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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] New post 10 Feb 2012, 04:21
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Is y an integer?

(1) y^3 is an integer
(2) 3y is an integer
[Reveal] Spoiler: OA
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Re: Is y an integer [#permalink] New post 10 Feb 2012, 06:07
Expert's post
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.
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Kudos [?]: 21 [0], given: 1

Re: Is y an integer [#permalink] New post 10 Feb 2012, 06:30
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 [#permalink] New post 10 Feb 2012, 06:45
Expert's post
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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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Is y an integer? (1) y^3 is an integer (2) 3y is an integer [#permalink] New post 25 May 2013, 03:42
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Re: Is y an integer [#permalink] New post 02 Jan 2014, 09:43
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 [#permalink] New post 02 Jan 2014, 10:23
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   [#permalink] 02 Jan 2014, 10:23
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