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 Flying Low (Posted on 2010-08-05)
Substitute each of the letters by a different digit from 0 to 9 to satisfy this system of alphametic equations. None of the numbers as well as none of the exponents contains a leading zero.
• (DID)W = FXA, and:
• (FLY)A = YLI, and:
• (LOW)O = ILT

 No Solution Yet Submitted by K Sengupta Rating: 2.0000 (1 votes)

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 UBASIC solution | Comment 4 of 5 |
`000100   dim used(9)000200   t=0000300   for l=1 to 9000400    if used(l)=0 then000500 :    used(l)=1000600 : for o=1 to 9000700 :  if used(o)=0 then000800 :    used(o)=1000900 : for w=1 to 9001000 :  if used(w)=0 then001100 :    used(w)=1001200 :    low=l*100+o*10+w001300 :    lt=l*10001400 : for i=1 to 9001500 :  if used(i)=0 then001600 :    used(i)=1001700 :001800 :    if low^o=i^(lt) then001900 :     for f=1 to 9002000 :      if used(f)=0 then002100 :        used(f)=1002200 :     for y=1 to 9002300 :      if used(y)=0 then002400 :        used(y)=1002500 :     for a=1 to 9002600 :      if used(a)=0 then002700 :        used(a)=1002800 :        fly=f*100+l*10+y002900 :        li=l*10+i003000 :        if fly^a=y^li then003100 :     for d=1 to 9003200 :      if used(d)=0 then003300 :        used(d)=1003400 :     for x=1 to 9003500 :      if used(x)=0 then003600 :        used(x)=1003700 :        did=d*101+i*10003800 :        xa=x*10+a003900 :        if did^w=f^xa then004000 :          ?did;w,f;xa004100 :          ?fly;a,y;li004200 :          ?low;o,i;lt004300 :        endif004400 :        used(x)=0004500 :      endif004600 :     next004700 :        used(d)=0004800 :      endif004900 :     next005000 :005100 :005200 :        endif005300 :        used(a)=0005400 :      endif005500 :     next005600 :        used(y)=0005700 :      endif005800 :     next005900 :        used(f)=0006000 :      endif006100 :     next006200 :    endif006300 :006400 :    used(i)=0006500 :  endif006600 : next006700 :    used(w)=0006800 :  endif006900 : next007000 :    used(o)=0007100 :  endif007200 : next007300 :    used(l)=0007400 :  endif007500   nextfinds`
` 343  6          7  18 729  8          9  24 256  5          4  20 `

These are the numbers in the original order given. With operators added:

343 ^ 6  =  7 ^ 18
729 ^ 8  =  9 ^ 24
256 ^ 5  =  4 ^ 20

 Posted by Charlie on 2010-08-05 12:59:53

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