All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Flying Low (Posted on 2010-08-05) Difficulty: 2 of 5
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution UBASIC solution | Comment 4 of 5 |
000100   dim used(9)
000200   t=0
000300   for l=1 to 9
000400    if used(l)=0 then
000500 :    used(l)=1
000600 : for o=1 to 9
000700 :  if used(o)=0 then
000800 :    used(o)=1
000900 : for w=1 to 9
001000 :  if used(w)=0 then
001100 :    used(w)=1
001200 :    low=l*100+o*10+w
001300 :    lt=l*10
001400 : for i=1 to 9
001500 :  if used(i)=0 then
001600 :    used(i)=1
001700 :
001800 :    if low^o=i^(lt) then
001900 :     for f=1 to 9
002000 :      if used(f)=0 then
002100 :        used(f)=1
002200 :     for y=1 to 9
002300 :      if used(y)=0 then
002400 :        used(y)=1
002500 :     for a=1 to 9
002600 :      if used(a)=0 then
002700 :        used(a)=1
002800 :        fly=f*100+l*10+y
002900 :        li=l*10+i
003000 :        if fly^a=y^li then
003100 :     for d=1 to 9
003200 :      if used(d)=0 then
003300 :        used(d)=1
003400 :     for x=1 to 9
003500 :      if used(x)=0 then
003600 :        used(x)=1
003700 :        did=d*101+i*10
003800 :        xa=x*10+a
003900 :        if did^w=f^xa then
004000 :          ?did;w,f;xa
004100 :          ?fly;a,y;li
004200 :          ?low;o,i;lt
004300 :        endif
004400 :        used(x)=0
004500 :      endif
004600 :     next
004700 :        used(d)=0
004800 :      endif
004900 :     next
005000 :
005100 :
005200 :        endif
005300 :        used(a)=0
005400 :      endif
005500 :     next
005600 :        used(y)=0
005700 :      endif
005800 :     next
005900 :        used(f)=0
006000 :      endif
006100 :     next
006200 :    endif
006300 :
006400 :    used(i)=0
006500 :  endif
006600 : next
006700 :    used(w)=0
006800 :  endif
006900 : next
007000 :    used(o)=0
007100 :  endif
007200 : next
007300 :    used(l)=0
007400 :  endif
007500   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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information