Question

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

Answer 1

Solution

To prove that a number is divisible by 6, we need to check if it is divisible by both 2 and 3. If it is, then it would be divisible by 6 also.

Let the positive integer be n.

Then, the next consecutive to n would be (n + 1) and (n + 2).

So, product of three consecutive integers

So, product of three consecutive integers = (n)(n + 1)(n + 2)

From Euclid's Division Lemma, a number can be represented as

a = bq + r

where, a, b and q are all positive integers and 0 ≤ r < b

Now, for a = n, and b = 2, r can be 0 or 1 only.

So, n can be written as either :

n = 2q

or, n = 2q + 1

Case - 1

n = 2q

Then, (n)(n + 1)(n + 2)

= (2q)(2q + 1)(2q + 2)

= 2(q)(2q + 1)(2q + 2)

= 2m [where, m is some integer equating to q(2q + 1)(2q + 2)]

And clearly, 2m is divisible by 2.

So, (n)(n + 1)(n + 2) is divisible by 2 for n = 2q (an even number)

Case - 2

n = 2q + 1

→ n(n + 1)(n + 2)

= (2q + 1)(2q + 1 + 1)(2q + 1 + 2)

= (2q + 1)(2q + 2)(2q + 3)

= (2q + 2)(2q + 1)(2q + 3)

= 2(q + 1)(2q + 1)(2q + 3)

= 2m

[where, m = (q + 1)(2q + 1)(2q + 3)]

Clearly, 2m is divisible by 2

So, n(n + 1)(n + 2) is divisible by 2 for n = 2q + 1 (an odd number)

Conclusion

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

$\rule{200}2$

Now we will have to check whether if (n)(n + 1)(n + 2) is divisible by 3 or not.

From Euclid's Division Lemma,

n = 3q

or

n = 3q + 1

or

n = 3q + 2

Case - 3

n = 3q

→ (n)(n + 1)(n + 2)

= 3q(3q + 1)(3q + 2)

= 3m

[For some integer m = q(q + 1)(q + 2)]

Clearly, 3m is divisible by 3

So, n(n + 1)(n + 2) is divisible by 3 for n = 3q.

Case - 4

n = 3q + 1

→ (n)(n + 1)(n + 2)

= (3q + 1)(3q + 1 + 1)(3q + 1 + 2)

= (3q + 1)(3q + 2)(3q + 3)

= (3q + 3)(3q + 2)(3q + 1)

= 3(q + 1)(3q + 2)(3q + 1)

= 3m

[where, m = (q + 1)(3q + 2)(3q + 1)]

Clearly, 3m is divisible by 3.

So, n(n + 1)(n + 2) is divisible by 3 for n = 3q + 1

Case - 5

n = 3q + 2

→ (n)(n + 1)(n + 2)

→ (3q + 2)(3q + 2 + 1)(3q + 2 + 2)

→ (3q + 2)(3q + 3)(3q + 4)

→ (3q + 3)(3q + 2)(3q + 4)

→ 3(q + 1)(3q + 2)(3q + 4)

→ 3m

[where, m = (q + 1)(3q + 2)(3q + 4)]

Clearly, 3m is divisible by 3

so, n(n + 1)(n + 2) is divisible by 3 for n = 3q + 2

Conclusion

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

Recall that if a number is divisible by both 2 and 3, it is divisible by 6 also.

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