[ July 21, 2002 ]
11 ways to arrange the ninedigits
Forwarded by Paul Gissing (email)
There are eleven ways in which the digits 1, 2, 3, ..., 9
can be arranged as a whole number plus a fraction
whose sum is 100.
Paul wrote that the question as presented to him
stated that there were 11 solutions.
One way is 
96 + (2148/537) = 100
How many others can you find ?
[ July 26-27, 2002 ]
Jean Claude Rosa (email) wrote that he found a partial
solution in the “Théorie des Nombres” from Edouard Lucas,
(18..; he hasn't the exact date of the edition)
JCR found the above given example but Edouard Lucas gave
only 7 more solutions and not 11.
Here they are
100 = 91 + (5742/638)
100 = 91 + (7524/836)
100 = 91 + (5823/647)
100 = 94 + (1578/263)
100 = 96 + (2148/537)
100 = 96 + (1428/357)
100 = 96 + (1752/438)
One day later JCR came up with the missing 4
thereby completing the initial puzzle.
100 = 3 + (69258/714)
100 = 81 + (5643/297)
100 = 81 + (7524/396)
100 = 82 + (3546/197)
JCR likes to ask a few follow-up questions :
Exist there also solutions if we replace 100
with another constant ?
Resolve the equation P=A+B/C
with P prime (or palprime), A and C also prime
(of course B must be composite by default)
Keeping the spirit of this WONplate A, B and C are written
by using once all the digits from 1 to 9.
[ August 12, 2002 ]
Here are already some solutions from JCR himself.
(P is prime and PP is palprime.)
103 = 97 + (2586/431)
639857 = 639851 + (42/7)
958367 = 958361 + (42/7)
No solutions for P if digitlength is 7 !
101 = 97 + (1852/463)
191 = 7 + (85192/463)
13831 = 13597 + (468/2)
94649 = 94531 + (826/7)
For reference goals and easy searching all the nine- & pandigitals
implicitly displayed in these topics are listed here.
Topic → 962148537, 915742638, 917524836, 915823647, 941578263,
962148537, 961428357, 961752438, 369258714, 815643297,
817524396, 823546197, 972586431, 639851427, 958361427,
971852463, 785192463, 135974682, 945318267
A000137 