Question

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

Answer 1

2,1and3 because 2*3*1=6

Answer 2

Let n be αny positive integer.

Thus, the three consecutive positive integers αre n, n + 1 αnd n + 2.

We know thαt αny positive integer cαn be of the form 6q, or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5. (From Euclid’s division lemmα for b = 6)

So,

For n = 6q,

$\implies \sf n(n+1)(n+2)= 6q(6q+1)(6q+2)$

$\implies \sf{ n(n+1)(n+2)= 6[q(6q+1)(6q+2)]}$

$\implies \sf n(n+1)(n+2)= 6m,$ which is divisible by 6. $\implies \sf [m= q(6q+1)(6q+2)]$

For n = 6q + 1,

$\implies \sf n(n+1)(n+2)= (6q+1)(6q+2)(6q+3)$

$\implies \sf{ n(n+1)(n+2)= 6[(6q+1)(3q+1)(2q+1)]}$

$\implies \sf n(n+1)(n+2)= 6m,$  which is divisible by 6. $\sf [m= (6q+1)(3q+1)(2q+1)]$

For n = 6q + 2,

$\implies \sf n(n+1)(n+2)= (6q+2)(6q+3)(6q+4)$

$\implies \sf n(n+1)(n+2)= 6[(3q+1)(2q+1)(6q+4)]$

$\implies \sf n(n+1)(n+2)= 6m,$ which is divisible by $\sf 6. [m= (3q+1)(2q+1)(6q+4)]$

For n= 6q+3,

$\implies \sf n(n+1)(n+2)= (6q+3)(6q+4)(6q+5)$

$\implies \sf n(n+1)(n+2)= 6[(2q+1)(3q+2)(6q+5)]$

$\implies \sf n(n+1)(n+2)= 6m,$ which is divisible by 6. $\sf [m= (2q+1)(3q+2)(6q+5)]$

For n= 6q+4,

$\implies \sf n(n+1)(n+2)= (6q+4)(6q+5)(6q+6)$

$\implies \sf n(n+1)(n+2)= 6[(3q+2)(3q+1)(2q+2)]$

$\implies \sf n(n+1)(n+2)= 6m,$ which is divisible by 6. $\sf [m= (3q+2)(3q+1)(2q+2)]$

For n= 6q+5,

$\implies \sf n(n+1)(n+2)= (6q+5)(6q+6)(6q+7)$

$\implies \sf n(n+1)(n+2)= 6[(6q+5)(q+1)(6q+7)]$

$\implies \sf n(n+1)(n+2)= 6m,$ which is divisible by 6. $\sf [m= (6q+5)(q+1)(6q+7)]$

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Hope it elps! :p