Question

Show that only one of the numbers $n$, $n + 2$ and $n + 4$ is divisible by 3.

Answer 1

We know that any positive integer of the form $3q$ or, $3q + 1$ or $3q + 2$ for some integer q and one and only one of these possibilities can occur. Here we need to prove for $n$, $n + 2$ and $n + 4$ in which by applying $n = 3q$, $n = 3q + 1$ and $n = 3q + 2$ then simplify the terms to find out all the possibilities when it is divisible by 3.

Complete step by step solution:
Any number of the form of $3q$, $3q + 1$or $3q + 2$, we have following cases:
Case I: When $n = 3q$
In this case we have,
$n = 3q$, which is divisible by 3,
Now again, $n = 3q$
$n + 2 = 3q + 2$
Hence, $n + 2$ leaves remainder 2 when divided by 3 hence, $n + 2$ is not divisible by 3.
Now again, $n = 3q$
$n + 4 = 3q + 4 = 3\left( {q + 1} \right) + 1$
Hence, $n + 4$ leaves remainder 1 when divided by 3 hence, $n + 4$ is not divisible by 3.
Thus, we can say that $n$ is divisible by 3 but $n + 2$ and n+4 is not divisible by 3.
Case II: When $n = 3q + 1$
In this case we have,
$n = 3q + 1$Here, $n$ is not divisible by 3 as n leaves remainder 1.
Now, $n = 3q + 1$
$n + 2 = \left( {3q + 1} \right) + 2 = 3q + 3 = 3\left( {q + 1} \right)$
Hence, $n + 2$is divisible by 3.
Again $n = 3q + 1$

$n + 4 = 3q + 1 + 4 = 3q + 5 = 3\left( {q + 1} \right) + 2$
Hence, $n + 4$ leaves remainder 2 when divided by 3, hence it is not divisible by 3.
Case III: When $n = 3q + 2$
In this case we have,
$n = 3q + 2$
Here, $n$ is not divisible by 3 as it leaves remainder 2.
Now, $n = 3q + 2$
$n + 2 = 3q + 2 + 2 = 3q + 4 = 3\left( {q + 1} \right) + 1$
Here, $n + 2$ leaves remainder 1 when divided by 3 hence, it is not divisible by 3.
Again, $n = 3q + 2$
$n + 4 = 3q + 2 + 4 = 3q + 6 = 3\left( {q + 2} \right)$
Here, $n + 4$ is divisible by 3.
Therefore, $n + 4$ is divisible by 3 but $n$ and $n + 2$ is not divisible by 3.
Thus, one and only one out of n, $n + 2$, $n + 4$ is divisible by 3.

Note:
The key point to prove the number $n$, $n + 2$ and $n + 4$ divisible by 3 is that we need to consider the $n = 3q$, $n = 3q + 1$ and $n = 3q + 2$ for the given values with respect to $n$ and a number is said to be divisible if does not contain remainder, if remainder exists then it is not divisible by the given number.