1. Given that $a$, $b$, $c$, and $d$ are prime numbers, and $a b c d$ is the sum of 77 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$
Given 77 consecutive positive integers, with the middle one being $x$. Then $a b c d=77 x=7 \times 11 x$, and $x \geqslant 39$.
Since $a, b, c, d$ are prime numbers, thus, $x$ can be decomposed into the product of two prime numbers, and the sum of these two prime numbers is minimized.

It is easy to verify that when $x$ is $49=7 \times 7$, the sum of these two prime numbers is the smallest.
Therefore, the minimum value of $a+b+c+d$ is $7+11+7+7=32$.