Question

show that the product of three consecutive natural numbers is divisible by 6​

Answer 1

$\Huge{\textbf{\textsf{\color{navy}{An}{\purple{sW}{\pink{eR}{\color{pink}{:-}}}}}}}$

Let three consecutive positive integers be, n, n + 1 and n + 2. When a number is divided by 3, the remainder obtained is either 0 or 1 or 2. ∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer. If n = 3p, then n is divisible by 3. If n = 3p + 1, ⇒ n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. If n = 3p + 2, ⇒ n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3. ⇒ n (n + 1) (n + 2) is divisible by 3. Similarly, when a number is divided 2, the remainder obtained is 0 or 1. ∴ n = 2q or 2q + 1, where q is some integer. If n = 2q ⇒ n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2. If n = 2q + 1 ⇒ n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2. ⇒ n (n + 1) (n + 2) is divisible by 2. Hence n (n + 1) (n + 2) is divisible by 2 and 3. ∴ n (n + 1) (n + 2) is divisible by 6. Read more on Sarthaks.com - https://www.sarthaks.com/2135/show-that-product-of-three-consecutive-natural-numbers-is-divisible-by-6

$\huge\boxed{\fcolorbox{green}{yellow}{ItZTanisha★᭄ꦿ᭄}}$

mark me brainliest