The product of any three consecutive natural numbers is divisible by 6

Question

Prove or find a counterexample: The product of any three consecutive natural numbers is divisible by 6.

Answer

Let n=1, Since xeN is arbitrary p(N) holds for all xeN.

(n).(n+1).(n+2),

(1).(1+1).(1+2)= 6

Thus,6|6

I don't understand what you're asking; are you having trouble understanding the argument made in your post, or are you checking the argument? — , Apr 10 '14 at 21:56

score 3 · Accepted Answer · answered Apr 10 '14 at 22:05

3

Since every other number is even, at least one (or two) of the three numbers $$n,n+1,n+2$$ will be even. Since every third number is divisible by three, exactly one of $$n,n+1,n+2$$ will have a factor of three. So in $$n\times (n+1)\times (n+2)$$ there will be at least one factor of two and exactly one factor of three, meaning it will be divisible by $$3\times 2=6$$

answered Apr 10 '14 at 22:05

I believe you mean "multiple" not "factor". – Joel Apr 10 '14 at 22:08

Kaj Hansen · Answer 2 · 2018-09-22T20:46:00.497

1

Hint: A number $x$ is divisible by $6 \iff x$ is divisible by both $2$ and $3$.

Given any $3$ consecutive integers $n$, $ \ n\!+\!1$, $ \ n \! + \! 2$, at least one will be divisible by $2$ and exactly one will be divisible by $3$. So why will their product be divisible by $6$?

edited Sep 22 '18 at 20:46

answered Apr 10 '14 at 22:57

Kaj Hansen

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The product of any three consecutive natural numbers is divisible by 6

2 Answers2