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Re: Is y an integer [#permalink]
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.

Re: Is y an integer [#permalink]
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.

Re: Is y an integer [#permalink]
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}.

Re: Is y an integer [#permalink]
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.

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