jmc

By the Euclidean algorithm, $$\begin{array}{rl}\text{gcd}(1729,1768)& =\text{gcd}(1729,1768-1729)\\ & =\text{gcd}(1729,39).\end{array}$$Since the sum of the digits of $1729$ is $19$, which is not divisible by $3$, it suffices to check whether or not $1729$ is divisible by $13$. We can find that it is by long division or noting that $12+1=\boxed{13}$ divides into $1729=12^3+1^3$ using the sum of cubes factorization.