Problems & Puzzles: Puzzles

Puzzle 1228 Extension to Puzzle 1227

Now we will ask for examples of primes with the most distinct primes concatenated.

For example Michael Branicky & Simon Cavegn both sent this one using 6 distinct primes

101193759613 -> 101193759612 = 2^2 * 3^2 * 11 * 19 * 37 * 59 * 61 * 101


Q. Send your largest example with the most distinct primes you can find (more that 6)?


Important note:

Paul Cleary sent his solutions to puzzle 1227 a little bit later (Saturday 5, 2025, 3:06 am). This is why I couldn't publish these result together with the rest; I usually post new results between 4-5 pm on each friday...The worst of this is that he already got two examples with seven distinct primes...!!! improving twice the example given above, gotten by Branicky & Cavegn...

1732317511   {17,3,2,31,7,5,11}   1732317510  --  2 * 3 * 5 * 7 * 11 * 17 * 31 * 14232547377911   {2,5,47,3,7,79,11}   2547377910  --  2 * 3^4 * 5 * 7 * 11^2 * 47 * 79

Later Paul wrote this:

 "Thanks for your patience. It’s been a hectic weekend, and I apologize for the slow reply.
After reviewing my work, I see that I misread the problem—d needed to be prime—and some of my submitted results aren’t. To preserve the integrity of your pages, please feel free to remove my submission. That error is entirely on me.Thanks for your understanding."

According with this all the submissions from Mr. Cleary to this puzzle and he previous one, Puzzle 1227, must be forgotten.

***

From 5-11 July, 2025. contributions came from Michael Branicky, Paul Rebert, Simon Cavegn, Oscar Volpatti,

***

Michael wrote:

Here is a solution with 7 distinct primes:
233118097213 = (23)(3)(11)(809)(7)(2)(13)-> 233118097212 = 2^2 x 3 x 7^2 x 11 x 13 x 23 x 149 x 809

Solution for 9 distinct primes:

79713347223511 = 79.7.13.3.47.2.23.5.11 -> 79713347223510 = 2 x 3^2 x 5 x 7 x 11 x 13^2 x 23 x 47 x 79 x 797

***

Paul wrote:

Q. Send your largest example with the most distinct primes you can find (more that 6)?
I found:
primes                    d                  d-1             = factorization
[5, 3, 13, 19, 2, 31] 531319231 531319230  = 2 * 3^3 * 5 * 13 * 19 * 31 * 257;
[7, 3, 2, 13, 41, 31] 732134131 732134130  = 2 * 3 * 5 * 7 * 13 * 31 * 41 * 211;

[2, 3, 19, 103, 7, 11] 2319103711 2319103710  = 2 * 3^4 * 5 * 7 * 11 * 19^2 * 103;
[7, 73, 2, 13, 3, 137] 7732133137 7732133136  = 2^4 * 3^2 * 7 * 13 * 59 * 73 * 137;
[113, 29, 5, 7, 2, 11] 1132957211 1132957210  = 2 * 5 * 7 * 11 * 29 * 113 * 449;
[379, 17, 31, 2, 7, 3] 3791731273 3791731272  = 2^3 * 3 * 7 * 17 * 31 * 113 * 379;

[2, 3, 79, 43, 1367, 7], 23794313677, 23794313676 = 2^2 * 3 * 7 * 43 * 61 * 79 * 1367;
[3, 2, 13, 37, 83, 163], 32133783163, 32133783162 = 2 * 3 * 13 * 37 * 83 * 163 * 823;
[2, 7, 97, 431, 79, 11], 27974317911, 27974317910 = 2 * 5 * 7 * 11^2 * 79 * 97 * 431;
[5, 7, 11, 353, 2, 631], 57113532631, 57113532630 = 2 * 3^2 * 5 * 7 * 11 * 37 * 353 * 631;
[7, 3, 83, 1231, 5, 31], 73831231531, 73831231530 = 2 * 3^2 * 5 * 7 * 31 * 37 * 83 * 1231;
[5, 2, 139, 37, 3, 991], 52139373991, 52139373990 = 2 * 3 * 5 * 11 * 31 * 37 * 139 * 991;

[2, 3, 13, 5, 39619, 31], 231353961931, 231353961930 = 2 * 3^2 * 5 * 7 * 13 * 23 * 31 * 39619;
[2, 3, 79, 43, 1367, 61], 237943136761, 237943136760 = 2^3 * 3 * 5 * 7 * 43 * 61 * 79 * 1367;
[2, 3, 97, 1279, 17, 19], 239712791719, 239712791718 = 2 * 3 * 17 * 19 * 97 * 997 * 1279;
[2, 3, 97, 1279, 17, 19], 239712791719, 239712791718 = 2 * 3 * 17 * 19 * 97 * 997 * 1279;


[2, 5, 47, 3, 7, 79, 11]  2547377911  2547377910   = 2 * 3^4 * 5 * 7 * 11^2 * 47 * 79;
[17, 3, 2, 3, 18, 5, 11]  1732317511  1732317510   = 2 * 3 * 5 * 7 * 11 * 17 * 31 * 1423;
[2, 5, 3, 7, 37, 353, 11] 25373735311 25373735310  = 2 * 3 * 5 * 7 * 11 * 29^2 * 37 * 353;
[2, 5, 37, 3, 7, 353, 11] 25373735311 25373735310  = 2 * 3 * 5 * 7 * 11 * 29^2 * 37 * 353;
[5, 2, 89, 193, 3, 7, 11] 52891933711 52891933710  = 2 * 3 * 5 * 7 * 11 * 31 * 43 * 89 * 193;
[7, 23, 5, 3, 337, 2, 11] 72353337211 72353337210  = 2 * 3^3 * 5 * 7 * 11 * 23 * 337 * 449;
[47, 3, 31, 41, 7, 2, 11] 47331417211 47331417210  = 2 * 3 * 5 * 7^4 * 11 * 31 * 41 * 47;
[61, 89, 5, 3, 23, 2, 31] 61895323231 61895323230  = 2 * 3 * 5 * 13 * 23 * 31 * 41 * 61 * 89;
[211, 5, 3, 7, 71, 2, 41] 21153771241 21153771240  = 2^3 * 3 * 5 * 7 * 41^2 * 71 * 211;
[3, 7, 5701, 2, 5, 89, 41], 375701258941, 375701258940 = 2^2 * 3 * 5 * 7 * 41 * 43 * 89 * 5701;
[2, 29, 3, 31, 67, 83, 43, 7], 2293316783437, 2293316783436 =  2^2 * 3 * 7 * 29 * 31 * 43 * 67 * 83 * 127;

***

Simon wrote:

11, 5, 2, 17, 7, 61, 31, 3, 19, 53, 11, 115217761313195311, 115217761313195310 = 2 * 3 * 5 * 7 * 11 * 17 * 19 * 31 * 53 * 61 * 947 * 1627

Yes 11 is repeated.
Reading the question as "the most primes, all distinct", then it will be basically exactly the same as Puzzle 1227 Q2.
Therefore I did interpret it as "the most distinct primes", allowing some non-distinct.

***

A little bit later, Oscar Volpatti sent the folllowing contributions:

As a first step, I performed an exaustive search up to p < 10^13:
generate every prime p in ascending order, try to cut p in every permitted way.
Numbers with d digits can be cut in 2^(d-1) ways, but most cuts can be discarded:  
every new piece must have non-zero leading digit and represent an unused prime which divides p-1.
The search provided four solutions with seven primes each: 
23.3.11.809.7.2.13
293.439.73.3.5.2.11
31.2.5.16217.7.3.41
3.701.43.89.2.37.11

In order to find larger solutions, I used the following trick.
Let solution p be the concatenation of np distinct primes, with product qp, using dp digits.
Split p into two blocks x and y, with x > y.
Let x be the concatenation of last nx primes, with product qx, using last dx digits of p.
Let y be the concatenation of remaining ny primes, with product qy, using first dy digits of p.
Such decomposition is possible for every solution p; in the worst case, choose nx = np-1 and ny = 1.
Then:
dp = dx+dy;
np = nx+ny;
qp = qx*qy;
p = y*10^dx + x;
p-1 = qp*t;
y*10^dx + (x-1) = qx*qy*t.
But qx and qy are coprime (no repeated prime), so last equation is equivalent to the following system of congruence relations: 
y*10^dx + (x-1) == 0 mod qx,
y*10^dx + (x-1) == 0 mod qy.
Given x, the congruence relation modulo qx allows to recover y among few candidates (often uniquely).
As an example, consider the following solution: 
p = 19.2.7.3.43.5.191.151
Pick x = 5.191.151, so dx = 7 and qx = 144205.
Solve congruence equation mod qx:
y = 23837 + 28841*k, with 0 <= k <= 179.
We obtain desired value y = 1927343 for k=66.
Search algorithm.
Consider every integer x in ascending order, try to cut x in every permitted way
(every new piece must have non-zero leading digit and represent an unused prime).
Solve congruence equation modulo qx.
For each candidate y, try to cut y in every permitted way and check if congruence modulo qy holds too.

Best solutions found so far.
5.13.11.3.2.29.7.8863.367.421 (np=10, dp=20)
3.7.3023.13.2.149.97.43.5.151 (np=10, dp=20)
2.433.36947.3.19.7.937.11903 (np=8, dp=21)
3.337.19.283.2.14559403.487 (np=7, dp=21)
35809.2.19.71.1801.65063.47 (np=7, dp=21)
3.19.83.2.11.7.13.15151423543 (np=8, dp=22)

Some primes have more than one permitted decomposition.
797.13.3.47.2.23.5.11
79.7.13.3.47.2.23.5.11
7.17.83.211.743.41.281.59
7.17.83.2.11.743.41.281.59

 

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