Question

prove that the product of any three consecutive positive integer is divisible by 6​

Answer 1

Answer:

take any 3 consecutive positive integer say 1,2,3

product of 1*2*3=6

so 6 is divisible by 6

take more numbers like 2,3,2

product of 2*3*2=12

so 12 is divisible by 6

please follow

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

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$\green{\underline \bold{Given :}}$

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• Three consecutive positive integers.

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$\red{\underline \bold{To \: Prove:}}$

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• Product of any three consecutive positive integer is divisible by 6

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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Let the three consecutive positive integers be x, x+1 and x+2.

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

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Therefore, n=3p or 3p+1 or 3p+2, where p is some integer.

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If n = 3p, then n is divisible by 3.

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

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

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So, we can say that one of the numbers among n,n+1 and n+2 is always divisible by 3 that is:

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⇛ n(n+1)(n+2) is divisible by 3.

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Similarly, whenever a number is divided by 2, the remainder obtained is either 0 or 1.

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

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

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

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So, we can say that one of the numbers among n, n+1 and n+2 is always divisible by 2.

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⇛ Since, n(n+1)(n+2) is divisible by 2 and 3.

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⇛ Hence, n(n+1)(n+2) is divisible by 6.

$\rule{200}5$