Aabhash777
If n is a positive integer, what is the value of n?
1. For any positive integer y, the product of all integers from y to y+n is divisible by 8.
2. n^2-5n+6=0
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 1For
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.
CombinedStatement 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