10. For a sports parade, the coach decided to line up the children in rows of 8, but 5 children were left over. Then he lined them up in rows of 10, but 3 places were left empty. It is known that there were no fewer than 100 and no more than 150 children. How many children were there?
Solution: $a=8 x+5=8(x+1)-3, a+3 \vdots 8$,

$$
\begin{array}{lc}
a=8 x+5=8(x+1)-3, & a+3 \vdots 8 \\
a=10 y-3, & a+3 \vdots 10,
\end{array} \Rightarrow a+3 \vdots 40, a+3=\underline{120,160}
$$

Answer: 117.

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