Problems & Puzzles: Puzzles

Puzzle 25.- Composed primes (by G.L. Honaker, Jr. )

"Find n-digit composite numbers abc... with exactly n prime factors

p1 x p2 x p3... such that a^p1 + b^p2 + c^p3... is a prime number".

i.e. 38 = 2 x 19 => 3^2 + 8^19 = 144115188075855881 (prime)


Solution

Patrick De Geest adds the following restrictions to the original puzzle of G.L Honaker, Jr. :

a) discard solutions with digit(s) 0 and 1 embedded

b) all the prime factors must be distinct (or powerfree).

And his solutions (sent at 20/12/98) for 3 or more digits are shown below:

Digits Solutions Choosen example
2 38 38 = 2 x 19 => 3^2 + 8^19 = 144115188075855881 (prime), G.L. Honaker, Jr.
3 434

595

962

434 = 2 x 7 x 31  -> 4^2 + 3^7 + 4^31  = 4611686018427390107 (prime), De Geest
4 8346

9338

8346 = 2 x 3 x 13 x 107 -> 8^2 + 3^3 + 4^13 + 6^107 = 182887402188115849169083086627158074049885861/ 734957094920281393977611551796677836891 = prime of length 84, De Geest                             
5 25662

96866

96866 = 2 x 7 x 11 x 17 x 37 -> 9^2 + 6^7 + 8^11 + 6^17 + 6^37  gives a prime of length 29 = 61886548790943230212281353681, De Geest  
6 539994 539994 = 2 x 3 x 7 x 13 x 23 x 43-> 5^2 + 3^3 + 9^7 + 9^13 + 9^23 + 4^43 =

77380115393458461552902543, is a prime of length 26, De Geest.


Finally Jeff Heleen solved this old puzzle (May 2003):

"using the given restrictions, for 7 digits there are 50 possible with none yielding a prime. For 8 digits there are 3 possible with none yielding primes. For 9 and 10 digits there are no possible.

This puzzle cannot go beyond 10 digit as numbers with more than 10 distinct factors have more digits in the result than factors."

***

 

 

 

 


Records   |  Conjectures  |  Problems  |  Puzzles