Problems & Puzzles:
Puzzles
Puzzle 245. As
13
13 is the earliest prime such that it’s leading digit
can take other five onedigit values and the resulting numbers are primes
too: (13, 23,
43, 53,
73, 83)
Question. Find the smallest
titanic prime as 13
Solution:
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 5tuplet by Norman Luhn and a CPAP5 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;
3,57,297,177,237,25111,231339,67419,273817,345111,2001753,912277,
5236153,9228627,10599391,2835261,60120003,14054037,27923629,41783347,
24590943,112161513,230484021,11446969.
***
(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
***
