and 12.

Answer: E.
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Re: x is divisible by 144. If ³√x is an integer, then which of the followi [#permalink] New post   01 Oct 2019, 06:53
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Bunuel wrote:
x is divisible by 144. If \(\sqrt[3]{x}\) is an integer, then which of the following is \(\sqrt[3]{x}\) definitely divisible by?

I. 4
II. 8
III. 12

A. I only
B. II only
C. III only
D. I and II only
E. I and III only


x is divisible by 144
--> x/144 = k, for some integer k
--> x = 144*k = 2^4*3^2*k

If \(\sqrt[3]{x}\) is an integer
--> \(\sqrt[3]{2^4*3^2*k}\) is an integer
--> k has to take values = \(2^2*3^1*a^3\) for any integer a
--> \(\sqrt[3]{2^4*3^2*k}\) = \(\sqrt[3]{2^4*3^2*2^2*3^1*a^3} = 12*a\)
--> It will always be divisible by \(4 & 12\)

IMO Option E
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Re: x is divisible by 144. If ³x is an integer, then which of the followi [#permalink] New post   16 Nov 2022, 04:55
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X must be divisible by 144.
X is also a perfect cube.
Therefore, least value of X is 12*12*12
minimum cube root of X is 12, which is divisible by 4 and 12. Answer E.
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Re: x is divisible by 144. If ³√x is an integer, then which of the followi [#permalink] New post   01 Oct 2019, 09:10
integer value which is divisible by 144 and \(\sqrt[3]{x}\) is an integer ; 12^3 ; \(\sqrt[3]{x}\) ; 12
which is divisible by 4 & 12
IMO E


Bunuel wrote:
x is divisible by 144. If \(\sqrt[3]{x}\) is an integer, then which of the following is \(\sqrt[3]{x}\) definitely divisible by?

I. 4
II. 8
III. 12

A. I only
B. II only
C. III only
D. I and II only
E. I and III only
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Re: x is divisible by 144. If ³√x is an integer, then which of the followi [#permalink] New post   07 Oct 2019, 20:02
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Bunuel wrote:
x is divisible by 144. If \(\sqrt[3]{x}\) is an integer, then which of the following is \(\sqrt[3]{x}\) definitely divisible by?

I. 4
II. 8
III. 12

A. I only
B. II only
C. III only
D. I and II only
E. I and III only



Since x must be a perfect cube, the smallest perfect cube for which x/144 = x/(3^2 x 2^4) is an integer is 3^3 x 2^6. Therefore, the cube root of that number is 3^1 x 2^2 = 12, which is definitely divisible by 4 and 12.

Answer: E
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Re: x is divisible by 144. If ³x is an integer, then which of the followi [#permalink] New post   23 Dec 2023, 22:29
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Re: x is divisible by 144. If ³x is an integer, then which of the followi [#permalink]
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