Question

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

75

likes373 views

Esmeralda

4.7

78 Answers

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.

Frequently asked questions (FAQs)

Math question: What is the vertex form of the parabola function 𝑦 = 2𝑥² + 4𝑥 + 3? (

+

What is the result of adding the vectors u = {-3, 5} and v = {7, -2}?

+

What is the square root of 2976?

+

New questions in Mathematics