Yuhang

• Jul 19th, 2005

 

By given condition, we have 
A+B+C = 13...(1) 
C+D+E = 13...(2) 
E+F+G = 13...(3) 
G+H+I = 13...(4) 
Add up (1)+(2)+(3)+(4), we get  
52=13*4 = A + B + C+ C + D + E+ E + F + G+ G + H + I = C+E+G+(A + B + C+ D +E + F + G + H + I) 
= C+E+G+(1+2+ 3+ 4+ 5+ 6+ 7+ 8+ 9) = C+E+G+(1+9)*9/2 = C+E+G + 45 
 
Hence,C+E+G = 52 - 45 = 7...(5) 
 
By (5), we have E = 7-C-G <= 7-3 = 4 (since, the smallest C,G are 1 or 2)  
By (2)-(5)., we get D-G = 6, or D = G+6...(6)  
Since D,G are between 1 and 9, 
so D= 7, 8,or 9, and we have G=1,2 or 3 
 
By (3)-(5)., we get F-C = 6, or F = C+6...(7)  
Since F,C are between 1 and 9, 
so F= 7, 8,or 9, and we have C=1,2 or 3 
 
By (6) ,(7) & (5): 
Case(i) When G =1, D = G+6 =7,  
and then C = 2, F = C+6=8, E = 7-C-G= 4 
Or C = 3, F = 9, E = 3 (invalid,since the same C,E value) 
 
Case(ii) When G =2, D = G+6= 8,  
and then C = 1, F = C+6=7, E= 7-C-G=4 
Or C = 3, F = 9, E = 2(invalid,since the same C,G value) 
 
Case(iii) When G =3, D = G+6=9, 
and then C = 1, F = C+6=7, E =7-C-G= 3(invalid,since the same E,G value) 
Or C = 2, F = C+6=8, E=7-C-G = 2(invalid,since the same C,E value) 
 
In these three possible cases of the values of G, we obtain that the only valid value of E is 4 

Ganesh

• Dec 14th, 2005

 

Also, to add, the values of ABCDEFGHI  can be expressed as

(1,3,9), (1,8,4), 2,4,7), (2,5,6)

md imran

 Profile Answers by md imran 

• May 12th, 2008

 

possible combo for 13 is
9+1+2
2+3+8
3+4+6
2+5+6

some digits are repeated  like 2 , 3 and 6..it means they are to be positioned at the junction
the logical pos comes out to be

9 1 2
      8
      3 4 6
            5
            2

therfore c = 3 from above fig

Zoetrinity

• Jun 17th, 2024

 

Lets reanalyze the given problem with the correct detailed approach to ensure the answer is accurate.
Given conditions:
1. (A + B + C = 13)
2. (C + D + E = 13)
3. (E + F + G = 13)
4. (G + H + I = 13)
Adding these equations together:
[(A + B + C) + (C + D + E) + (E + F + G) + (G + H + I) = 52]
Simplify:
[A + B + C + C + D + E + E + F + G + G + H + I = 52]
[A + B + 2C + D + 2E + F + 2G + H + I = 52]
We know the sum of digits 1 through 9 using n(n+1)/2 = (9+1)*9/2 = 45:
[A + B + C + D + E + F + G + H + I = 45 ]
Thus:
[(A + B + C + D + E + F + G + H + I) + (C + E + G) = 52 ]
[45 + (C + E + G) = 52]
[ C + E + G = 7]
Now:
[E = 7 - C - G].........5
Substituting [E = 7 - C - G] into equations 2 and 3 which both contains E, we derive:
[C + D + E = 13].......2
[C + D + (7 - C - G) = 13]
[D - G = 13 - 7]
[D = 6 + G].......….6
[ E + F + G = 13].......3
[(7 - C - G) + F + G = 13]
[F - C = 13 - 7]
[F = 6 + C ]...........7
To satisfy conditions, both (D) and (F) must be digits from 1 to 9.
From equation 6 & 7, D and F must be greater than 6 but less than or equals 9 (ie 7, 8, or 9)
So C and G can only be digits 1, 2 or 3.
We examine these possibilities in the following case study:
### Case Analysis:
1. **Case 1: (G = 1)**
- (D = 6 + G = 7)
- Valid (D)
- (C = 2) or (C = 3) (keeping within bounds of digit constraints)
- For (C = 2):
- (E = 7 - C - G = 4)
- (F = 6 + C = 8)
- All digits are unique and valid
- For (C = 3):
- (E = 7 - C - G = 3) (invalid as duplicates occur for C & E)

2. **Case 2: (G = 2)**
- (D = 6 + G = 8)
- Valid (D)
- (C = 1) or (C = 3)
- For (C = 1):
- (E = 7 - C - G = 4)
- (F = 6 + C = 7)
- All digits are unique and valid
- For (C = 3):
- (E = 7 - C - G = 2) (invalid as duplicates occur for G & E)
3. **Case 3: (G = 3)**
- (D = 6 + G = 9)
- Valid (D)
- (C = 1) or (C = 2)
- For (C = 1):
- (E = 7 - C - G = 3) (invalid as duplicates occur for G & E )
- For (C = 2):
- (E = 7 - C - G = 2) (invalid as duplicates occur for C & E)
**Valid configurations:**
- **When (G = 1) and (C = 2)**
- (E = 4)
- (D = 7)
- (F = 8)
- **When (G = 2) and (C = 1)**
- (E = 4)
- (D = 8)
- ( F = 7)
Thus, in all valid scenarios, the value of (E) is consistently **4**.
Therefore, the numerical digit E represents is: 4