Problems & Puzzles: Puzzles

Puzzle 22.- Primes & Persistence

In the sequence 679, 378, 168, 48, 32, 6 each term is the product of the decimal digits of the previous one. Neil Sloane defines the "persistence" of a number as the steps (five in the example) before the number collapses to a single digit. (p. 262, Ref. 2) 

So, we ask : Find the least primes with "persistence" k, such that 1<=k<=12 

---

Solution

Patrick De Geest found (18 Sep 1998) the following solutions: 

        k       least prime 
        --      ---------------- 
        1       2 [2] 
        2       29 [18][8] 
        3       47 [28][16][6] 
        4       277 [98][72][14][4] 
        5       769 [378][168][48][32][6] 
        6       8867 [2688][768][336][54][20][0] 
        7       186889 [27648][2688][768][336][54][20][0] 
        8       2678789 [338688][27648][2688][768][336][54][20][0] 
        9       26899889[4478976][338688][27648][2688][768][336][54] 
                                   [20][0] 

*** 

Jud Mc Cranie has found (19/09/98) the solution for k=10 & 11: 

     10     3778888999[438939648][ 4478976] etc. 
    11      277777788888989[4996238671872][438939648] etc. 
      
Now we only need the solution for k=12.

Interesting Links to this puzzles, sent by De Geest:

 http://www.astro.virginia.edu/~eww6n/math/MultiplicativePersistence.html 
 http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=003001 
 http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=046500

***

Shyam Sunder Gupta comments: "The solution mentioned for k=1 is wrong. The least prime with persistence k=1 is 11 . In fact 2 is the least prime with persistence k=0 . The least prime with persistence k=12 is greater than 10^50."

The solutions should be rearranged the following way:

        k       least prime 
        --      ---------------- 
        0       2 [itself] 
        1       11 [1] 
        2       29 [18][8] 

***