I recently came across this question from a non-calculator exercise. The units and tens place value digits I can see as $0$ and $0$ since in $12!$ we have $10*5*2=100$ but is there a way to find $a$ without needing to calculate almost the entire $12!$? As it is not supposed to require a calculator there is perhaps something I am missing here? I am thinking something to do with primes??
The question;
$12!=4a90016bc$. Find the missing digits clearly showing your working
Expanding the comment of dxiv,
Let $n = a_k 10^k + \ldots +10a_1 + a_0$ (where each $a_i$ are digits) then see that $$n - (a_k + a_{k-1} + \ldots a_1 + a_0) = a_k(10^k -1) + \ldots + a_1(10-1) $$ Since $$10^m -1 = 9(10^{m-1} + 10^{m-2} + \ldots +10 + 1)$$ we see that $9 \mid 10^m-1$ for any $m$.
So $$9 \mid n -(a_k + a_{k-1} + \ldots a_1 + a_0)$$.
Here $n = 12!$ and $S =a_k + \ldots + a_0 = 20 + a + b+ c$. Since you already know that $b=c=0$ we have $S = 20 + a$. Since $9 \mid 12!$ we must have $9 \mid 20+a$.
Since $0\le a < 10$ we have $a =7$
Alternatively:
Here you can see a divisibility rule of $11$. Using that we see that $$ 11 \mid 18 -a - b + c$$ which implies that $$ 11 \mid 18 -a$$