Problems & Puzzles: Puzzles

Puzzle 1193 6666666666 Loungrides found (when?) that 6666666666 is only 10-digit rep-digit number such that the sum of itself and all of its truncations plus one is prime. See this curio: 6666666666+666666666+66666666+6666666+666666+66666+6666+666+66+6+1=7407407401. [Loungrides]   Q. Find a larger prime alike the given above. Use only rep-digit integers using the digit 6 more that 10 times at the start.

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From 19 to 25 October, 2024, contributions came from Emmanuel Vantieghem, Vicente Felipe Izquierdo, Michael Branicky, Gennady Gusev, Ken Wilke, JM Rebert

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Emmanuel wrote:

These are the  n  > 10  such that the repunit with  n  times  6  added by its truncations plus 1  gives a prime :
 20, 27, 35, 42, 141, 146, 153, 182, 218, 358, 386, 440 (the next one is > 1000  and sets us up with prime proving).

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Vicente wrote:

We can obtain this sum in the following way:

s = 1+ Sum(6 (10^k-1)/9,{k,0,10}]

resolving:

s= 1 + 1234567900 x 6

If we generalize, then

s = 1 + 1234567900 x d, where 0<=d<=9

For d = 7 and 10 digits, we have a Prime greater then with digit 6: 8641975301 

With the digit 6, the lenght for which the sum is prime, is:20, 27, 35, 42,141,146,153,182,218,...

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Michael wrote:

Testing repdigits with all 6's up to length n = 10000, the sum of itself and all of its truncations plus one is prime

for n = 1, 2, 3, 5, 6, 10, 20, 27, 35, 42, 141, 146, 153, 182, 218, 358, 386, 440, 2693, 5907, 6748, 7281, 9897.

 

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Gennady wrote:

The sum 1+k+kk+kkk+...+'k..k' (n times) = k*((10^(n+1)-9*n-10)/81)+1 = s(k,n)

Question from puzzle is: find prime s(6,n) for n>10.

I have checked all s(6,n) for n<=13000. s(6,n) is prime for n:

1, 2, 3, 5, 6, 10, 20, 27, 35, 42, 141, 146, 153, 182, 218, 358, 386, 440, 2693, 5907, 6748, 7281, 9897.

Also, s(k,n) is a prime number for different k for the following n (n<=3000):

k:  n

1: 12, 58

2: 116, 128, 366, 1181

3: 12, 20, 100, 312

4: 19, 53, 54, 87, 769, 1188, 2974

5: 20, 1278

7: 14, 158, 716

8: 170, 900

9: 70, 102, 670, 882

 

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Ken wrote:

Comment: Since any summand composed of only k 6’s has the form 2*(10^k-1)/3 ,
the sum 1 + 6 + 66 + * * * + 66 * * * 6 (k times) is given by
m = 1+2*{ ( 10^(k + 1) - 1)/9 – (k+1) }/3. (1)
Using the ECM (elliptic curve) program from UBASIC, suitably modified to accommodate
integers of the form (1), we obtain the following results: {the notation (740)(n) means
that the block of digits 740 is repeated n times followed by the remaining digits in the
number |
k prime
1 7
2 73
3 739
5 74071 = (740)(1) 71
6 740737 = (740)(1) 737
10 7407407401=(740)(3) 1
20 74074074074074074061 = (740)(6) 61
27 7407407407407407407407407723 = (740)(8) 723
35 74074074074074074074074074074074061 = (740)(11) 51
42 740740740740740740740740740740740740740713 = (740)(13) 713
I found no other prime values for 42 &lt; k &lt;141. Oddly enough one can easily show that
for k==4( mod 5) must be divisible by 5.
Using Dario Alpern’s Integer Factorization Calculator,
(www.alpertron.com.ar/ECM.HTM),
I was able to extend the above table to k&lt;201 with the following additional primes
found:

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Rebert wrote:

n p_n

1 7

2 73

3 739

5 74071

6 740737

10 (740)_(3)1 = 7407407401

20 (740)_(6)61 = 74074074074074074061

27 (740)_(8)723

35 (740)_(11)51

42 (740)_(13)713

141 (740)_(46)647

146 (740)_(47)73977

153 (740)_(50)639

182 (740)_(59)73953

218 (740)_(71)73929

358 (740)_(118)7169

386 (740)_(127)73817

440 (740)_(145)73781

2693 (740)_(896)72279

5907 (740)_(1967)736803

6748 (740)_(2248)2909

7281 (740)_(2425)735887

9897 (740)_(3297)734143

25917 (740)_{8637}723463

 

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