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 odd, even expressions
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mingshum
Senior Member

USA
29 Posts

Posted - 02/24/2009 :  11:39:08  Show Profile  Reply with Quote
What is the fastest way to solve this SAT prep problem?

If x, y, and z are positive integers such that the value of x+y is even and the value of (x + y) + x + z is odd, which of the following must be true?

A) x is odd
B) x is even
C) If z is even, then x is odd.
D) If z is even, then xy is even.
E) xy is even.

I eliminated choices A,B and E.

Answer is C.
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Haven
Senior Member

USA
28 Posts

Posted - 02/24/2009 :  12:56:58  Show Profile  Reply with Quote
On way of looking at this problem is to substitute letters for number values "O" for odd and "E" for even.

The problem tells you that the value of x + y = a even number. So that means that x and y must both be odd or x and y must both be even.

The problem also tells you that (x + y) is squared. Since (x + y) is an even number it can be reasoned that (x + y) will be an even number also.

Now with this information you can write an equation substituting possible number values with letters

(x + y)+ x + z = odd

even + odd + z = odd In this case z must be even when x is odd

or

even + even + z = odd In this case z must be odd when x is even

I hope that this explanation was of some help to you.
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mingshum
Senior Member

USA
29 Posts

Posted - 02/24/2009 :  13:42:08  Show Profile  Reply with Quote
very clear explanation...thanks
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