$0\cdot1\cdot2\cdot3\cdot4\cdot5 = 0$, but it works for any six consecutive whole numbers. What is the math behind this?
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2It actually works for any $5$ consecutive whole numbers. See answers below. – Bram28 Apr 05 '18 at 01:32
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See The product of $n$ consecutive integers is divisible by $n$ factorial. In your case, the product of $6$ consecutive integers is divisible by $,6!=720,$. – dxiv Apr 05 '18 at 01:40
4 Answers
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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.
$$\color{red}{0},1,2,3,4,5$$ $$1,\color{red}{2},3,4,\color{red}{5},6$$ $$\color{red}{2},3,4,\color{red}5,6,7$$ $$3,\color{red}{4},\color{red}5,6,7,8$$ $$\vdots$$
Andrew Li
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Either there's a number ending in $0$ or a $5$ and there's always an even number in there.
Cameron Williams
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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 $lcm(2,3,4,5,6)=60$
Tiago Emilio Siller
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The number divisible by $2$ and the number divisible by $5$ might be the same. – lhf Apr 05 '18 at 01:44