Advanced Quant -- DS question - yes no vs. value

[versatile801]

Is the integer n odd?
(1) (n)^2 - 2n is not a multiple of 4.
(2) n is a multiple of 3.

The explanation in the third chapter of DS says that A is the answer for the above question - given below

(1) SUFFICIENT: n^2 - 2n = n(n - 2). If n is even, both terms in this product will be even, and the product will be divisible by 4. Since n2 - 2n is not a multiple of 4, we know that the integer n cannot be even"”it must be odd.

I know this is Yes / No Q, but if i want to put value 0 or 2 ( both are even ) in place of n, the result is not divisible by 4 again.

so, in above case if i take 0 or 2 in place of n , the result is not going to be divisible by 4 as it happens with the odd number.

Please explain - chapter 3, principle 2 Q - rephrasing

thanks

[nakul.maheshwari000]

If I understand this correctly, it only says "N" has to be an integer. It does not say the multiple has to be an integer.

n = 2
Then, n2 - 2n = 2.
4 X 1/2 = 2

n = 0
Then, n2-2n = 0
4 X 0 = 0

One of those tricky "integer" language questions.

[jnelson0612]

Is the integer n odd?
(1) (n)^2 - 2n is not a multiple of 4.
(2) n is a multiple of 3.

The explanation in the third chapter of DS says that A is the answer for the above question - given below

(1) SUFFICIENT: n^2 - 2n = n(n - 2).

I know this is Yes / No Q, but if i want to put value 0 or 2 ( both are even ) in place of n, the result is not divisible by 4 again.

so, in above case if i take 0 or 2 in place of n , the result is not going to be divisible by 4 as it happens with the odd number.

Actually, if you plug in either 0 or 2 for n the result will be divisible by 4. Let's check:
n=0, then 0(0-2) = 0
n=2, then 2(2-2) = 0

zero IS in fact a multiple of 4; it is a multiple of all integers.