Problems & Puzzles:
Puzzles
Puzzle 75. Prime numbers and the
number 2000
I couldn’t resist the temptation of calculating
some prime numbers related to the number 2000 (the
conspicuous year coming). Below you can found six
claims. I guess that the first three claims are
ending absolute results, while the other three for sure
may be improved.
1. Smallest prime with D = 2000 digits:
N = 10^1999 + 7321 = 1(0)_{1995}
7321_{1995}
7321, SOD = 14
(Actually a strong pseudoprime)
2. Largest prime with D = 2000 digits:
N = 10^2000 9297= (9)_{1996}
0703_{1996}
0703, SOD = 17964
(Actually a strong pseudoprime)
3. Smallest prime such that SOD = 2000:
N = 3*10^222 +(10^2221) –10^125 = 3(9)_{96}
8(9)_{125} , D = 223
(Absolute prime verified with APRTCLE)
4. Smallest calculated prime such that D
= SOD = 2000:
N=10^1999+9*10^(1643+222)+(10^2211)*10+1 =
1(0)_{133} 9 (0)_{1643}
(9)_{221} 1_{133} 9 (0)_{1643}
(9)_{221} 1
(Actually a strong pseudoprime)
5. Smallest calculated prime
k*2^n +
1 such that SOD=2000
N = 75727*2^1458 +1, D=444
(Absolute prime verified with Proth.exe)
6. Smallest calculated prime
k*2^n 
1 such that SOD=2000
N = 139119*2^1457  1, D=444
(Absolute prime verified with Proth.exe)
Questions:
1.Would you like to improve the claims 4, 5 &
6?
2.Can you redo the exercise for 1, 2, 3 & 4 using
palprimes?
3.I'll be very glad to add here other curio suggestions
related with primes & the number 2000.
Solution
Chris Nash found "the smallest prime with
2000 digits and digit sum 2000. Here is his email sent at
29/11/99, solving completely the 4th claim of this
puzzle:
"the smallest prime with 2000 digits
and digit sum of 2000. It is 1(0)_{1776}3(9)_{7}8(9)_{96}8(9)_{117}.
It is probable prime, and as yet not proven, but could be
provable I am certain with a little extra factoring work
on N+1. It is certainly the smallest such prime. It was
discovered as follows. First the number
X = 1(0)_{1776}3(9)_{222}
= 10^1999 + 4*10^222  1
was constructed. It has sum of digits 2002. Then a search
was done for
primes of the form
X  10^(223n)  10^(223k)
using PrimeForm and searching from n=1 to 223, k=1 to
223. Note that all numbers of this form have 2000 digits
and digit sum 2000, finally note the first prime found by
this search will be smaller than any other. The result
followed shortly afterwards  n=9, k=106, giving the
closed form
N = 10^1999 + 4*10^222  10^214  10^117  1."
Latter he added: "Looking at it again, I
thought it very interesting (at least for Americans!)
that this 2000 digit prime with sum of digits 2000 had
precisely 1776 zeroes in it  it is also a '4th of July'
prime as well as a 'millenium' prime..."
Chris Nash has also found smaller
primes of the form k*2^n+/1 such that SOD=2000:
33401*2^1373+1
67893*2^13711
Both with 418 digits.
"I used a modified version of PrimeForm which
first tested the sum of digits, then trial factoring,
finally primality proof... I'm going to attempt n=1350,
but I doubt I will find an answer, the probability here
is very very close to zero... there may be a smaller
solution than these, but it will take a lot of searching
I think!"
