If x and y are integers, what is the value of y?


(1) y^x = y
\(y^x-y=0......y(y^{x-1}-1)=0\)
So either y=0 or \(y^{x-1}=1\), that is x is 1 or y is 1 if \(x\neq{0}\)
Insuff

(2) x > 1
Nothing about y..
Insufficient

Combined
Either y=0, or y=1,
And when x is odd, y can be 0, 1 or -1 for example y^3=y....y^3-y=0...y(y-1)(y+1)=0
Insufficient

E

St2: alone is not sufficient but it gives information about X; X is positive and more than 1

Now coming back to St1: y^x = y This is only possible when y=1

No matter what is the value of X (eg. 2,3,4,5..100), y^x = y will always hold true.

My ans is C; please correct me if I am missing something!

\(x,y\,\,{\text{ints}}\)

\(? = y\)

\(\left( {1 + 2} \right)\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {2,0} \right)\,\,\,\, \Rightarrow \,\,\,? = 0\,\, \hfill \\
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,{\text{?}}\,{\text{ = }}\,{\text{1}}\,\, \hfill \\
\end{gathered} \right.\)

The correct answer is therefore (E).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Yes you are missing something....
What if x is even y^2=y, here y can be 1 or 0
And when x is odd, say 3..y^3=y... Here y can be 0, 1 or -1

Hi, chetan2u!
Just to correct your typo: when y^x = y and x=2, you mean y can be 1 or 0 (not -1), of course.
Regards,
Fabio.
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Thank you Fabio..
Of course y cannot be negative ..
It was a typo and I meant odd value of X..
Example (-1)^3=-1 so y can be -1 too

Make sense and much clear to me now. Thank you for the explanation