Question 355187
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We make use of these facts:

even + even = even
even + odd = odd
odd + odd = even

even × even = even
even × odd = even
odd × odd = odd

Let's examine the three terms:

6y^2 will always be an even integer regardless of what y is 

9y will be an even integer if y is even and it will be an odd integer if y is odd.

17 is odd regardless.

We want the sum of these to be even.  6y^2 is always even and 17 is
always odd.  So the sum of the first and third terms 6y^2 + 17 is an
even number plus an odd number, which will always be an odd number.  
So in order for the sum of the three term to be even we must have the
middle term 9y odd, so that the odd number 6y^2 + 17 will be added to
another odd number to make an even sum.  So to make the middle term 9y
odd we must choose y to be odd.  Therefore the answer is (e).

Edwin</pre>