1. Given three positive integers $a$, $b$, and $c$ whose squares sum to 2011, and the sum of their greatest common divisor and least common multiple is 388. Then the sum of the numbers $a$, $b$, and $c$ is $\qquad$ .
Let the greatest common divisor of $a$, $b$, and $c$ be $d$.
Then $a = d a_{1}$, $b = d b_{1}$, $c = d c_{1}$.
Assume without loss of generality that $a_{1} \geqslant b_{1} \geqslant c_{1}$, and the least common multiple of $a_{1}$, $b_{1}$, and $c_{1}$ is $m$.
From the problem, we have
$$
a_{1}^{2} + b_{1}^{2} + c_{1}^{2} = \frac{2011}{d^{2}}, \quad 1 + m = \frac{388}{d}.
$$

Thus, $d^{2} \mid 2011$, $d \mid 388$, and $d < 45$, $388 = 2^{2} \times 97$, and $d$ is odd.
Therefore, $d = 1$.
Hence, $m = 387 = 9 \times 43$, and $a_{1}$, $b_{1}$, and $c_{1}$ are all less than 45, and they are divisors of $m$, which are $1, 3, 9, 43$.
Upon inspection, $a_{1} = 43$, $b_{1} = c_{1} = 9$.
Therefore, $a + b + c = 61$.