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Is a an integer? [#permalink]
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31 Oct 2014, 12:35
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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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31 Oct 2014, 12:36
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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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13 Apr 2016, 12:21
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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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13 Apr 2016, 12:34
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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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18 Apr 2016, 12:23
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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18 Aug 2016, 08:12
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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18 Aug 2016, 08:17
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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18 Aug 2016, 08:26
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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30 Jan 2017, 00:03
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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14 Jan 2018, 02:37
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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14 Jan 2018, 05:22
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.




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