Problems & Puzzles: Puzzles

 Puzzle 936. Palprimes such that... Paul Cleary sent the following nice puzzle, The numbers 10000900001 and 30011111003 are the smallest and largest Palindromic primes who’s digit sum is equal to the numbers length or number of digits and who’s digit sum is also a palindromic prime>7.   Q. Find the next smallest and largest numbers that have the same property.

Contriburtions came from Emmanuel Vantighem and Pierandrea Formusa

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Emmanuel wrote:

The next solutions should should  be found among the 101-digits numbers.

The smallest I could find is :
10000000000000000000000000000000000000000000489799799798400000000000000000000000000000000000000000001
The biggest :
99999400001000000000000000000000000000000000000000100000000000000000000000000000000000000010000499999

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Pierandrea wrote:

- next smallest is 10002520001

PARI/GP code to get the above solution:
is(n)=n==eval(concat(Vecrev(Str(n))))&&isprime(n) && sumdigits(n)==length(Str(n)) && sumdigits(n)==eval(concat(Vecrev(sumdigits(n)))) && sumdigits(n)>7
forprime(p=10000900002,10500000000,if(is(p),print(p);break));

- next largest is 30000500003

Python code to get the above solution:
from sympy import prevprime,isprime

def pal(l):
return(l == l[::-1])

def SOD(num):
#it returns Sum Of Digits
s=str(num)
ris=0

for k in s:
ris += int(k)

return(ris)

n=prevprime(30011111003)

while True:
if pal(str(n)) and SOD(n)==len(str(n)) and isprime(SOD(n)) and pal(str(SOD(n))) and SOD(n)>7:
print(n)
break
n=prevprime(n)

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Jan van Delden wrote on Dec 29, 2018:

Smallest solutions for prime p are of the form:

1{0}[k]R(n)d(n){0}[k]1 where d is odd and R(n) is the reverse of n.

p  d n k

5  3 0 1

7  1 2 1

11 9 0 4

13 7 2 4

17 1 52 5

19 3 52 6

23 3 72 8

29 7 64 11

31 7 92 12

37 7 662 14

41 9 87 17

43 5 981 17

47 9 972 19

53 9 885 22

59 9 789 25

61 9 7981 25

67 7 7976 28

71 9 9984 30

73 7 8996 31

79 9 9988 34

83 7 79993 35

89 7 99994 38

97 9 789991 41

For instance we have p=5 10301 and p=13 1000027200001

A larger solution:

1009 9 99999999899999999999999999999999999999999999999999999995 446

I also searched for solutions that are maximal in the sense that d(n)={9}[k]e, with e the 'last' digit, but I only found p=11 (p<10000).

Largest solutions are of the form:

for prime p are of the form:

n{0}[k]d{0}[k]R(n) where d is odd and R(n) is the reverse of n.

p  n k d

5  1 1 3

7  3 2 1

11 30011 0 1

13 31101 1 1

17 700001 2 1

19 710001 3 1

23 90100000001 0 1

29 9400000000001 1 1

31 96 13 1

37 9800000001 8 1

41 990100001 11 1

43 992000000000001 6 1

47 993100000000001 8 1

53 9970000000000001 10 1

59 9991000000000000000001 7 1

61 9992000000001 17 1

67 9995000001 23 1

71 99970000000000000001 15 1

73 9998000000000001 20 1

79 9999100100000000000000000000000000001 2 1

83 9999400001 31 1

89 9999701 37 1

97 9999920001 38 1

For instance for p=7 we have: 3001003 and for p=11 we have 301111103

A larger solution:

1009 99999999999999999999999999999999999999999999999999999997000100000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000001 260 1

I also searched for solutions that are maximal in the sense that n={9}[k]e, with e the 'last' digit and d=1. I found p=7,31,4409 (p<10000).

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