Number Properties

[Bobogee]

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

[rijul007]

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

[sandeep_chads]

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.