EgmatQuantExpert wrote:
If \(x, y\) and \(z\) are positive integers such that \(x^4y^3 = z^2\), is \(x^9 - y^6\) odd?
(1) \(\frac{x^4y^3}{(x^2 + y^2)}\) can be written in the form \(4k + 3\), where \(k\) is a positive integer
(2) \(z = x + y\)
Statement 1 gives that \(\frac{x^4y^3}{(x^2 + y^2)}\) is odd integer
Case 1) X= odd, y= odd, makes denominator \((x^2 + y^2)\) = O+O=E and numerator is odd which makes the expression Not an integer. Not possible
Case 2) X= Even, y= Even, makes denominator \((x^2 + y^2)\) = E+E=E and the numerator a multiple of two even numbers and hence even. The multiples of 2 in numerator and denominator can cancel out to make the simplified expression an odd integer. Possible. Here \(x^9 - y^6\)= E-E =E.Case 3) one of x,y is E and other O, makes denominator \((x^2 + y^2)\) = E+O= O, the numerator is even as it is a multiple of even number, then it cant be odd. Not possible
Statement 1 is sufficient.
Statement 2 gives \(z = x + y\) and we already have \(x^4y^3 = z^2\).
Case 1) X= O, y= O, gives Z =E, and \(x^4y^3 = z^2\) gives Z=O*O=O. Not possible.
Case 2) X= E, y= E, gives Z =E, and \(x^4y^3 = z^2\) gives Z=E*E=E. Possible. Here \(x^9 - y^6\)= E-E =E.Case 3) one of x,y is E and other O gives Z=O+E=O, and \(x^4y^3 = z^2\) gives Z=E*O=E. Not possible.
Statement 2 is sufficient.
The answer is D.Please hit +1 Kudos if you liked the answer. 
02 Jun 2019, 23:21
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