Problems & Puzzles: Puzzles

Puzzle 208.  Happy new Year 2003!

Better late than never. Please all of you have my very best wishes for a new prime year (2003) like this we are just starting.

Now, is there a better way of celebrating it, than posing some interesting puzzles about this prime number?

I will start with one puzzling question; you are invited not only to solve it but to send other interesting questions about this prime number, OK?

Q1. What is the least prime number p such that the period of 1/p, in its decimal expansion, is equal to 2003?

Solution: Solutions came from Johann Wiesenbauer, Phil Carmody and J. K. Andersen. All of them provided the same solution: the prime number 2231588810593399.

Other interesting questions related to the prime 2003  became from Johann Wiesenbauer and Jon Wharf. See below.

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Wiesenbauer obtained it factoring 10^2003-1 using a " a small ECM-program written in DERIVE 5".

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Carmody simply wrote "probably found by Harvey Dubner, but I'm not sure.".

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J.K. Andersen wrote "found in 1 minute with Michael Scott's factor.exe (modified to allow larger numbers)".

He also reported that Carmody wrote to the primenumbers list "that the factor 2231588810593399 was known and is in the file http://groups.yahoo.com/group/primenumbers/files/Factors/Base10.zip"

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So two of our readers re-discovered a number that was already produced by Harvey Dubner (for other purposes?)

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Regarding the follow up questions about 2003 we have:

(1997, 1999, 2003) is a so-called prime triple of the form (p,p+2,p+6). The one before had been (1871, 1873, 1877) and the next one will be (2081, 2083, 2087). If you look at the "distances" 126 and 84, respectively, between the triples, are they considered to be "large" or "small" when compared with an expected value obtained by heuristic reasoning?

What is the largest set of primes such that the average of the whole set is
2003, but the average of any subset is not?

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J. K. Andersen sent (3/2/03)the following follow-up puzzle about 2003:

Smarandache(2003)

Let N be the decimal concatenation of the natural numbers to 2003: N = 12345678910111213...20022003. Unfortunately N is not a prime. Let us try to do something about that.

Split N into two or more parts all of them prime numbers, which concatenate to N. The splits may be anywhere in N except before a 0. Primes are allowed to occur more than once. Probable primes are acceptable for questions 1, 2, 3.

Questions

1. Minimize the number of primes used in a split.
2. Maximize the smallest of the primes in a split.
3. In how many ways can the splits be made?
4. Find a split with proven primes.
5. Now try the other way: Concatenate N with one or more freely chosen primes to make a prime. Prove the prime.

I like question 5 because it may at first seem infeasible to find a prime of this size which can be proven. I have not done it but know how to.

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Faride Firoozbakht wrote (May 2003):

A nice property of 2003:

If we set 1111 between each two successive prime numbers from 2 until 2003,we obtain a number with 2231 digits which is prime number

(n=211113111151111...19971111199911112003 is prime).

Later (August, 2003) she added:

2^13+3^13+5^13+...+2003^13 is prime.

Previous prime with this property is 1361 and next prime is 3851. So such property has happened 642 years ago, and will be happen after 1848 years.

I couldn't find another prime number p such that 2^p+3^p+5^p+...+2003^p is prime.

Is 13 (or any other number) unfortunate in your culture? If it is so, what is the reason?

Another property singular for the prime 2003, sent by Faride is the following one:

"2003 is the only prime of the form 2*10^p+p,where p and p+4 is prime."

Proof: If p and p+4 is prime then p is 3 or p must be equal to 1 (mod 3). Thus if p > 3 , 2*10^p+p is equal to 0 (mod 3) and it can not be prime.

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