Proposed by Alan Frisch, Christopher Jefferson, Ian Miguel, Toby Walsh
The original fractions puzzle is specified as follows. Find 9 distinct non-zero digits that satisfy:
A D G
-- + -- + -- == 1
BC EF HI
where BC
is shorthand for 10B+C
, EF
for 10E+F
and HI
for 10H+I
.
A simple generalisation is as follows. Find 3n non-zero digits satisfying: $ 1 = \sum_{i \in 1..n} x_i / y_iz_i $
where $y_iz_i$ is shorthand for $10y_i+z_i$ and the number of occurrences of each digit in $1..9$ is between $1$ and $ceil(n/3)$.
Since each fraction is at least $1/99$, this family of problems has solutions for at most $n <= 99$. An interesting problem would be to find the greatest $n$ such that at least one solution exists. A further generalisation might specify that the fractions sum to $ceil(n/3)$.