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.
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!"
***
On Jan 9, 2019 Pierandrea Formusa wrote:
Find below my answer to your Puzzle
75, with regard to question 3 (3.I'll
be very glad to add here other curio suggestions related with primes
& the number 2000. ).
We can express 2000 like the sum of a
prime and the square of a prime and the cube of a prime and the
square of the square of a prime in a prime number of ways (13), with
all primes different. And 13 appear only 2 times considering all
solutions and 2 is the only even prime.
Below the 13 solutions:
p1 
p2 
p3 
p4 
p1 
p2^2 
p3^3 
p4^4 
sum 
1433 
19 
5 
3 
1433 
361 
125 
81 
2000 
953 
29 
5 
3 
953 
841 
125 
81 
2000 
113 
41 
5 
3 
113 
1681 
125 
81 
2000 
563 
5 
11 
3 
563 
25 
1331 
81 
2000 
419 
13 
11 
3 
419 
169 
1331 
81 
2000 
227 
19 
11 
3 
227 
361 
1331 
81 
2000 
59 
23 
11 
3 
59 
529 
1331 
81 
2000 
911 
11 
7 
5 
911 
121 
343 
625 
2000 
863 
13 
7 
5 
863 
169 
343 
625 
2000 
743 
17 
7 
5 
743 
289 
343 
625 
2000 
503 
23 
7 
5 
503 
529 
343 
625 
2000 
191 
29 
7 
5 
191 
841 
343 
625 
2000 
71 
31 
7 
5 
71 
961 
343 
625 
2000 
***
On Jan 16, 2019, Paul Cleary wrote:
No.5. 89393068 2^1272+1, sod =2000 and 391 digits.
No.6. 8567418*2^12721, sod = 2000 and 390 digits.
I found approx. 50 or so other solutions between 444
digits and 390, they do seem to have dried up though smaller than 390
digits.
***
