Quora Question: Can you arrange the digits 1-9 to make a nine-digit number
such that the first digit is divisible by one, the first two digits are
divisible by two, the first three divisible by three and so on?
My solution:
ea=.&.>
at=.>10#.ea(a=.362880 9$1 to 9){.ea{n=.>:perm 9
{:"1 at#~*./"1[0
I guess to simplify it, I would filter the possibilities at each step.
first digit divisible by 1:
N1=: 1+i.9
Second digit divisible by 2:
N2=: ;(10*N1)+each (<2*1+i.4)-.each N
Third digit divisible by 3:
digits=: 10&#.inv
N3=: (#~ 0=3|]);(10*N2)+each (<1+i.9) (-. digits) each N2
An
If we do a bit of work by ourselves? Consider that the 5 has to be in fifth
position (starting from 1), and that there must be even digits (four of
them) on the even positions. The other four odd digits go to the remaining
odd positions. Then there are only 576 possibilities left.
odds =: (i.24) A
Wow! Raul & Ben both provided solutions to my Quora problem that were
radically different from my original solution, and radically different from
each other. Both solutions reduced the execution time & space by several
orders of magnitude, when compared with my original solution. Also, both
used di
