In number theory, prove that the product of any three consecutive integer is divisible by six.?

Answer 1

We want to prove that n(n + 1)(n + 2) is divisible by 6 for all natural numbers n.

Proof by induction:

Base case: The formula holds true for n = 1, because
n(n + 1)(n + 2) = 1(2)(3) = 6, which is obviously divisible by 6.

Induction Hypothesis: Assume the formula holds true for n = k.
That is, assume k(k + 1)(k + 2) is divisible by 6.

{We want to prove that (k + 1)(k + 2)(k + 3) is divisible by 6}

But, what does (k + 1)(k + 2)(k + 3) equal?

(k + 1)(k + 2)(k + 3) = (k + 3)(k + 1)(k + 2)

Distribute the (k + 3) over the rest. This gives us

= k(k + 1)(k + 2) + 3(k + 1)(k + 2)

k(k + 1)(k + 2) is divisible by 6 because of our induction hypothesis.

3(k + 1)(k + 2) is divisible by 6 because (k + 1)(k + 2), the product of two consecutive numbers, is divisible by 2 (i.e. product of an odd and even number is an even number).

The sum of two expressions divisible by 6 is divisible by 6, so it follows that (k + 1)(k + 2)(k + 3) is divisible by 6, showing the formula holds true for n = k + 1.

Therefore, by the principle of mathematical induction,
n(n + 1)(n + 2) is divisible by 6 for all natural numbers n.

Answer 2

At least 1 integer must be divisible by 2, and exactly 1 integer will be divisible by 3. It is either the same integer (divisible by 6) or the product of these 2 integers must be divisible by 6.

Answer 3

Adding on to Archknight you get:

m^3 + 3m^2 + 2m

If you notice all the exponents multiply up to 6.

Well if you took m(m+1)(m+2)(m+3) you will notice that all the exponents multiply up to 24 which would give you the divisor for any 4 consecutive integers. So if we can just use proof to provide the reason why the multiplication of the exponents give you the divisor then we're done....I hope this helps.

Answer 4

Sorry, but I do not have an answer. I just wish to try to help you get one. Nobody as yet has provided a mathematical proof of this. Sure, it is obviously true, but without that proof, you really haven't done anything. I am sure there is one out there but I am unable to think of it on my own right now. I at least know how it starts.

'm' is a non-negative integer

A product of 3 consecutive integers would be:
m(m+1)(m+2)

Good luck with the rest of the proof!

Answer 5

dear God, this is so simple!

123
ie, 1*2*3=6, which is divisible by 6
2*3*4=24, divisible by 6
3*4*5=60, divisible by 6

Answer 6

enable First integer = x then 2d integer = x+a million and 0.33 integer =x+2 Product =x(x+a million)(x+2) evaluate a hypothesis that x(x+a million)(x+2) is a diverse of three try approach of induction For x=a million Product =x(x+a million)(x+2) =a million*2*3 =6 a diverse of three enable us to assume it real for x=n i.e n(n+a million)(n+2) is a diverse of three To teach that it is likewise real for x= n+a million for n+a million x(x+a million)(x+2)=(n+a million)(n+a million+a million)(n+a million+2)=(n+a million)(... =(n+a million)(n+2)(n+3)=(n+a million)(n+2)n+(n+a million)(n+2... if (n+a million)(n+2)n is a diverse of three then (n+a million)(n+2)(n+3)=(n+a million)(n+2)n+(n+a million)(n+2)... =diverse of three+n+a million)(n+2)×3 hence (n+a million)(n+2)(n+3) is a diverse of three hence the assumed hypothesis is right that x(x+a million)(x+2), the made of three consecutive integers is divisible via 3