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Is y an integer? (1) y^3 is an integer (2) 3y is an integer
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10 Feb 2012, 05:21
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Is y an integer? (1) y^3 is an integer (2) 3y is an integer
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Is y an integer? (1) y^3 is an integer (2) 3y is an integer
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10 Feb 2012, 07:07
Is y an integer?(1) y^3 is an integer > y is either an integer (..., 1, 0, 1, 2, ...) or \(\sqrt[3]{integer}\), for example \(\sqrt[3]{2}\). Not sufficient. Notice here that y cannot be some reduced fraction like 1/3 or 13/5, because in this case y^3 won't be an integer.(2) 3y is an integer > y is either an integer or integers/3, for example: 1/3, 2/3, ... Not sufficient. Notice here that y cannot be some irrational number like \(\sqrt{2}\) or \(\sqrt[3]{integer}\), because in this case 3y won't be an integer.(1)+(2) Since both y^3 and 3y are integers then, as discussed above, y must be an integer. Sufficient. Answer: C.
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Re: Is y an integer
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10 Feb 2012, 07:30
Bunuel wrote: Is y an integer?
(1) y^3 is an integer > y is either an integer (..., 1, 0, 1, 2, ...) or \(\sqrt[3]{integer}\), for example \(\sqrt[3]{2}\). Not sufficient. Notice here that y can not be some reduced fraction like 1/3 or 13/5, because in this case y^3 won't be an integer.
B) 3y is an integer > y is either an integer or integers/3, for example: 1/3, 2/3, ... Not sufficient. Notice here that y can not be some irrational number like \(\sqrt{2}\) or \(\sqrt[3]{integer}\), because in this case 3y won't be an integer.
(1)+(2) Since both y^3 and 3y are integers then, as discussed above, y must be an integer. Sufficient.
Answer: C. Bunnel can you explain here by putting in some values for y. I am still unable to trace how by using both I can find that y is an integer.



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Re: Is y an integer
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10 Feb 2012, 07:45
subhajeet wrote: Bunuel wrote: Is y an integer?
(1) y^3 is an integer > y is either an integer (..., 1, 0, 1, 2, ...) or \(\sqrt[3]{integer}\), for example \(\sqrt[3]{2}\). Not sufficient. Notice here that y can not be some reduced fraction like 1/3 or 13/5, because in this case y^3 won't be an integer.
B) 3y is an integer > y is either an integer or integers/3, for example: 1/3, 2/3, ... Not sufficient. Notice here that y can not be some irrational number like \(\sqrt{2}\) or \(\sqrt[3]{integer}\), because in this case 3y won't be an integer.
(1)+(2) Since both y^3 and 3y are integers then, as discussed above, y must be an integer. Sufficient.
Answer: C. Bunnel can you explain here by putting in some values for y. I am still unable to trace how by using both I can find that y is an integer. Sure. Generally \(\sqrt[3]{integer}\) is either an integer itself or an irrational number, it can not be some reduced fraction like 1/2 or 2/3. (The same way as \(\sqrt{integer}\) is either an integer itself or an irrational number). From (1) \(y=integer\) or \(y=\sqrt[3]{integer}\); From (2) \(y=integer\) or \(y=\frac{integer}{3}\); So, from (1)+(2) \(y=integer\). Because if from (1) \(y=\sqrt[3]{integer}\), for example if \(y=\sqrt[3]{2}\), then \(3y=integer\) won't hold true for (2): \(3y={3*\sqrt[3]{2}}\neq{integer}\). The same way: if from (2) \(y=\frac{integer}{3}\), for example if \(y=\frac{1}{3}\), then \(y^3=integer\) won't hold true for (1): \(y^3=(\frac{1}{3})^3\neq{integer}\). Hope it's clear.
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Re: Is y an integer
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02 Jan 2014, 10:43
Bunuel wrote: subhajeet wrote: Bunuel wrote: Is y an integer?
(1) y^3 is an integer > y is either an integer (..., 1, 0, 1, 2, ...) or \(\sqrt[3]{integer}\), for example \(\sqrt[3]{2}\). Not sufficient. Notice here that y can not be some reduced fraction like 1/3 or 13/5, because in this case y^3 won't be an integer.
B) 3y is an integer > y is either an integer or integers/3, for example: 1/3, 2/3, ... Not sufficient. Notice here that y can not be some irrational number like \(\sqrt{2}\) or \(\sqrt[3]{integer}\), because in this case 3y won't be an integer.
(1)+(2) Since both y^3 and 3y are integers then, as discussed above, y must be an integer. Sufficient.
Answer: C. Bunnel can you explain here by putting in some values for y. I am still unable to trace how by using both I can find that y is an integer. Sure. Generally \(\sqrt[3]{integer}\) is either an integer itself or an irrational number, it can not be some reduced fraction like 1/2 or 2/3. (The same way as \(\sqrt{integer}\) is either an integer itself or an irrational number). From (1) \(y=integer\) or \(y=\sqrt[3]{integer}\); From (2) \(y=integer\) or \(y=\frac{integer}{3}\); So, from (1)+(2) \(y=integer\). Because if from (1) \(y=\sqrt[3]{integer}\), for example if \(y=\sqrt[3]{2}\), then \(3y=integer\) won't hold true for (2): \(3y={3*\sqrt[3]{2}}\neq{integer}\). The same way: if from (2) \(y=\frac{integer}{3}\), for example if \(y=\frac{1}{3}\), then \(y^3=integer\) won't hold true for (1): \(y^3=(\frac{1}{3})^3\neq{integer}\). Hope it's clear. Unable to understand from the explanation provided...... how from (1) + (2) > y=Integer ??? Can you pls provide some alternate solution/explanation.



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Re: Is y an integer? (1) y^3 is an integer (2) 3y is an integer
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02 Jan 2014, 11:23
subhajeet wrote: Is y an integer?
(1) y^3 is an integer (2) 3y is an integer Statement I is insufficient y ^ 3 = 1 y = 1 (YES y is an integer) y ^ 3 = 2 y = 2^1/3 (y is not an integer) Statement II is insufficient 3y = 3 y = 1 (YES y is an integer) 3y = 1 y = 1/3 (y is not an integer) Combining is sufficient (Usually your approach should be algebraic here) y^3 = p 3y = q If 3y = q then the only problem which makes y not an integer is that y is a fraction which is ruled out by the first statement as Fraction ^ 3 can never be an integer. Similarly the number being a cube root (problem in the first statement) is ruled out by the second statement as 3(Surd) cannot be an integer. Hence the answer is C
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Re: Is y an integer? (1) y^3 is an integer (2) 3y is an integer
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29 Nov 2014, 08:00
Hi there,
I have doubt here. As per option 1 it is given as "y^3 is an integer" which means y power, but why are we discussing cube root of y.
Switching from y^3 to cube root of y would change our answer completely. Either the question is wrong or the analysis is wrong.
Please do comment.
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Re: Is y an integer? (1) y^3 is an integer (2) 3y is an integer
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29 Nov 2014, 09:06



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Re: Is y an integer? (1) y^3 is an integer (2) 3y is an integer
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27 Oct 2016, 07:23
subhajeet wrote: Is y an integer?
(1) y^3 is an integer (2) 3y is an integer i got to C. 1 > what if sqrt3(y) is a noninteger? not sufficient 2 > y can be an integer, or can be a fraction, for ex. 1/3. not sufficient. 1+2 if y^3 is an integer and 3y is an integer i don't think there is such a number that would be a noninteger.. thus, sufficient.



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Re: Is y an integer? (1) y^3 is an integer (2) 3y is an integer
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