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Is x an integer? (1) x^3 is an integer (2) 3x is an integer

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Is x an integer? (1) x^3 is an integer (2) 3x is an integer [#permalink] New post   15 Feb 2018, 03:45
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Is x an integer?


(1) x^3 is an integer

(2) 3x is an integer
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Bunuel
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Is x an integer? (1) x^3 is an integer (2) 3x is an integer [#permalink] New post   05 Jun 2018, 02:32
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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.
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Re: Is x an integer? (1) x^3 is an integer (2) 3x is an integer [#permalink] New post   19 Dec 2020, 17:57
Bunuel wrote:
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 ?
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Re: Is x an integer? (1) x^3 is an integer (2) 3x is an integer [#permalink] New post   20 Dec 2020, 01:48
Expert Reply
abhinavsodha800 wrote:
Bunuel wrote:
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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Re: Is x an integer? (1) x^3 is an integer (2) 3x is an integer [#permalink] New post   20 Dec 2020, 05:14
Bunuel wrote:
abhinavsodha800 wrote:
Bunuel wrote:
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:
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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.
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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Re: Is x an integer? (1) x^3 is an integer (2) 3x is an integer [#permalink] New post   06 Sep 2023, 00:37
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Re: Is x an integer? (1) x^3 is an integer (2) 3x is an integer [#permalink]
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