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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 [#permalink]

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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
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 39686
Is y an integer? (1) y^3 is an integer (2) 3y is an integer [#permalink]

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10 Feb 2012, 07:07
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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.

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Re: Is y an integer [#permalink]

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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.

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 [#permalink]

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10 Feb 2012, 07:45
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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.

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? (1) y^3 is an integer (2) 3y is an integer [#permalink]

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25 May 2013, 04:42
Bumping for review and further discussion.
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Re: Is y an integer [#permalink]

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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.

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 [#permalink]

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02 Jan 2014, 11:23
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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 [#permalink]

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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.

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Re: Is y an integer? (1) y^3 is an integer (2) 3y is an integer [#permalink]

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29 Nov 2014, 09:06
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.

Posted from my mobile device

Everything is correct.

$$y^3=integer$$ is the same as $$y=\sqrt[3]{integer}$$.
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Re: Is y an integer? (1) y^3 is an integer (2) 3y is an integer [#permalink]

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05 Jan 2016, 15:05
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Re: Is y an integer? (1) y^3 is an integer (2) 3y is an integer [#permalink]

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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 non-integer? 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 non-integer..
thus, sufficient.
Re: Is y an integer? (1) y^3 is an integer (2) 3y is an integer   [#permalink] 27 Oct 2016, 07:23
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