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Question

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


Solution

Let the three consecutive positive integers be $$n$$, $$n+1$$ and $$n+2$$.

Whenever a number is divided by $$3$$, the remainder obtained is either $$0,1$$ or $$2$$.

Therefore, $$n=3p$$ or $$3p+1$$ or $$3p+2$$, where $$p$$ is some integer.


If $$n=3p$$, then $$n$$ is divisible by $$3$$.

If $$n=3p+1$$, then $$n+2=3p+1+2=3p+3=3(p+1)$$ is divisible by $$3$$.

If $$n=3p+2$$, then $$n+1=3p+2+1=3p+3=3(p+1)$$ is divisible by $$3$$.

So, we can say that one of the numbers among $$n, n+1$$ and $$n+2$$ is always divisible by $$3$$ that is:

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

Similarly, whenever a number is divided by $$2$$, the remainder obtained is either $$0$$ or $$1$$.

Therefore, $$n=2q$$ or $$2q+1$$, where $$q$$ is some integer.

If $$n=2q$$, then $$n$$ and $$n+2=2q+2=2(q+1)$$ is divisible by $$2$$.

If $$n=2q+1$$, then $$n+1=2q+1+1=2q+2=2(q+1)$$ is divisible by $$2$$.

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

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

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

Mathematics

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