Problems & Puzzles: Puzzles

Puzzle 589. Palindromes multiplying consecutive primes

Long time ago (2001?) my friend Patrick De Geest published the complete list of palindromes obtained by multiplying K consecutive primes.

K p1 pk Product
2 2 3 6
2 7 11 77
2 17 19 323
2 191 193 36863
2 1051 1061 1115111
2 1934063 1934071


3 7 13 1001
4 5 13 5005
5 7 19 323323

See: &

Perhaps is time to get at least one single new result.

Q1. Can you get a new result?



Contributions came from Jan van Delden


Jan wrote:

If the product has an even  number of digits, 11 must be a factor, so it is easy to search for solutions. In this case the number of digits of the product must be greater than 11826 (a limit reached in just 15 minutes), but I sincerely doubt there is a solution of this form, because the maximum number of corresponding digits in the head/tail of all products I found is merely 3. So the situation seems to become worse with increasing number of digits.
If the product has an odd number of digits, the situation is more complex. To limit the search I only investigated the situation where the product is nearly a square, i.e. it consists of a product of two consecutive primes, since the given solutions in this case are of this form (so they look the most promising). As I found it actually is, because for every n=2k+1 where no solution is found, there exist nearly solutions in the sense that k-1 digits in the head and tail correspond.
For instance for n=21 I found 13422495703 13422495727 with product 180163391 219 193361081. [The search for n=21 is not finished yet!]


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