[ *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)