S10 - 5,k Go the easy route and take N = k Clearly then k 2 k 3 k 4 k k are all composite numbers These observations give one way

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Solution to Problem 10 Congratulations to this week’s winner Daniel Reeves There were two other correct solutions received, from Mike Fitzpatrick and from a Bradley mathematics professor and runner extraordinaire. You are trying to find a number N for which N + 1, N + 2, N + 3,. .., N + k are all composite. What number divides N + 2? The only thing you know about this number is that, if N were divisible by 2, then N + 2 would be, too. If N were divisible by 3, then N + 3 would be, too. Continuing, if N were divisible by r, then N + r would be, too. How can you construct a number which is divisible by 2, 3, 4,
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Unformatted text preview: 5,. ..,k? Go the easy route and take N = k!. Clearly, then, k! + 2, k! + 3, k! + 4,. .., k! + k are all composite numbers. These observations give one way of "constructing" new primes. Suppose you have a collection of primes, say, p 1 , p 2 ,..., p t . Then the number p 1 p 2 ...p t + 1 must contain a "new" prime not in your set (in fact, usually a big one) as a factor. Although I wouldnt recommend it as an efficient way of locating primes, its a way which is guaranteed to work....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.

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S10 - 5,k Go the easy route and take N = k Clearly then k 2 k 3 k 4 k k are all composite numbers These observations give one way

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