**Explanation:-**

Solution:

Number of terms in the series = 10.

(We can get it easily by pointing the number of zeros in power of terms.

In 1st term number of zero is 1, 2nd term 2, and 3rd term 3 and so on.)

10107,Writtenas,(7+3)(4×2+2)7Theremainderwilldependon,327So,remainderwillbe21010007,remainder=210100007,remainder=1So, we get alternate 2 and 1 as remainder, five times each.So, required remainder is given by(2+1+2+1+2+1+2+1+2+1)7=157$$\begin{array}{rl}& \frac{{10}^{10}}{7},\phantom{\rule{thinmathspace}{0ex}}\text{Written}\phantom{\rule{thinmathspace}{0ex}}\text{as},\\ & \frac{{\left(7+3\right)}^{\left(4\times 2+2\right)}}{7}\\ & \text{The}\phantom{\rule{thinmathspace}{0ex}}\text{remainder}\phantom{\rule{thinmathspace}{0ex}}\text{will}\phantom{\rule{thinmathspace}{0ex}}\text{depend}\phantom{\rule{thinmathspace}{0ex}}\text{on},\\ & \frac{{3}^{2}}{7}\\ & \text{So,}\phantom{\rule{thinmathspace}{0ex}}\text{remainder}\phantom{\rule{thinmathspace}{0ex}}\text{will}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}2\\ & \frac{{10}^{1000}}{7},\phantom{\rule{thinmathspace}{0ex}}\text{remainder}=2\\ & \frac{{10}^{10000}}{7},\phantom{\rule{thinmathspace}{0ex}}\text{remainder}=1\\ & \text{So, we get alternate 2 and 1 as remainder, five times each}\text{.}\\ & \text{So, required remainder is given by}\\ & \frac{\left(2+1+2+1+2+1+2+1+2+1\right)}{7}\\ & =\frac{15}{7}\end{array}$$