V OM - I - Task 2

Investigate when the sum of the cubes of three consecutive natural numbers is divisible by $18$.
Let $ a - 1 $, $ a $, $ a + 1 $, be three consecutive natural numbers; the sum of their cubes

can be transformed in the following way:

Since one of the numbers $ a - 1 $, $ a $, $ a + 1 $ is divisible by $ 3 $, then one of the numbers $ a $ and $ (a + 1) (a - 1) + 3 $ is also divisible by $ 3 $. Therefore, the sum $ S $ of the cubes of three consecutive natural numbers is always divisible by $ 9 $.
The sum $ S $ is thus divisible by $ 18 = 9 \cdot 2 $ if and only if it is divisible by $ 2 $. Since $ S = 3a^3 + 6a $, $ S $ is even if and only if $ a $ is even, i.e., when $ a - 1 $ is odd. Hence, the conclusion:
The sum of the cubes of three consecutive natural numbers is divisible by $ 18 $ if and only if the first of these numbers is odd.