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01 Jul 2015, 00:50
Kudos
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Hello everybody,
lets assume we have 6 consecutive integers multiplied, for instance (n-3)(n-2)(n-1)n(n+1)(n+2). Now I want to perform a prime factorization in order to determine possible factors of this expression. Of course I cannot determine all factors but at least some of which I know for sure that they must be included.
Since we have 6 consecutive integers, at least one number within that set must be divisible by 1,2,3,4,5, and 6. Next, I wanted to prime factorize these numbers:
1 = 1
2 = 2
3 = 3
4 = 2*2
5 = 5
6 = 2*3
As you can see, some of the prime factors overlap. In total we have:
1,2,2,2,2,3,3,5
Can I conclude that all the above stated factors are included in the set or do I have to eliminate the overlaps? If all the factors are included, then any combination of these factors must also be a factor of the expression, for instance: 2*2*2*3 = 24 must be a factor of (n-3)(n-2)(n-1)n(n+1)(n+2)
lets assume we have 6 consecutive integers multiplied, for instance (n-3)(n-2)(n-1)n(n+1)(n+2). Now I want to perform a prime factorization in order to determine possible factors of this expression. Of course I cannot determine all factors but at least some of which I know for sure that they must be included.
Since we have 6 consecutive integers, at least one number within that set must be divisible by 1,2,3,4,5, and 6. Next, I wanted to prime factorize these numbers:
1 = 1
2 = 2
3 = 3
4 = 2*2
5 = 5
6 = 2*3
As you can see, some of the prime factors overlap. In total we have:
1,2,2,2,2,3,3,5
Can I conclude that all the above stated factors are included in the set or do I have to eliminate the overlaps? If all the factors are included, then any combination of these factors must also be a factor of the expression, for instance: 2*2*2*3 = 24 must be a factor of (n-3)(n-2)(n-1)n(n+1)(n+2)
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01 Jul 2015, 02:09
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nicok06
Yes, you can - you don't want to eliminate the overlaps. The general rule is this: the product of k consecutive positive integers is always divisible by k!
So, for example, if you multiply six consecutive integers, the product will always be divisible by 6! = 6*5*4*3*2*1, and of course by any factor of 6! as well.
The reason this is all true is because multiples are equally spaced - since multiples of 6 are 6 apart, then in any list of six consecutive integers, we must always have exactly one multiple of 6. Similarly, since multiples of 5 are 5 apart, we always must have at least one multiple of 5 (and possibly two of them) among any six consecutive integers, and so on.
29 Jan 2018, 00:19
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Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
