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.