Example 2: Now there is a number whose total is unknown. When reduced by 5 repeatedly, there is no remainder. When reduced by 715 repeatedly, there is a remainder of 10. When reduced by 247, there is a remainder of 140. When reduced by 391 repeatedly, there is a remainder of 245. When reduced by 187 repeatedly, there is a remainder of 109. What is the total number?
Let the total number be $x$, then we have the system of congruences $\left\{\begin{array}{l}x \equiv 0(\bmod 5), \\ x \equiv 10(\bmod 715), \\ x \equiv 140(\bmod 247), \\ x \equiv 245(\bmod 391), \\ x \equiv 109(\bmod 187) .\end{array}\right.$
Since $5,715,247,391,187$ are not pairwise coprime, but we have
$(5,715)=5|(10-0),(715,247)=13|(140-10)$,
$(715,187)=11|(109-10),(391,187)=17|(245-109)$.
Thus, $[5,715,247,391,187]=[5,143,19,23,17]$, and $5,143,19,23,17$ are pairwise coprime, so the original system of congruences has a solution. The equivalent system of congruences is $\left\{\begin{array}{l}x \equiv 0(\bmod 5), \\ x \equiv 10(\bmod 143), \\ x \equiv 140 \equiv 7(\bmod 19), \\ x \equiv 245 \equiv 15(\bmod 23), \\ x \equiv 109 \equiv 7(\bmod 17) .\end{array}\right.$
At this point, $M=5 \cdot 143 \cdot 19 \cdot 23 \cdot 17=5311735$.
$M_{1}=\frac{M}{5}, M_{2}=\frac{M}{143}, M_{3}=\frac{M}{19}, M_{4}=\frac{M}{23}, M_{5}=\frac{M}{17}$.
Thus, $M_{1}^{\prime} \equiv 3(\bmod 5), M_{2}^{\prime} \equiv 49(\bmod 143), M_{3}^{\prime} \equiv-1(\bmod 19)$,
$$
M_{4}^{\prime} \equiv 1(\bmod 23), M_{5}^{\prime} \equiv-7(\bmod 17) .
$$

Therefore, $x \equiv 3 \cdot 1062347 \cdot 0+49 \cdot 37145 \cdot 10+(-1) \cdot 279565 \cdot 7+1 \cdot 230945 \cdot 15+$ $(-7) \cdot 312455 \cdot 7 \equiv 10020(\bmod 5311735)$.

This is the solution we seek.