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

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

Question: Is y an Integer?

Statement 1: \(y^3\)is an integer

If \(y^3 = 1\), y = 1 i.e. YES an Integer
If \(y^3 = 2\), \(y = 2^{1/3}\) i.e. NOT an Integer

NOT SUFFICIENT

Statement 2: \(3y\)is an integer

If \(3y = 1\), y = 1/3 i.e. NOT an Integer
If \(3y = 3\), \(y = 1\) i.e. YES an Integer

NOT SUFFICIENT

Combining the two statements

\(y^3\) is an Integer i.e. y is definitely not a fraction
and \(3y\) is an Integer

i.e. y must be an Integer

Sufficient

Answetr: Option C

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