5. Given the set $A_{n}=\left\{x \mid 2^{n}<x<2^{n+1}\right.$, and $\left.x=7 m+1, n 、 m \in \mathbf{N}_{+}\right\}, A_{6}$ the sum of all elements in $A_{6}$ is ( ).
A. 792
B. 890
C. 891
D. 990
5. C.

Since $6^{6}<x<2^{7}$, i.e., $64<x<128$, and $x=7m+1$, then $x$ takes the values $71, 78, 85, 92, \cdots, 120$, 127, a total of 9 numbers, and the sum is $\frac{9(71+127)}{2}=891$.