Prove that the product of three consecutive positive integer is divisible by 6.

Solution: True. Justification: Let a, a + 1 be two consecutive positive integers. By Euclid’s division lemma, we have a = bq + r, where 0 ≤ r < b For b = 2 , we...

Solution: True. Justification: At least one out of every three consecutive positive integers is divisible by 2. Therefore, The product of three consecutive positive integers is divisible by 2. At least one out of every...

Solution: By Euclid’s division lemma, we have a = bq + r;  0 ≤ r < b For a = n and b = 3, we have n = 3q + r, …(i) where q is...

Solution: Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer. By Euclid’s division lemma, we have a = bq + r; 0 ≤...

Let a = n^3 - n ⇒ a = n.(n^2 - 1) ⇒ a = n.(n-1).(n+1)  [∵(a^2 - b^2) = (a-b)(a+b)] ⇒ a = (n-1).n.(n+1)  ...(i) We know that, 1. If a number is...

n2 - 1 is divisible by 8, if n is an odd integer 

Let, (n – 1) and n be two consecutive positive integers ∴ Their product = n(n – 1) = n 2 −n We know that any positive integer is of the form 2q...

Another solution  would be: n^3 - n = n (n^2 - 1) = n (n - 1) (n + 1) Whenever a number is divided by 3, the remainder obtained is...

True, because n(n+1)(n+2) will always be divisible by 6, as least one of the factors will be divisible by 2 and at one of the factors will be divisible by...