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:

***

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]

***

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