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3x3 Grid Alphamagic (Posted on 2009-09-17) Difficulty: 3 of 5
Substitute each of the capital letters by a different digit from 0 to 9, such that each of the columns, each of the rows and each of the two main diagonals of this 3x3 grid have the common sum LRK. It is known that L is nonzero and, none of the numbers in the nine cells contains a leading zero.

CL    RO    OR
EK    HS    AA
KE    KB    SH

Note: LRK is equal to 100*L + 10*R + K.

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Analytical Solution (and secret revealed) Comment 2 of 2 |
1) B = 0, because every other letter either begins one of the 9 numbers or the sum

2) The sum must be over 100 and less than 300, so L is 1 or 2

3) CL + EK + KE = LRK, so L + E = 10 (based just on looking at the unit digits)
   Since L is 1 or 2, E must be 8 or 9.

4) CL + RO + OR = CL + HS + SH = CL + EK + KE,
   so RO + OR = HS + SH = EK + KE,
   so R+O = E+K = H+S
   
   since E is 8 or 9, this sum must be 9 or 11 or 12 or 13.
   It can't be 10, because L + E = 10,
   and the maximum is 13 because (4+5+6+7+8+9)/3 = 13
   
   The only possibilities for E&K are 
   
   E K 
   - - 
   8 1 
   8 3 
   8 4 
   8 5 
   9 2 
   9 3 
   9 4    
   
5) RO + HS + KB = KE + KB + SH = LRK
    Based just on looking at the unit digits,
    Since B = 0, O + S = K or 10 + K
           and  E + H = 10 + K (it can't equal K because E > K)
           
6) EK + HS + AA = LRK,
     so based just on looking at the unit digits, S+A = 10   

7) Let's work out the values for L, H, S & A, based on E & K

   E K  L=10-E  H=10+K-E S=E+K-H A = 10-S
   - -  ---------  ------------ ----------- ----------
   8 1   2          3              6             4
   8 3   2          5              6             4
   8 4   2          6              6             4      X
   8 5   2          7              6             4      
   9 2   1          3              8             2      X
   9 3   1          4              8             2
   9 4   1          5              8             2
   
 7) Three of the rows are inconsistent, giving us just 4 possibilities.  Let's add in values for O & R

   E K L H S A O=10+K-S R=E+K-O
   -  - - -  - -   ------------ -----------
   8 1 2 3 6 4   5              4            X
   8 3 2 5 6 4   7              4            X
   8 5 2 7 6 4   9              4            X
   9 3 1 4 8 2   5              7      
   9 4 1 5 8 2   6              7       

   (Hmm.  If I were more clever, I could have eliminated E = 8 earlier, By combining equations, I could have arrived at
  R = 3E - 20 and A = 20-2E, so E cannot be 8)   

 8) At any rate, two of the rows are inconsistent, so the only possibilities are

   E K L H S A O R B C 
   -  - -  - -  -  -  - - -
   9 3 1 4 8 2  5 7 0 6
   9 4 1 5 8 2  6 7 0 3
   
 9) Using the diagonals for the first time, we can rule out the first row based on
 
    KE + HS + OR = LRK, so E + S + R = K (mod 10) 
   
 10) Therefore, 01234 56789 = BLACK HORSE, which hardly seems a coincidence.
 
     This checks out, as 
     31  76  67
     94  58  22
     49  40  85 = 174
   

Edited on September 18, 2009, 5:41 pm

Edited on September 19, 2009, 1:23 am
  Posted by Steve Herman on 2009-09-18 17:33:05

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