By direct proof, how can you prove that “The sum of any three consecutive even integers is always a multiple of 6”.

Let’s denote the three consecutive even integers as n, n+2, and n+4. The sum of these three integers is n + (n+2) + (n+4) = 3n + 6. We can factor out a 3 from this expression to get 3(n + 2). Since n is an even integer, n + 2 is also an even integer. Therefore, the expression 3(n + 2) is a multiple of 6 because it’s the product of 3 and an even integer. Hence, we have proved that the sum of any three consecutive even integers is always a multiple of 6.