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Is a an integer?

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Is a an integer?  [#permalink]

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New post 31 Oct 2014, 13:35
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Is a an integer?

(1) a^3 is an integer.
(2) The cube root of a is an integer.

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Re: Is a an integer?  [#permalink]

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New post 13 Apr 2016, 13:34
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thalcantero wrote:
I still don't get why the first statement is insufficient, could someone explain a little further please? :|


Is a an integer?

(1) a^3 is an integer --> a will be an integer if a^3 is a perfect cube, for example, if a^3 is 0, 1, 8, 27, ... but a can also be the cube root from an integer (which is not a perfect cube), for example if a is \(\sqrt[3]{2}\), \(\sqrt[3]{3}\), \(\sqrt[3]{4}\), ... Not sufficient.

(2) The cube root of a is an integer. \(\sqrt[3]{a}=integer\) --> \(a=integer^3=integer\). Sufficient.

Answer: B.

Hope it's clear.
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Re: Is a an integer?  [#permalink]

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New post 31 Oct 2014, 13:36
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From Statement 1, a could definitely be an integer since any integer cubed will remain an integer. However, a could also be a cube root. If we cube the cube root of 2, the result is 2. Insufficient. Statement 2, however, guarantees that a must be an integer. a = cubert(a) * cubert(a) * cubert(a). If cubert(a) is an integer, then a must be an integer as well. Thus, the correct answer is B
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Re: Is a an integer?  [#permalink]

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New post 13 Apr 2016, 13:21
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I still don't get why the first statement is insufficient, could someone explain a little further please? :|
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Re: Is a an integer?  [#permalink]

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New post 18 Apr 2016, 13:23
Thanks a lot for the explanation, it is pretty clear now.
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Re: Is a an integer?  [#permalink]

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New post 18 Aug 2016, 09:12
Bunuel : I am not able to understand this part of your explanation ''but a can also be the cube root from an integer (which is not a perfect cube), for example if a is 2√3, 3√3''.
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Re: Is a an integer?  [#permalink]

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New post 18 Aug 2016, 09:17
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narendran1990 wrote:
Bunuel : I am not able to understand this part of your explanation ''but a can also be the cube root from an integer (which is not a perfect cube), for example if a is 2√3, 3√3''.


Not sure what to explain... If a is \(\sqrt[3]{2}\), \(\sqrt[3]{3}\), or \(\sqrt[3]{4}\), then a^3 will be 2, 3, or 4 respectively.
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Re: Is a an integer?  [#permalink]

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New post 18 Aug 2016, 09:26
I am not good with some basics of exponents, so I couldn't understand your explanation. It is clear now.

Thank you.
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Re: Is a an integer?  [#permalink]

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New post 30 Jan 2017, 01:03
Bunuel wrote:
narendran1990 wrote:
Bunuel : I am not able to understand this part of your explanation ''but a can also be the cube root from an integer (which is not a perfect cube), for example if a is 2√3, 3√3''.


Not sure what to explain... If a is \(\sqrt[3]{2}\), \(\sqrt[3]{3}\), or \(\sqrt[3]{4}\), then a^3 will be 2, 3, or 4 respectively.


Only this made me understand the problem Thanks!
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Re: Is a an integer?  [#permalink]

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New post 14 Jan 2018, 03:37
JusTLucK04 wrote:
Is a an integer?

(1) a^3 is an integer.
(2) The cube root of a is an integer.


The answer is B

From Statement 1 it is clear that cube of an integer is always an integer but when we take cube of root(2) then also the result will be integer hence insufficient
Statement 2 on the other hand clearly tells us that the cube root of the number is always an integer .Therefore the number a must be an integer hence sufficient .
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Re: Is a an integer?  [#permalink]

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New post 14 Jan 2018, 06:22
arvind910619 wrote:
JusTLucK04 wrote:
Is a an integer?

(1) a^3 is an integer.
(2) The cube root of a is an integer.


The answer is B

From Statement 1 it is clear that cube of an integer is always an integer but when we take cube of root(2) then also the result will be integer hence insufficient
Statement 2 on the other hand clearly tells us that the cube root of the number is always an integer .Therefore the number a must be an integer hence sufficient .


hello, it seems that practice is the only way to ensure that we will not make any mistakes in thinking, is that so?
it is because I have not worked on any math problems for a long time. When I try to find the answer, I always worry that I will miss something.
Re: Is a an integer? &nbs [#permalink] 14 Jan 2018, 06:22
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