# 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.)

• There are no letters here. If you mean the letters in the names of the countries represented by the flags, are the country names in English, the country's native tongue, or some other language? Apr 2, 2016 at 13:46
• Perhaps the letters are "PLUS", and "IS EQUAL TO"? Apr 2, 2016 at 18:18

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.