Quantitative Aptitude MCQ - Numbers and Algebra
y 4 3 2 1 -1 -3 -4
x 6 5 4 3 1 -1 -2
Hence, minimum value of
And maximum value of
(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!.
Factors of a2 are 1. a and a2.
Factors of ab are 1, a, b and ab.
Factors of a3 are 1. a. a2 and a3.
