Why is the one's digit of the product of 6 consecutive positive whole numbers is always zero?

Question

$0\cdot1\cdot2\cdot3\cdot4\cdot5 = 0$, but it works for any six consecutive whole numbers. What is the math behind this?

Answer 1

That's because if you have six consecutive numbers, there will always be an even number or a five or $0$ within those consecutive numbers.

$$\color{red}{0},1,2,3,4,5$$ $$1,\color{red}{2},3,4,\color{red}{5},6$$ $$\color{red}{2},3,4,\color{red}5,6,7$$ $$3,\color{red}{4},\color{red}5,6,7,8$$ $$\vdots$$

Answer 2

In every six consecutive numbers, there is a multiple of $2$ and a multiple of $5$.

Answer 3

Either there's a number ending in $0$ or a $5$ and there's always an even number in there.

Answer 4

Given $6$ consecutive numbers, at least one is divisible by $2$ and at least one is divisible by $5$. So the product is divisible by $10$.

Furthermore, at least one is divisible by $3$ and at least one is divisible by $4$. Then the product of six consecutive numbers is always divisible by $lcm(2,3,4,5,6)=60$