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, 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, p₁, p₂,..., p_t. Then the number p₁p₂...p_t + 1 must contain a "new" prime not in your set (in fact, usually a big one) as a factor. Although I wouldn't recommend it as an efficient way of locating primes, it's a way which is guaranteed to work.

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Page last updated 30 March 1997.