Question

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

Answer 1

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let the no. be (x) , (x + 1) ,(x + 2).

A number which is divided by 3, will be having the remainder 0 or 1 or 2.

x = 3n or (3n + 1) or (3n + 2) if x = 3n, then x is divisible by 3

if x = 3n + 1 ,then x + 2 = 3n + 1 + 2 = 3n + 3

=> x = 3(n + 1) is divisible by 3

if x = 3n + 2, then x + 1 = 3n + 2 + 1

=> 3n + 3 = 3(n + 1)

so, we can say that one of the numbers n, n + 1 and n + 2 is always divisible by 3.

n (n + 1) (n + 2) is divisible by 3.

similarly, when a no. is divisible by 2, remainders obtained is 0 or 1.

Thus, x = 2r or (2r + 1)

if x = 2r, then x = 2r and (x + 2) then,2r + 2

=> 2(r + 1) are divisible by 2

if x = (2r + 1), then x + 1 = 2r + 1 + 1 = 2(r + 1)is divisible by 2.

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

x (x + 1) (x + 2) is divisible by 2.

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

n (n + 1) (n + 2) is divisible by 6.

hope it helps you...✨✨

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Answer 2

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let the no. be (x) , (x + 1) ,(x + 2).

A number which is divided by 3, will be having the remainder 0 or 1 or 2.

x = 3n or (3n + 1) or (3n + 2) if x = 3n, then x is divisible by 3

if x = 3n + 1 ,then x + 2 = 3n + 1 + 2 = 3n + 3

=> x = 3(n + 1) is divisible by 3

if x = 3n + 2, then x + 1 = 3n + 2 + 1

=> 3n + 3 = 3(n + 1)

so, we can say that one of the numbers n, n + 1 and n + 2 is always divisible by 3.

n (n + 1) (n + 2) is divisible by 3.

similarly, when a no. is divisible by 2, remainders obtained is 0 or 1.

Thus, x = 2r or (2r + 1)

if x = 2r, then x = 2r and (x + 2) then,2r + 2

=> 2(r + 1) are divisible by 2

if x = (2r + 1), then x + 1 = 2r + 1 + 1 = 2(r + 1)is divisible by 2.

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

x (x + 1) (x + 2) is divisible by 2.

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

n (n + 1) (n + 2) is divisible by 6.