Q:

Geographic alphametic riddle with European flags

Every letter stands for a digit in base-10 representation, different letters stand for different digits, and leading digits are always non-zero.
 $\quad$ PLUS $\quad$ 
$\quad$ IS EQUAL TO $\quad$ 
Which digit does each letter represent? 
(Please present the full analysis how these digits can be determined.) 

A:

The answer is

 FINLAND + IRELAND = DENMARK: 2518017 + 5398017 = 7916034

There are a few things we'll make use of to narrow down the letters as much as possible.

 Because of A + A = A and N + E = N, one must be 0 and the other must be 9. These will form the two main branches. For both, we'll work to prove that F is equal or greater than 5, because from F + I = D there are only two ways for D to have a value less than 5. After restricting most letters to a handful of digits, we brute force for N because its value has an avalanche of effects. From N + N = R we derive R, which we then use in combination with I + R = E to derive I. F and D follow next from the restrictions put on D and F + I = D. Finally, from D we can derive K, which leaves us to validate that L and M can take the two remaining values.

From A + A + carry = A we can deduce

 1) A = 0; carry = 0, i.e., N = 1, 2, 3 or 4 (letter before A must not result to a carry digit)
 2) A = 9; carry = 1, i.e., N = 5, 6, 7, or 8.

From N + E + carry = N we can deduce

 1) E = 0; carry = 0, i.e., L = 1, 2, 3, or 4
 2) E = 9; carry = 1, i.e., L = 5, 6, 7, or 8

Overall, we have two cases which we need examine.

 1) A = 0; E = 9; N = 1/2/3/4; L = 5/6/7/8
 2) A = 9; E = 0; N = 5/6/7/8; L = 1/2/3/4

First case.

 Since A = 0, there is no carry for L + L = M. Therefore, L can't be 5 because that would make M = 0. This also means that M is even, i.e., M = 2/4/6/8. To summarise, A = 0; E = 9; N = 1/2/3/4; L = 6/7/8; M = 2/4/6/8; K = 2/4/6/8 (because D + D = K).

 Assume D = 3. Because of F + I = D, we require the digits 1, 2, 3, which only leaves N = 4 as the possible value. Since D + D = K results to no carry, N + N = R necessitates R = 8. Since 2, 4 and 8 have been used up, M and K are only left with 6, which can't satisfy both. Contradiction. Assume D = 4. This requires the digits 1, 3 and 4, which results to N = 2. However, N + N = R results to R = 4, which is a contradiction. Therefore, D = 6/7/8 (can't have D = 5 because of D + D = K and K can't be 0). This also means that N + N = R has a carry digit, which makes R odd. Finally, because N + E = N goes over 9, I + R = E has a carry digit, i.e., I + R + 1 = 9.

 So far we have A = 0; E = 9; N = 1/2/3/4; D = 6/7/8; R = 2*N+1; I = 8 - R; K = 2*D mod 10; M = 2*L mod 10. Brute force for N.

N = 1: R = 3. I = 5. D can be 6/7/8. It can't be 6 because it'd require F = 1 and it can't be 8 because it'd require F = 3. Therefore, F = 2; D = 7; K = 4. This only leaves 6 and 8 for L and M and we have a solution with L = 8 and M = 6. SOLUTION FOUND.

N = 2: R = 5; I = 3. D can't be 6 because it'd require F = 3 and it can't be 8 because it'd require F = 5. Therefore, F = 4; D = 7; K = 4 = F. Contradiction.

N = 3: R = 7; I = 1. D can't be 7 and it can't be 8 because it'd require F = 7. Therefore, F = 5; D = 6; K = 2. This leaves us with 6 or 8 for L. It can't be 6 because M would be 2 and it can't be 8 because M would be 6. Contradiction.

N = 4: R = 9 = E. Contradiction.  

Second case.

 Because A = 9, L + L = M has a carry, so M is odd. We follow a similar approach to above to excluse the possibility that D is 3 or 4. For D = 3, we also require the digits 1 and 2, which leaves L = 4, but L + L + carry = 9 = A. Contradiction. For D = 4, we get L = 3; M = 7; K = 8. Since N + E = N results to no carry, I + R = 10. Since we assumed that F and I are 1 and 3 in some order, this means that R has to be either 7 or 9, which is impossible. So, we get F = 6/7/8 again. This also means that N + N = R has a carry and R must be odd. Since N + E doesn't go over 9, I + R = E has no carry digit. However, since E = 0, it means that I + R add up to 10. This means that F + I has a carry digit, i.e., F + I + 1 = D.

 So far we have A = 9; E = 0; N = 5/6/7/8; D = 6/7/8; R = 2*N+1 mod 10; I = 10 - R; K = 2*D mod 10; M = 2*L+1 mod 10. Let's brute force for N.

N = 5: R = 1; I = 9 = A. Contradiction.

N = 6: R = 3; I = 7. D can only be 8, but because of F + I + 1, F would have to be 0. Contradiction.

N = 7 -> R = 5; I = 5. Contradiction.

N = 8: R = 9 = A. Contradiction.  

