n is a natural number - Junior Cycle Mathematics - Question 2 - 2015

Let the three consecutive natural numbers be:

• $n$, $n + 1$, $n + 2$.

Calculating their sum, we have:

$ext{Sum} = n + (n + 1) + (n + 2) = 3n + 3$

This can be factored as:

$3(n + 1)$

Since $3(n + 1)$ is obviously divisible by 3, this proves that the sum of any 3 consecutive natural numbers is divisible by 3.

Let the four consecutive natural numbers be:

• $n$, $n + 1$, $n + 2$, $n + 3$.

Calculating their sum, we get:

$ext{Sum} = n + (n + 1) + (n + 2) + (n + 3) = 4n + 6$

Now, examining divisibility by 4, we note:

• The term $4n$ is divisible by 4.
• The term $6$ is not divisible by 4.

Hence, $4n + 6$ can be expressed as:

$4n + 6 = 4k + 2 ext{ where } k ext{ is an integer.}$

This means that $4n + 6$ is not divisible by 4 since it yields a remainder of 2. Therefore, the statement is true; the sum of any 4 consecutive natural numbers is never divisible by 4.