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 odd, even expressions

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T O P I C    R E V I E W
mingshum Posted - 02/24/2009 : 11:39:08
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.
2   L A T E S T    R E P L I E S    (Newest First)
mingshum Posted - 02/24/2009 : 13:42:08
very clear explanation...thanks
Haven Posted - 02/24/2009 : 12:56:58
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


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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