Trial and error may be more expeditious, but pure logic can give a unique answer on the assumptions that (1) each distinct letter represents a distinct decimal digit and (2) MAC > 99, MAAR > 999, and JOCKO > 9999.

\(\displaystyle 100M +10A + C + 1000M + 100A + 10A + R = 10000J + 1000O + 100C + 10K + O \implies\)

\(\displaystyle (C + R) + 10(A + A) + 100(M + A) + 1000M = O + 10K + 100C + 1000O + 10000J.\)

It may seem that you have one equation to find 7 unknowns. But in fact, you have as well over 14 inequalities and a restriction to integer solutions.

\(\displaystyle C + R = O + 10x, \ x = 0 \text { or } x = 1.\)

\(\displaystyle 10(x + A + A) = 10(K + 10y), \ y = 0 \text { or } y = 1.\)

\(\displaystyle \therefore x + 2A = K + y. \)

\(\displaystyle 100(y + M + A) = 100(C + 10z), \ z = 0 \text { or } z = 1.\)

\(\displaystyle y + M + A = C + 10z.\)

\(\displaystyle 1000(z + M) = 1000O + 10000J.\)

\(\displaystyle z + M = O + 10J.\)

\(\displaystyle \text {But } z \le 1, \ J \ge 1, \ O \ge 0, \text { and } M \le 9 \implies z + M \le 10 \text { and } 10 \le O + 10J.\)

\(\displaystyle \therefore z = 1,\ J = 1, O = 0, \ \text { and } M = 9.\)

\(\displaystyle O = 0 \implies C + R = O + 10x = 10x \implies R = 10 - C C \text { or } R = - C, \text { the latter being impossible.}\)

\(\displaystyle \therefore x = 1 \text { and } R = 10 - C.\)

Moreover, A, C, K, and R are integers > 1 and < 9 because O is zero, J is one, and M is 9.

\(\displaystyle \text {Furthermore, } y + M + A = C + 10z \implies y + 9 + A = C + 10 \implies y + A = C + 1.\)

\(\displaystyle y = 1 \implies 1 + A = C + 1 \implies A = C, \text { which is impossible.}\)

\(\displaystyle \therefore y = 0 \text { and } C < A - 1 \implies 3 \le A \le 8 \text { and } 2 \le C \le 7.\)

\(\displaystyle x + 2A = K + 10y \implies 1 + 2A = K \implies A = \dfrac{K - 1}{2}.\)

\(\displaystyle K \text { is even} \implies A \text { is not an integer, which is impossible.}\)

\(\displaystyle \therefore K = 3, 5, \text { or } 7.\)

\(\displaystyle K \le 5 \implies A \le 2, \text { which is impossible.}\)

\(\displaystyle \therefore K = 7 \implies A = 3 \implies C = 2 \implies R = 8.\)

\(\displaystyle MAC= 932,\ MAAR = 9338, \text { and } JOCKO = 10270.\)

Let’s check

\(\displaystyle 932 + 9338 = 10270. \ \checkmark\)