Prove that product of three conscutive division divisible by three?
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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.
hence n (n + 1) (n + 2) is divisible by 3.
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.
hence n (n + 1) (n + 2) is divisible by 3.
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Answer:
If n = 3k + 2, then n + 1 = 3k + 2 + 1 = 3k + 3 = 3(k + 1) which is again divisible by 3. So we can say that one of the numbers among (n, n + 1 and n + 2) is always divisible by 3. Therefore the product of numbers n(n+1)(n+2) is always divisible by 3.
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