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 Pandigital and Doubly Pandigital (Posted on 2009-06-02)
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.

 See The Solution Submitted by K Sengupta No Rating

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 computer solution | Comment 1 of 2
`   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

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