Can a product of 4 consecutive natural numbers end in 116

Question

So i was given this question with two parts:

(a) Prove that the product of two consecutive even numbers is always divisible by 8.

(b) Can a product of 4 consecutive natural numbers end in 116?

For part (a) here is my solution

Let the first even number be $2n$. The second will be $2n + 2$.

$$2n(2n + 2)$$

$$= 4n^2 + 4n$$

$$= 4(n^2 + n)$$

$$= 4n(n + 1)$$

If n is even:

$$= 4(2k)(n + 1)$$

$$= 8k(n + 1)$$

This is divisible by 8.

If n is odd

$$= 4(2k + 1)(2k + 1 + 1)$$ $$= 4(2k + 1)(2k + 2)$$ $$= 4(2k + 1)2(k + 1)$$ $$= 8(2k + 1)(k + 1)$$

This is divisible by 8

How do you do part (b)?

Answer 1

Part a: you are forgetting that $n$ could be odd.

Part b: among four consecutive numbers you surely find two consecutive even numbers. If the product can be written $p=1000x+116$, note that $1000$ is divisible by $8$ and, by part a, $p$ is divisible by $8$. So…

Answer 2

I recognize the merit of the exercise. But it could be mentioned that the product of four consecutive natural numbers ends in 0 or in 4. The latter occurs when the first of the four numbers ends in 1 or 6.