Question

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

Answer 1

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here is your answer
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➡Let us three consecutive  integers be, n, n + 1 and n + 2.

➡Whenever a number is divided by 3 the remainder obtained is either 0 or 1 or 2.

➡let 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, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.

➡If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

➡So that n, n + 1 and n + 2 is always divisible by 3.

➡ n (n + 1) (n + 2) is divisible by 3.
 
➡Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1.

➡∴ n = 2q or 2q + 1, where q is some integer.


➡If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) a
re divisible by 2.

➡If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

➡So that n, n + 1 and n + 2 is always divisible by 2.

➡ n (n + 1) (n + 2) is divisible by 2.
 
➡But n (n + 1) (n + 2) is divisible by 2 and 3.
 
➡∴ n (n + 1) (n + 2) is divisible by 6.
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