Among any three consecutive positive integers one is a multiple of 3

Solution 1:

• One of the numbers in $(1;2;3)$ is divisible by $3$.

• Suppose one of the numbers in $(n;n+1;n+2)$ is divisible by $3$.

• Now consider $(n+1;n+2;n+3$).

  • If $n$ was divisible by $3$, then so is $n+3$. So that is OK.
  • If either of $n+1$ or $n+2$ was divisible by $3$, then they still are now. So in that case we're also good.

• Again by using induction and with a similar argument, you prove that this statement also holds for all non-positive integers.

By induction, the statement is now proven, for all triples of consecutive integers. Thus in particular for $(8q;8q+1;8q+2)$.

The same arguments can be used to prove that out of any $N$ consecutive integers, one is a multiple of $N$.

Solution 2:

I have seen many answers to related questions that in the context of this question, would be along the lines of "consider all possible values of $q$ modulo $3$". Here is a different approach.

If we multiply everything together modulo $3$, we get:

$$8q(8q + 1)(8q + 2) \equiv 2q^3 + q\pmod 3$$

By Fermat's Little Theorem,

$$2q^3 + q \equiv 2q + q \equiv 0\pmod 3$$

This implies that $3$ divides $(8q)(8q +1)(8q+2)$.

Since $3$ is a prime and it divides $(8q)(8q +1)(8q+2)$, then by Euclid's Lemma, it divides at least one of $8q$ and $(8q +1)(8q+2)$.

If it divides $8q$, then we are done. If not, it must divide $(8q+1)(8q+2)$. Again, using Euclid's lemma, it must divide at least one of $(8q+1)$ and $(8q +2)$.

It follows that at least one of the three must be divisible be $3$.