x,y are odd and z is even then ((x^2+y^2)z^2)/8 is Select.
A) even  B) odd  C) fraction  D) fractiWQ  E) either even or odd
As x, y are odd x2+y2x2+y2 is always even. Now if z is a multiple of 4, then z2z2 is divisible by 8, then the equation is even. if z is a not a multiple of 4, but only a multiple of 2, then z2z2 is not completely divisible as it contains only two 2's and other two is cancelled in x2+y2x2+y2 which results in an odd number.   (32+52)428=34×168=34×2(32+52)428=34×168=34×2   (32+52)628(32+52)628 = 34×368=17×9   Answer:either even or odd
<result>E</result>
x,y are odd and z is even then ((x^2+y^2)z^2)/8 is