Find three different digits that no matter in what order they are arranged, the three digit number formed is the product of two primes.
100 for D1=0 to 7
200 for D2=D1+1 to 8
300 for D3=D2+1 to 9
400 Good=1
500 B$=cutspc(str(D1))+cutspc(str(D2))+cutspc(str(D3))
600 for Ii=1 to 6
700 N=val(B$)
800 P=2
900 T=N:Ct=0
1000 while P<=sqrt(T)
1100 if T@P=0 then
1200 :Ct=Ct+1:T=T//P
1300 :else
1400 :P=nxtprm(P)
1500 wend
1510 if T>1 then Ct=Ct+1
1600 if N>99 and Ct<>2 then Good=0:cancel for:goto *NotGood
1700 gosub *Permute(&B$)
1800 next Ii
1900 if Good then print D1,D2,D3
2000 *NotGood
2100 next D3
2110 next D2
2120 next D1
2200 end
2300
2400 *Permute(&A$)
2500 local I,J,X$,Fl
2600 X$=""
2700 for I=len(A$) to 1 step -1
2800 L$=X$
2900 X$=mid(A$,I,1)
3000 if X$<L$ then cancel for:goto *Outta
3100 next I
3200 *Outta
3300 if I=0 then
3400 :for J=1 to len(A$)\2
3500 :X$=mid(A$,J,1)
3600 :mid(A$,J,1)=mid(A$,len(A$)-J+1,1)
3700 :mid(A$,len(A$)-J+1,1)=X$
3800 :next J
3900 :else
4000 :for J=len(A$) to I+1 step -1
4100 :if mid(A$,J,1)>X$ then cancel for:goto 4300:endif
4200 :next J
4300 :mid(A$,I,1)=mid(A$,J,1)
4400 :mid(A$,J,1)=X$
4500 :for J=1 to (len(A$)-I)\2
4600 :X$=mid(A$,I+J,1)
4700 :mid(A$,I+J,1)=mid(A$,len(A$)-J+1,1)
4800 :mid(A$,len(A$)-J+1,1)=X$
4900 :next J
5000 return
finds 1, 7 and 8.
The zero was included among the digits even though it would lead to two permutations with a leading zero; even so, none were found with that allowance.
A modification to print the prime factors of the six permutations gives
178: 2 89
187: 11 17
718: 2 359
781: 11 71
817: 19 43
871: 13 67
|
|
Posted by Charlie
on 2005-03-21 15:15:26 |
Exhaustive Search via computer program -- spoiler
