# Thread: it is really imp plz solve

1. ## it is really imp plz solve

A+B+C+D=D+E+F+G=G+H+I=17 where each letter represent a number from 1 to 9. Find
out what does letter D and G represent if letter A=4

2. Originally Posted by bharathlece
A+B+C+D=D+E+F+G=G+H+I=17 where each letter represent a number from 1 to 9. Find
out what does letter D and G represent if letter A=4
Let me just answer the title.

No, it is not impossible to solve. But the solution reqiures too much trial and error, too much reliance on chance. A computer can easily get the answer, but a person,....

Also, the puzzle is like the game of sudoku. I hate sudoku. Too much trial and error.

3. hey , man this is the most common asked question in indian puzzle talent show dnt say that u hate some or u like some . give me replay if u got the answer dnt give me some other xyz replay about which u like and which u dnt like . dnt waste ur time by giving replay nor dnt waste my time in reading them. if it hurt u u deserves it coz ur giving replay like dat to me

4. Originally Posted by bharathlece
hey , man this is the most common asked question in indian puzzle talent show dnt say that u hate some or u like some . give me replay if u got the answer dnt give me some other xyz replay about which u like and which u dnt like . dnt waste ur time by giving replay nor dnt waste my time in reading them. if it hurt u u deserves it coz ur giving replay like dat to me
Umm, okay, so you have no sense of humor. I can play that.

If that is common in your place, how come you cannot solve it?

I never was hurt. I was just teasing you.

And I hate sudoku.

And your English is worse than mine.

(Moderators, I know you have senses of humor.)

5. Originally Posted by bharathlece
A+B+C+D=D+E+F+G=G+H+I=17 where each letter represent a number from 1 to 9. Find
out what does letter D and G represent if letter A=4
The solution is impossible to get, since you have,
$A+B+C+D=17$
$D+E+F+G=17$
$G+H+I=17$
$A+B+C+D+E+F+G+H+I+(D+G)=51$
But these letters are the numbers 1,2,...,9 in some order thus,
$(1+2+3...+9)+(G+H)=51$
Thus,
$G+H=6$
Then in the third equation we have that $I=11$ which is a violation because it cannot be more then 9.

6. Originally Posted by ThePerfectHacker
The solution is impossible to get, since you have,
$A+B+C+D=17$
$D+E+F+G=17$
$G+H+I=17$
$A+B+C+D+E+F+G+H+I+(D+G)=51$
But these letters are the numbers 1,2,...,9 in some order thus,
$(1+2+3...+9)+(G+H)=51$
Thus,
$G+H=6$
Then in the third equation we have that $I=11$ which is a violation because it cannot be more then 9.
Perhaps the A, B, ..., I aren't all distinct? He didn't actually say that they were, but that was my first inclination as well.

Still, we have 4 conditions on 9 variables (counting the condition that all A, B,..., I are between 0 and 10 as a condition). I don't see how to solve it uniquely.

-Dan

7. Originally Posted by topsquark
Perhaps the A, B, ..., I aren't all distinct? He didn't actually say that they were, but that was my first inclination as well.

Still, we have 4 conditions on 9 variables (counting the condition that all A, B,..., I are between 0 and 10 as a condition). I don't see how to solve it uniquely.

-Dan
Linear Algebra does not work here. Because we assume the elements are in some field. The integers do not form a field. Thus, the fact that there are 3 equations and 9 variables does not mean we cannot have a unique solution.

8. Further, they if they are not distinct then it solves easily for example,
A=4
B=1
C=3
D=9
E=1
F=1
G=6
H=2
I=9