When 11 and 13 are together

$\text{11 divides m + 13n} \\ \implies \dfrac{m + 13n}{11} \in Z^+ \\ \dfrac{m + 11n + 2n}{11} \in Z^+ \\ \dfrac{11n}{11} + \dfrac{m + 2n}{11} \in Z^+ \\ \implies \text{11|(m+2n), ie 11 divides (m+2n)} \\ \implies 11 | 6(m+2n) \\ \boxed{\implies 11 | (6m + 12n)} \\ \text{13|(m + 11n)} \\ \implies \dfrac{m+11n}{13} \in Z^+ \\ \dfrac{13n + m - 2n}{13} \in Z^+ \\ \dfrac{13n}{13} + \dfrac{m-2n}{13} \in Z^+ \\ \implies 13 | (m-2n) \\ \boxed{\implies 13 | 6m - 12n}\\ \text{11 | (6m + 12n) }\\ \text{11 | (6m + 11n + 1n)} \\ \text{11 | (6m + n)} \\ \text{13 | 6m - 12n} \\ \text{13 | 6m - 13n + n} \\ \text{13 | (6m + n)} \\ \implies 143 | (6m+n) ~~ \boxed{\text{Since HCF(13, 11) = 1, (11)(13) = 143}} \\ \implies 6m + n = 143k, ~ k \in Z \\ 6m + n + 5n = 143k + 5n \\ 6m + 6n = 143k + 5n \\ 6(m+n) = 144k + 5n-k \\ \implies 6|(144k + 5n-k) \\ \implies 6|(144k+6n +(-n-k)) \\ \implies 6|(-(n+k))\\ \implies 6|(n+k) \\ \implies n+k \ge 6 \\ \text{let n = 5, k = 1} \\ 6(m+n) = 143k + 5n = 138k + 5(n+k) \ge 138 + 5\times 6 = 168 \\ \implies m + n \ge \dfrac{168}{6} = 28 \\ \therefore min(m+n) = 28$