1. Good post? |

## Number Properties

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

1. xy + xz is an even integer
2. y + xz is an odd integer
SPOILER: A

2. 1 out of 1 members found this post helpful. Good post? |
xy + z is an odd integer

This means
CASE 1
xy is odd and z is even

x and y => odd
z=> even

CASE 2
xy is even and z is odd

either one of x and y is even or both of them are even
z=> odd

Ques=> Is x even?

Statement 1:
xy + xz is an even integer
If we take case 1
xy is odd
hence to make xy+xz even xz too should be odd
this cant be possible because z is even

now we consider case 2
xy is even
so in order to make xy+xz to be even .. xz too should be even
we know that z is odd
hence x has to be even

This statement is Sufficient

Statement 2
y + xz is an odd integer

for y+xz to be odd
i)
y is odd and
xz is even

let us take case 1 again
xy is odd and z is even
but this isnt possible because z is even

now lets look at case 2
xy is even and z is odd

y+xz is odd

y is odd
now if xy is even and y is odd
x has to be even

ii)
y is even
xz is odd

case 1
xy is odd and z is even
not possible becaouse z xz is odd

case 2
xy is even and z is odd
xz is odd
x has to be odd

looking at the conditions .. x could either be even or odd

Insufficient

Option
SPOILER: A

3. Good post? |
I had another way of solving the equation:

I will only prove with one option:

Given - xy + z = odd
Option 2 - y + xz = odd

Let us create a table for available options for x,y,z for both the equations:

 Given if x = even if x = odd (option1) if x is odd (option2) x even odd odd y can be anything odd even z odd even odd Option2 x even odd odd y odd even odd z can be anything odd even

If read carefully, text in bold and text in italics are the same for both the cases. Therefore the equation holds true for x = odd

Also, we can have x = even, y = odd, and z = odd with the given equations therefore not sufficient.

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