If x, y and z are integers and xy + z is an odd integer

[stock.mojo11]

If x, y and z are integers and xy + z is an odd integer, is x an even integer?

A. xy + xz is an even integer
B. y + xz is an odd integer

Need a fast solution.

[andrew.k.john]

Hi StockMojo. Here's how I would approach this question

Fact(1): xy+xz is even.

We also know that xy+z is odd. This means that Odd/Even characteristic of z changes when multiplied by x. If z is odd, then xz must be even for the expression xy+xz to be even and xy+z to be odd. Likewise, if z is even, then xz must be odd. However, is z is even, the xz WILL ALSO be even, so this can never be true. So z is odd and xz is even, which means x is even. (1) is SUFFICIENT.

Fact(2): y+xz is odd.

If y+xz and xy+z are both odd, the sum of the two must be even. so xy+xz+y+z is even. This can be factored to be (y+z)(x+1) is even. Either (y+z) or (x+1) must therefore be even. Now, let's test some cases.

Case 1: (y+z) is even, both y and z are even. This cannot happen because if y and z are both even, this violates our original fact that xy+z is odd.

Case 2: (y+z) is even, both y and z are odd. If both y and z are odd, then x MUST be even for the original facts to hold.

Case 3: (y+z) is odd, y is even, z is odd. If this is true, then x must be odd since (x+1) must be even. Let's test this out against our original facts. xy+z will be odd, since (odd)(even) + (odd) = (odd). y+xz will be odd, since (even) + (odd)(odd) = (odd). Here is an example where x is odd, which makes (2) INSUFFICIENT.

Although that is the proof for Fact(2), it is too long to reasonably arrive at during the test. The best choice is to test each case with real numbers for y and z, and solve for possible x's. This would effectively cut down the time.

Using this strategy, here are the two cases I would test.

Case 1: y=1, z=1. In this case, x+1 is odd, so x is even.

Case 2: y=2, z=1. In this case, x+2 is odd and 2x+1 is also odd. The second expression does not narrow the choices for x, but the first one does. x is odd.

They contradict, so we know Fact(2) is insufficient.

[kevincan]

If x, y and z are integers and xy + z is an odd integer, is x an even integer?

A. xy + xz is an even integer
B. y + xz is an odd integer

(1) Since xy + z is odd and xy +xz is even, subtracting we can deduce that z-xz = z(1-x) is odd. Thus 1 - x is odd, so x must be even. SUFF

(2) SUbtracting, we see that xy - y + z - xz = y(x - 1) - z(x - 1) =(y - z)(x - 1) is even. If y - z is odd, x - 1 is even and thus x is odd. On the other hand, if y - z is even, x is even. NOT SUFF

[RonPurewal]

long post on this problem here:

post21653.html#p21653

[imanemekouar]

I looked at the link , but I dont uderstand t why you did the parity things.And when you did how did you conclude in statement 1 that the promptis YES. Can you explain further.

[ivivi]

Another way is to plug-in numbers
x y z xy+z( given as odd) xy+xz(is even) y+xz(is odd)
3 2 3 9 15 11
3 3 2 11 15 9
3 2 2 8
3 3 3 12
2 2 3 7 10 8
2 3 2 8
2 2 2 6
2 3 3 9 12 9

As you can see 1 is suff and 2 is not.
Apologize for the missing spaces, which got removed when I posted.

[RonPurewal]

I looked at the link , but I dont uderstand t why you did the parity things.And when you did how did you conclude in statement 1 that the promptis YES. Can you explain further.

hi - if you have questions about that thread, please post them over there, not here.

"parity" just means whether the number is odd or even (in the same way that "sign" means either positive or negative). it just makes things easier to write - it's easier to write, and read, "same parity" than to write/read "either both odd or both even".