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Bunuel
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If x is both the cube and the square of an integer and x is between 2 Updated on: 04 Jul 2019, 04:30
Originally posted by Bunuel on 05 Jun 2017, 07:44.
Last edited by Sajjad1994 on 04 Jul 2019, 04:30, edited 1 time in total.
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C
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If x is both the cube and the square of an integer and x is between 2 and 200, what is the value of x?

(A) 8
(B) 16
(C) 64
(D) 125
(E) 169

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C
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If x is both the cube and the square of an integer and x is between 2 05 Jun 2017, 08:22
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Bunuel
If x is both the cube and the square of an integer and x is between 2 and 200, what is the value of x?

(A) 8
(B) 16
(C) 64
(D) 125
(E) 169
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Check the options only a 10 sec answer...

\(a^2 = 64\) , So, \(a = 8\)
\(b^3 = 64\) , So, \(b = 4\)

So, the value of x will be (C) 64
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If x is both the cube and the square of an integer and x is between 2 05 Jun 2017, 08:26
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Bunuel
If x is both the cube and the square of an integer and x is between 2 and 200, what is the value of x?

(A) 8
(B) 16
(C) 64
(D) 125
(E) 169
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My approach was checking all the alternatives. In a few seconds, you find out that 64 = 8^2 = 4^3.
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If x is both the cube and the square of an integer and x is between 2 05 Jun 2017, 08:59
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Bunuel
If x is both the cube and the square of an integer and x is between 2 and 200, what is the value of x?

(A) 8
(B) 16
(C) 64
(D) 125
(E) 169
Show more

The question confused me a bit. Anyways, checking answer options is the best way to solve this.

(A) 8 ==> \(\sqrt{8}\) is not an integer. Incorrect.

(B) 16 ==> Cuberoot(16) is not an integer. Incorrect.

(C) 64 ==> Matches. Correct!

(D) 125 ==> \(\sqrt{125}\) is not an integer. Incorrect.

(E) 169 ==> Cuberoot(169) is not an integer. Incorrect.
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If x is both the cube and the square of an integer and x is between 2 05 Jun 2017, 09:35
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clear 64
8^2 = 64
4^3 = 64

ans C
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If x is both the cube and the square of an integer and x is between 2 05 Jun 2017, 11:00
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If a number is square of an integer, then the powers of all prime numbers in its prime factorisation Must be divisible by 2.
If a number is cube of an integer, then the powers of all prime numbers in its prime factorisation Must be divisible by 3.

Combining above two statements, it follows that if a number is square of an integer as well as cube of an integer, then powers of all prime numbers in its prime factorisation Must be Divisible by both 2 & 3 - or Divisible by 6.

Check options. A) 8 = 2^3. Powers are not multiples of 6, Eliminated
B) 16 = 2^4. Powers are not multiples of 6, Eliminated
C) 64 = 2^6. Here it is, powers are multiples of 6
D) 125 = 5^3. Powers are not multiples of 6, Eliminated
E) 169 = 13^2. Powers are not multiples of 6, Eliminated

Hence C answer
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If x is both the cube and the square of an integer and x is between 2 24 Jan 2021, 07:14
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