In the following Alphametic addition problem, each letter represents a unique digit. It is known that when the digit represented by $$A$$ is even and the digit represented by $$N$$ is odd, then the Alphametic addition problem has two different solutions. Find these two solutions.
$$\text{AMTI} + \text{NMTC} = \text{FINAL}$$

Two four-digit numbers add to less than 20000, so $$F = 1$$. In the hundreds column $$2M$$ plus the carry gives the odd digit $$N$$, so there is a carry of 1 from the tens column; then $$2T$$ plus the units carry ends in the even digit $$A$$, so the units carry is 0 and $$2T = A + 10$$, while $$I + C = L$$. Thus $$(A, T)$$ is $$(2, 6)$$, $$(4, 7)$$, $$(6, 8)$$ or $$(8, 9)$$, and the thousands column gives $$A + N + c = I + 10$$, where $$c$$ is the hundreds carry. Checking the cases, $$A = 2$$ and $$A = 4$$ give nothing, $$A = 6$$ gives $$6485 + 9482 = 15967$$, and $$A = 8$$ gives $$8293 + 5294 = 13587$$, $$8692 + 3695 = 12387$$ and $$8794 + 5792 = 14586$$. So under the stated conditions the solutions are these four additions (the paper asks for two).