Example 5 (2003 British Mathematical Olympiad)
Given that $34!=295232799 c d 96041408476186096435 a b 000000$, find the values of the digits $a, b, c, d$.

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Given that $34!=k \times 11^{3} \times 7^{4} \times 5^{7} \times 3^{15} \times 2^{32}=k \times 11^{3} \times 7^{4} \times 3^{15} \times 2^{25} \times 10^{7}$,
so $b=0, a \neq 0$.
Considering the scenario when the last 7 zeros are removed, we know that the number can be divided by $2^{25}$, and also by 8, meaning the last three digits $\overline{55 a}$ can be divided by 8 (since 1000 can be divided by 8), thus $a=2$.

Since 34! can be divided by 9, the sum of its digits can also be divided by 9, so, $141+c+d \equiv 0(\bmod 9)$, which means $c+d \equiv 3(\bmod 9)$.

And 34! can also be divided by 11, so the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions can be divided by 11, thus, $80+d \equiv 61+c(\bmod 11)$, which means $8+d \equiv c(\bmod 11)$,

When $c+d=3,8+d=c$, there is no solution;
When $c+d=12,8+d=c$, there is no solution;
When $c+d=12, d=c+3$, there is no solution;
When $c+d=3, d=c+3$, we get $c=0, d=3$.
In conclusion, $a=2, b=0, c=0, d=3$.