Is the value of x(y + 1) even?
Statement 1: x and y are prime numbers.
Note that prime numbers are only divisible by 1 and itself. Also, 2 is the only even prime number.
So, for x(y+1); if x is even and y is odd
2(5+1) = 2*6 = 12, or
2(7+1) = 2*8 = 16
In all cases, the expression is even
Buf if x is odd and y is even
5(2+1) = 5*3 = 15, or
7(2+1) = 7*3 = 21
In all cases, the expression is odd.
The information in this statement is not enough to determine if x(y + 1) is even or odd. Statement 1 is, therefore, NOT SUFFICIENT.
Statement 2: y>7
This means that y can be between 8 and infinity, and x can be any number between 0 and infinity. Thus, this gives room for multiple cases or variations that can make the expression even or odd. Hence, statement 2 is NOT SUFFICIENT.
Combining both statements together:
x and y are prime integers, and y>7. This means that y will always be an odd prime integer.
So if x is even and y is (odd>7)
x (y+1) = 2(11+1) = 2*12 = 24, or
x(y+1) = 2(13+1) = 2*14 = 28.
If x is odd and y is (odd>7)
x(y+1) = 3(11+1) = 3*12 = 36, or
x(y+1) = 5(13+1) = 5*14 = 70
(y+1) = even and any number multiplied by an even number will be even, so, x(y+1) is even.
Hence, both statements combined together are SUFFICIENT.
Thus, answer = option C
