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01 Oct 2019, 00:01
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
67% (01:57) correct
33%
(02:03)
wrong
based on 885
sessions
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Date
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Not Attempted Yet
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
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
14 Mar 2021, 01:55
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Bunuel
\(x=144k=2^4*3^2k\)
\(\sqrt[3]{x}=\sqrt[3]{144k}=\sqrt[3]{2^4*3^2k}=2\sqrt[3]{2*3^2k}=integer\)
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.
General Discussion
01 Oct 2019, 06:53
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Bunuel
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
