Problems & Puzzles: Puzzles

Puzzle 245.  As 13

13 is the earliest prime such that itís leading digit can take other five one-digit values and the resulting numbers are primes too: (13, 23, 43, 53, 73, 83)

Question. Find the smallest titanic prime as 13


Three puzzlers informed that the solution to this puzzle may not to came but after several years of PC computing: J. K. Andersen, Luke Pebody and Faride Firoozbakht.

Andersen wrote:

3 of the 9 possible numbers will always be divisible by 3. Then we need 6 simultaneous titanic primes. The largest known number of simultaneous titanic primes is 5. The only such cases I know is a 5-tuplet by Norman Luhn and a CPAP-5 by Jim Fougeron. I estimate 6 would take more than 30 GHz years and I am not trying.

Pebody wrote:

There is no number n smaller than 2*3*5*7*11*13*17*19=223092870 such that k*10^499+n is prime for all k in {1,2,4,5,7,8} or for all k in {1,3,4,6,7,9}. I expect there will be before around 10^30. Can't find one though

Faride wrote:

Let a(n) be the smallest number m such that m < 10^n and all the six numbers 10^n+ m,(Mod[m,3]+2)*10^n+ m, 4*10^n+ m,(Mod[m,3]+5)*10^n+ m ,7*10^n+ m &(Mod[m,3]+ 8)*10^n+ m are primes. In fact you want 10^999 + a(999).

I haven't yet been able to find even smallest m such that the first three numbers of the above six numbers are primes for n=999. I think we aren't able to find a(999) without using a supper computer.

So far, I only had been able to find a(n) for n=1,2,...,24 as follows;



(Note by CR) If the Faride's results are plotted in Excel and a trend 'potential' function is asked, we obtain that m is approximately equal to 0.5*n^6; this means that for n=999 m=5x10^17, approximately




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