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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

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    $\begingroup$ Hint: $\;12!\,$ is a multiple of $\,9\,$. $\endgroup$
    – dxiv
    Commented Sep 3, 2021 at 8:21

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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 $$ 11 \mid 18 -a$$

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    $\begingroup$ (+1) The downvote is undeserved. $\endgroup$
    – ultralegend5385
    Commented Sep 3, 2021 at 8:47
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    $\begingroup$ I welcome downvoters to explain the downvotes $\endgroup$
    – Infinity_hunter
    Commented Sep 3, 2021 at 8:52
  • $\begingroup$ I think the downvote might be because the downvoter thinks the question lacks context. I disagree, but still think that you should have left some things as an exercise for the OP, instead of getting to the answer. $\endgroup$
    – ultralegend5385
    Commented Sep 3, 2021 at 8:54
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    $\begingroup$ @ultralegend5385 Maybe, but I answered because OP has shown some effort in solving the question. $\endgroup$
    – Infinity_hunter
    Commented Sep 3, 2021 at 9:04

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