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abhinavsodha800
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Is x an integer?

(1) x^3 is an integer --> x is either an integer (..., -1, 0, 1, 2, ...) or \(\sqrt[3]{integer}\), for example \(\sqrt[3]{2}\). Not sufficient. Notice here that x cannot be some reduced fraction like 1/3 or 13/5, because in this case x^3 won't be an integer.


(2) 3x is an integer --> x is either an integer or integers/3, for example: 1/3, 2/3, ... Not sufficient. Notice here that x cannot be some irrational number like \(\sqrt{2}\) or \(\sqrt{integer}\), because in this case 3x won't be an integer.

(1)+(2) Since both x^3 and 3x are integers then, as discussed above, x must be an integer. Sufficient.

Answer: C.

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Hope it helps.
i didn't understood the first statement explanation . It says x^3 should be an integer and you are taking (integer)^1/3 in this if i will take x as 2^1/3 with power 3 [(2^1/3)^3 ] it will result into integer only. Then how come x is decimal here ? How come x^3 is same as x^1/3 ?

\(x^3 =integer\);

Take the cube root: \(x=\sqrt[3]{integer}\).

If that integer is cube of an integer, then x will be an integer. For example, if that integer = ..., -8, -1, 0, 1, 8, 27, ..., then x = -2, -1, 0, 1, 2, 3, ...
If that integer is NOT cube of an integer, then x will NOT be an integer. For example, if that integer = 2, , then \(x=\sqrt[3]{2}\).

Hope it's clear.
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Bunuel
abhinavsodha800
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Is x an integer?

(1) x^3 is an integer --> x is either an integer (..., -1, 0, 1, 2, ...) or \(\sqrt[3]{integer}\), for example \(\sqrt[3]{2}\). Not sufficient. Notice here that x cannot be some reduced fraction like 1/3 or 13/5, because in this case x^3 won't be an integer.


(2) 3x is an integer --> x is either an integer or integers/3, for example: 1/3, 2/3, ... Not sufficient. Notice here that x cannot be some irrational number like \(\sqrt{2}\) or \(\sqrt{integer}\), because in this case 3x won't be an integer.

(1)+(2) Since both x^3 and 3x are integers then, as discussed above, x must be an integer. Sufficient.

Answer: C.

Similar questions to practice:
https://gmatclub.com/forum/if-x-is-a-pos ... 65976.html
https://gmatclub.com/forum/if-x-is-a-pos ... 88994.html
https://gmatclub.com/forum/what-is-the-v ... 07195.html
https://gmatclub.com/forum/if-sqrt-4a-is ... 06886.html
https://gmatclub.com/forum/if-8-0-5y-3-0 ... 06606.html
https://gmatclub.com/forum/is-s-an-odd-i ... 06562.html
https://gmatclub.com/forum/if-d-is-a-pos ... 04421.html
https://gmatclub.com/forum/if-z-is-a-pos ... 01464.html
https://gmatclub.com/forum/if-x-is-a-pos ... 01918.html
https://gmatclub.com/forum/if-y-is-a-pos ... 08287.html
https://gmatclub.com/forum/is-z-an-integer-191421.html
https://gmatclub.com/forum/is-y-an-integ ... 27344.html
https://gmatclub.com/forum/if-x-is-an-in ... 34837.html
https://gmatclub.com/forum/is-a-even-133175.html
https://gmatclub.com/forum/if-n-is-an-in ... 37875.html
https://gmatclub.com/forum/is-s-an-odd-i ... 27400.html
https://gmatclub.com/forum/if-n-p-q-p-an ... 01475.html
https://gmatclub.com/forum/if-x-is-a-pos ... 90619.html
https://gmatclub.com/forum/x-2-5y-49-is- ... 84779.html
https://gmatclub.com/forum/is-x-an-integer-200487.html

Hope it helps.
i didn't understood the first statement explanation . It says x^3 should be an integer and you are taking (integer)^1/3 in this if i will take x as 2^1/3 with power 3 [(2^1/3)^3 ] it will result into integer only. Then how come x is decimal here ? How come x^3 is same as x^1/3 ?

\(x^3 =integer\);

Take the cube root: \(x=\sqrt[3]{integer}\).

If that integer is cube of an integer, then x will be an integer. For example, if that integer = ..., -8, -1, 0, 1, 8, 27, ..., then x = -2, -1, 0, 1, 2, 3, ...
If that integer is NOT cube of an integer, then x will NOT be an integer. For example, if that integer = 2, , then \(x=\sqrt[3]{2}\).

Hope it's clear.
Thanks for the reply . When i solved this question i couldn't even imagine that i would have think so much deep into it. So can i consider this as a thumb rule whenever i see x^3 i should also consider this case of checking x = cube root of integer
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