Clan x86
General Forums => Academic / School => Math and Other Problems => Topic started by: Ender on May 23, 2007, 11:06:09 am
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Prove that there exists a string of n consecutive composite numbers for any number n. In other words, prove that there are strings of numbers as long as you like that haven't a single prime in them.
Proof will be posted next Wednesday unless requested otherwise.
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Consider for example, n!+9, n!+10, n!+11, n!+12, ..., n! + n
where n is any natural number >= 9.
We have a sequence of n-9 consecutive non-prime integers. Since n-9 is unbounded, the desired result follows.
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Yep =) that's right.
The factorial is the key move. In the form n! + a, if you can factor a out and make the sum strictly a product of numbers, then n! + a obviously has to be composite. Oh, and I probably should have clarified in my post of the problem that "composite" implies we're doing this over the natural numbers.
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Oh, and I probably should have clarified in my post of the problem that "composite" implies we're doing this over the natural numbers.
I didn't know what a composite number was, but In other words, prove that there are strings of numbers as long as you like that haven't a single prime in them.
made the question quite clear.
I like how the solution isn't very fussy. :)