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Consider $n > 1$ consecutive integers $a, a+ 1, . . . , a+ (n−1)$. Prove that the product thereof is always a multiple of $n!. (Think about a number of the form C y x .)

vitamin d
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Semjoon
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    Welcome to MSE. You'll get a lot more help if you show that you have made a real effort to solve the problem yourself, even if you haven't made much progress. What are your thoughts? What have you tried? How far could you get? Where are you stuck? This question will likely be closed if you don't add more context. Please respond by editing the question body. Clarifications don't belong in the comments. – saulspatz Feb 20 '21 at 14:15
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    https://math.stackexchange.com/questions/12067/the-product-of-n-consecutive-integers-is-divisible-by-n-without-using-the – Joffan Feb 20 '21 at 14:34
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    Combinatoric hint: the product could be written as $(a+n-1)!/(a-1)!$. – Joffan Feb 20 '21 at 14:36

1 Answers1

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(Not a native English speaker, so please pardon my mistakes)

Sorry, but I can't think of anything using C y x. But the proof below may be the most common one.

Consider a mod n, (a + 1) mod n, ... (a + (n - 1) ) mod n. (1)

Obviously, these values are all different. There are n values.

On the other hand, the possible results of (a number mod n) include 0, 1, ..., n-1. There are also n values.

So we can say that the values in (1) take up 0, 1, ..., n-1. Therefore, one of them must be 0, so there must be a number in a, a+1, ..., a+n-1 that is a multiple of n. So the product must be a multiple of n.

FreeLieAlgebra
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  • The task is to proof thet the product is divisible by $n!$ (the factorial of $n$) , not by $n$. – Peter Feb 21 '21 at 07:44