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 5x5 grid have the common sum 275. None of the numbers in the 25 cells contains a leading zero.

**RL** **AA** **CN** **BD** **RC**

**AE** **MC** **ML** **EA** **RB**

**UA** **MB** **EE** **LR** **AU**

**BR** **EU** **LM** **LN** **UE**

**BN** **LD** **NC** **UU** **BM**

(1) D = 0 (the only character not to start a number)

(2) From column 3, we can conclude that C + M + E + L + N = 25

(3) Since the digits 0 to 9 add to 45, we can conclude that

R + A + U + B = 20

(3a) From row 30, since R + A + U + B + E = 5 (mod 10)

and R + A + U + B = 20, then E equals 5

(4) Using (2) and the diagonal starting with RL, we can conclude that

the initial digits, R + M + E + L + B = 25.

(5) Combining (2) and (4), C + N = R + B

(6) From column 4, A + R + N + U = 5 (mod 10)

From row 2, A + C + B + U = 5 (mod 10)

(6a) Therefore, C + B = R + N (mod 10)

(7) Combining (5) and (6a), we can conclude that

2*C = 2*R (mod 10)

So C = R (mod 5) and N = B (mod 5)

(8) From column 1,

10*(R+A+U+B+B) + (L+E+A+R+N) = 275

But R+A+U+B = 20

So 10+B + (L+E+A+R+N) = 75

L+E+A+R+N is between 15 and 35,

So B is 4 or 6.

So N is 9 or 1

And L, E,A,R,N are either (1,2,3,4,5) or (5,6,7,8,9)

(in some order).

So L,A,R,N are either (2,3,4) or (6,7,8)

(in some order).

(9) Applying the same logic to column 5 indicates that

C,B,U,M are either (2,3,4) or (6,7,8)

(in some order)

and R is 4 or 6

and C is 1 or 9

(10) So, unless I've made mistakes, the only possibilities are

D = 0, E = 5

and either

(i) B = 4, R = 6, C = 1, N = 9,

(L,A) = (2,3) or (3,2), (U,M) = (7,8) or (8,7)

or

(i) R = 4, B = 6, N = 1, C = 9,

(U,M) = (2,3) or (3,2), (L,A) = (7,8) or (8,7)

8 possibilities, but I don't have time to check.

*Edited on ***October 11, 2009, 3:37 pm**

*Edited on ***October 11, 2009, 3:39 pm**