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=∑i∈1..nxi/yizi
where yizi is shorthand for 10yi+zi 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).