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."
***
�
�
�
�
|