Determine all possible triplet(s) (P, Q, N), where each of P and Q is a
five digit positive base ten integer with P > Q and P*Q = N, such that:
- P and Q together contain each of the digits from 0 to 9 exactly once, and:
- N contains each of the digits from 0 to 9 exactly once, and:
- N2 contains each of the digits from 0 to 9 exactly twice. Neither N nor N2 can contain any leading zero.
5 dim Dct(9)
10 Pq="1023456789":H=Pq
15 loop
20 Ps=left(Pq,5):Qs=right(Pq,5)
25 if Ps>Qs and left(Qs,1)<>"0" then
30 :P=val(Ps):Q=val(Qs):N=P*Q
40 :Ns=cutspc(str(N))
50 :if len(Ns)=10 then
55 :Good=1
60 :for I=1 to 9
65 :if instr(I+1,Ns,mid(Ns,I,1))>0 then Good=0:endif
70 :next
75 :if Good then
80 :Ns2=cutspc(str(N*N))
85 :if len(Ns2)=20 then
90 :for I=0 to 9:Dct(I)=0:next
100 :for I=1 to 20
105 :inc Dct(val(mid(Ns2,I,1)))
110 :if Dct(val(mid(Ns2,I,1)))>2 then Good=0:endif
120 :next
130 :if Good then print P;Q;N;Ns2:endif
132 :endif
200 gosub *Permute(&Pq)
210 if left(Pq,1)="0" then goto 800
250 endloop
800 end
(The permute subroutine is described elsewhere on the site.)
This finds:
62037 54981 3410856297 11633940678784552209
as P, Q, N and N^2. No other solutions were found.
|
|
Posted by Charlie
on 2009-06-02 14:02:33 |
computer solution
