If (a, n)! is defined as product of n consecutive numbers starting from a, where a and n are both natural numbers, and if H is the HCF of (a, n)! and n!, then what can be said about H?
| A. | h = a! |
| B. | h = n! |
| C. | h ≥ n! |
| D. | h ≥ a * n |
Answer : Explanation : (a. n)! = product of n consecutive natural numbers starting from 'a' which is atleast divisible by n!. (n)! = product of n consecutive natural numbers. For n = 2 : (a. n)! = a(a + 1) and n! = 2 a(a + 1) is divisible by 2!. For n = 3 : (a n)! = a(a + 1)(a + 2) and n! = 6. One of the factors of a(a + 1)(a + 2) is divisible by 3 and other by 2. Thus, proceeding in this manner, (a. n)! and n! have HCF = n! ∴ H = n!. |
|
|
Option: A Explanation : Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. |
