If n is a positive integer, what is the value of n?

Let's start with statement 2, as that statement is simpler of the two statements.

Statement 2

\(n^2 - 5n + 6 = 0\)

(n-2) (n-3) = 0

So n can be either 2 or it can be 3

As we are getting two values of n, we can rule out. B and D.

Statement 1

For any positive integer y, the product of all integers from y to y+n is divisible by 8.

Let's see when can this happen.

If y is odd, we need a minimum of four integers, including y, to be sure that the product will be divisible by 8.

y = 1, the product of all integers from y to y+n will be divisible by 8, when we have at least 3 twos (i.e. \(2^3\)) in the product.

So 1 * 2 * 3 * 4

Let's take another value

y = 3; the minimum value of n should be 3.

Ex. 3 * 4 * 5 * 6

Now if y is even, the minimum value of n should be 2.

Note, this is the minimum required value, any value greater than the minimum value also satisfies.

As n can take any value \(\geq{3} \) we can have multiple possible values. Hence we can rule out A.

Combined

Statement 2 limits the value of n to either 2 or to 3. Now as the nature of y is not known, we need to take the maximum of 2 and 3 to be sure and for Statement 1 to always hold true.

Hence the value of n = 3

The statements combined can answer the question.

Option C