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Alphametic puzzle

Posted on 2006-04-24
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Last Modified: 2008-08-20
            X Y Z             X Y Z
         +    A B            -  A B
           -------            -------
         C D E F            B G A

find X,Y,Z,G.


i;m not able to fully solve it ..
plz help
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Question by:shilpi84
8 Comments
 
LVL 2

Expert Comment

by:mellowmarquis
ID: 16523556
Should there be an equals sign between those two equations?
Are the symbols meant to be indivudual numbers or algrebraic expressions?

ie: xyz = 123
or: xyz= x (times) y (times) z

cheers,
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Expert Comment

by:A. Cristian Csiki
ID: 16523568
hy there a solution for your puzzle is here:
http://www.tkcs-collins.com/cgi/alpha_solve.cgi
good luck
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Expert Comment

by:A. Cristian Csiki
ID: 16523574
sorry, in my last comment, there is the solution for the sum, checking now for diff
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Author Comment

by:shilpi84
ID: 16523647
there is no = sign between the two
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Author Comment

by:shilpi84
ID: 16523662
x has got to be 9 since xyz + ab=cdef which is a 4 letter. word max carry from y + a = 1 so x=9
hence c=1, d=0.
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Expert Comment

by:A. Cristian Csiki
ID: 16523691
b,c,d, and x are always like this: b=8, c=1, d=0, x=9 and a=f...
leave me a lil bit to give you the rest of the values
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Expert Comment

by:A. Cristian Csiki
ID: 16524310
the solutions:

b=8  ;  c=1  ; d=0 ; x=9 ; z=4 ; f=2 ; a=6 ; y=5 ; g=8 ; e=2;
b=8  ;  c=1  ; d=0 ; x=9 ; z=4 ; f=2 ; a=6 ; y=6 ; g=9 ; e=3;
b=8  ;  c=1  ; d=0 ; x=9 ; z=5 ; f=3 ; a=7 ; y=4 ; g=6 ; e=2;
b=8  ;  c=1  ; d=0 ; x=9 ; z=5 ; f=3 ; a=7 ; y=5 ; g=7 ; e=3;
b=8  ;  c=1  ; d=0 ; x=9 ; z=5 ; f=3 ; a=7 ; y=6 ; g=8 ; e=4;
b=8  ;  c=1  ; d=0 ; x=9 ; z=5 ; f=3 ; a=7 ; y=7 ; g=9 ; e=5;
b=8  ;  c=1  ; d=0 ; x=9 ; z=6 ; f=4 ; a=8 ; y=4 ; g=5 ; e=3;
b=8  ;  c=1  ; d=0 ; x=9 ; z=6 ; f=4 ; a=8 ; y=5 ; g=6 ; e=4;
b=8  ;  c=1  ; d=0 ; x=9 ; z=6 ; f=4 ; a=8 ; y=6 ; g=7 ; e=5;

fits both equations.
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Accepted Solution

by:
jitendra_wadhwani earned 2000 total points
ID: 16525041
Adding a two digt number  to a 3 digint number is resulting 4 digit number...which will be between 1001 to 1198 so C = 1;

Tow digit number can be max 99 so the 3 digit number is greated than or equal to 901

so X = 9 Confirm;


if YZ >  A B then X Y Z - A B should result X _ _ but it is resulting B _ _ means
YZ < AB and in this case B can be nothing but 8 only

so till now

X = 9; B=8; C = 1;

XYZ can be max 999 and min 901
AB can be min 10 and max 99 so addition wil not cross


  9 Y Z
  + A 8
=====
1 D E F


  9 Y Z
  - A 8
=====
  8 G A


NOw D can be only 0 or 1 as 5 digit number will be between 1001 to 1198

Since c is already 1 so D = 0;

Till now

X = 9; B=8; C = 1; D = 0;

Remaining Y Z A E F G are among 2,3,4,5,6,7,

Take the addition case

if z=2 then F should be 0 so which is not possible so Z is not 2
if z = 3 then F will be 1 again not possible

if z = 4 then F will be 2 possible

 sub case then A will be 6

remaining Y E F among 3,5,7
Check othenr if fails increasy Z and try again

and so on you will reach

 
D=0 ; C=1 ; E=2; F=3 ;Y=4 ; Z=5 ;G=6 ; A=7 ; B=8  ;   X=9 ;  
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