|
Answer» Answer:there are two methods to solve the problem which are discussed below.Method 1:Let us considera = n3\xa0– na = n (n2\xa0– 1)a = n (n + 1)(n – 1)Assumtions:1. Out of three (n – 1), n, (n + 1) one must be even, so\xa0a\xa0is divisible by 2.2. (n – 1) , n, (n + 1) are consecutive integers thus as proved\xa0a\xa0must be divisible by 3.From (1) and (2)\xa0a\xa0must be divisible by 2 × 3 = 6Thus, n³ – n is divisible by 6 for any positive integer n.Method 2:When a number is divided by 3, the possible remainders are 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where r is some integer.Case 1:\xa0Consider n = 3pThen n is divisible by 3.Case 2: Consider n = 3p + 1Then n – 1 = 3p + 1 –1⇒ n -1 = 3p is divisible by 3.Case 3:\xa0Consider n = 3p + 2Then n + 1 = 3p + 2 + 1⇒ n+1 = 3p + 3⇒ n+1 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.Similarly, when a number is divided by 2, the possible remainders are 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.Case 1: Consider n = 2qThen n is divisible by 2.Case 2: Consider n = 2q + 1Then n–1 = 2q + 1 – 1n – 1 = 2q is divisible by 2 andn + 1 = 2q + 1 + 1n +1 = 2q + 2n+1= 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.∴ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.Therefore, as per the divisibility rule of 6, the given number is divisible by six.n3\xa0– n =\xa0n (n – 1) (n + 1) is divisible by 6.
|
