47.1. Find the sum of all three-digit numbers divisible by 7.

$(9-10$ grade $)$

6
47.1. There are a total of 900 three-digit numbers. Among them, multiples of seven will be $\left[\frac{900}{7}\right]=128$ numbers. It is easy to find that the first such number is 105, and the last one is 994. Thus, we need to find the sum of 128 terms of an arithmetic progression with the first term 105 and the last term 994. We get:

$$
S_{128}=\frac{105+994}{2} \cdot 128=1099 \cdot 64=70336 .
$$