Pin the tail on the factorial

By Legendre's Formula, there are

$\left\lfloor\frac{33}{5}\right\rfloor + \left\lfloor\frac{33}{25}\right\rfloor = 7$

factors of $5$ in $33!,$ and there are

$\left\lfloor\frac{33}{2}\right\rfloor + \left\lfloor\frac{33}{4}\right\rfloor + \left\lfloor\frac{33}{8}\right\rfloor + \left\lfloor\frac{33}{16}\right\rfloor + \left\lfloor\frac{33}{32}\right\rfloor = 31$

factors of $2$ in $33!.$ So, there are exactly $7$ factors of $10$ in $33!.$ Since the expression currently ends with $6$ zeroes, we must have $c = 0.$

If there are $31$ factors of $2$ in $33!,$ we also know that $2^{14} \mid 33!.$ Since a positive integer is divisible by $2^{14}$ if and only if its last $14$ digits are divisible by $2^{14},$

$\dfrac{\overline{94401ab0000000}}{2^{14}}$

must be an integer. To simplify this fraction, we can write

$\dfrac{\overline{94401ab0000000}}{2^{14}} = \dfrac{10^7 \cdot \overline{94401ab}}{2^{14}} = \dfrac{5^7 \cdot \overline{94401ab}}{2^7}.$

This means that $\overline{94401ab}$ must be divisible by $2^7 = 128.$ Using long division, we can find that $\overline{94401ab} = 9440100 + \overline{ab} \equiv 100 + \overline{ab} \pmod{128},$ so $\overline{ab} \equiv 28 \pmod{128}.$ Because $\overline{ab}$ is a two-digit number, we must have $\overline{ab} = 28.$

In conclusion, $\overline{abc} = \boxed{280}.$

The prime factorization of $33!$ is

$2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^4 \cdot 11^3 \cdot 13^2 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$

Since it's return in base representation of $10$ , so the last $7$ digits of $33!$ are all $0$ 's. That's because $33!$ is written in base $10$ representation and has a total of $7$ factors of $5$ as the largest prime factor of $10$ is $5$

Removing the last seven digits means the remaining 7 digits is $\overline{94401ab}$ , thus $c=0$

Because we're dividing $33!$ by $10^7 = 2^7 \cdot 5^7$ , we are left with

$2^{24} \cdot 3^{15} \cdot 7^4 \cdot 11^3 \cdot 13^2 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$

So we want to find the last two digits of the above expression: $\bmod{100}$

$2^{24} \equiv (2^4)^6 \equiv 16^6 \equiv 16$

$3^{15} \equiv (3^5)^3 \equiv 43^3 \equiv 7$

And apply the similar congruences to the rest of the terms, we have

$\equiv 16 \cdot 7 \cdot 1 \cdot 31 \cdot 69 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$

$\equiv 28$

So $a=2, b=8$ . Hence, $\overline{abc} = \boxed{280}$

Notice that the order of $5$ in $33!$ is $7$ because $5, 10, 15, 20, 30$ each contribute one factor and $25$ contributes two. Therefore the number ends in $7$ zeros (the order of $2$ being much bigger than that of $5$ ), so $c = 0$ .

To determine the remaining digits, we will generalize the familiar rule that the number is divisible by four precisely when its last two digits are divisible by four.

Lemma $2^n$ divides last $n$ digits of $A$ if and only if $2^n$ divides $A$ .

Proof Write $A = B \cdot 10^n + C$ . This means that $A \equiv C \pmod{2^n}$ and the lemma follows.

As there are many factors of $2$ in $33!$ (certainly more than $16$ ) and we only used $7$ of them for the last $7$ zeros, it's clear that $2^7 = 128$ divides $33!$ . The lemma then says that also $\overline {94401ab}$ is divisible by $128$ . So it's enough to divide $9440199$ by $128$ to get $9440199 = 128 \cdot 73751 + 71$ . This means that the number we are looking for is $9440128$ and consequently the answer is $\overline {abc} = 280$ .

First I made a prime factorization of 33!

33! = $31 \times 29 \times 23 \times 19 \times 17 \times 13^{2} \times 11^{3} \times 7^{4} \times 5^{7} \times 3^{15} \times 2^{31}$

Knowing that the last 6 digits of the 33! is zero, so I eliminated all those zeroes by removing $2^{6} \times 5^{6}$ = $10^{6}$ to make our last 3 digits be $\overline{abc}$

Now the reduced prime factorization :

$31 \times 29 \times 23 \times 19 \times 17 \times 13^{2} \times 11^{3} \times 7^{4} \times 5 \times 3^{15} \times 2^{25}$

since $2 \times 5$ = 10. We can easily say that our last digit $\overline{c}$ = 0. Again we remove $2 \times 5$ to make our last 3 digits be 1 $\overline{ab}$ . :

$31 \times 29 \times 23 \times 19 \times 17 \times 13^{2} \times 11^{3} \times 7^{4} \times 3^{15} \times 2^{24}$

In solving for digit $\overline{b}$ , I multiplied all the last digits in the prime factorization (eg . $2^{24}$ = 6):

$1 \times 9 \times 3 \times 9 \times 7 \times 9 \times 1 \times 1 \times 7 \times 6$ = 642, 978

So it means our digit $\overline{b}$ is equal to 8.

Lastly in solving digit $\overline{a}$ , I use the divibility rules of 8. Since our last 3 digit number should be 1 $\overline{a}$ 8. The only possible values of digit $\overline{a}$ to make 1 $\overline{a}$ 8 be a divisible of 8 is 2 and 6. So I divided each by 8.

$\frac{128}{8}$ = 16 $\frac{168}{8}$ = 21

Clearly from this, digit $\overline{a}$ is must be 2, since 8 is equal to $2^{3}$ and we if we try to eliminate $2^{3}$ from $2^{24}$ there would be still $2^{21}$ which makes the last 3 digit still an even number.

So, $\overline{abc}$ = $\boxed{280}$

Put it in the calculator... 33! = 8683317618811886495518194401280000000. Thus, abc = 280.

There are five multiples of 5 and one multiple of 25 in 33!. Therefore, there are 7 zeros at the end. c=0.

Then, we observe that 9440000 is divisible by 128. Therefore, in order to make 94401ab divisible by 128. ab=28.

Main Idea : Convert the factorial as a product of primes.

The primes before $31$ are : $2,3,5,7,11,13,17,19,23,29,31,33$

The highest power of a prime $p$ contained in a factorial is given by:

$\lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor+ ... + \lfloor \frac{n}{p^k} \rfloor , \ \ p^k \leq n$

Keeping the above facts in find we write: $\ \ 33!=2^{31} \cdot 3^{16} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^2 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$ $\\33!=10^7(2^{24}\cdot 3^{15} \cdot 7^{4} \cdot 11^{3} \cdot 13^2 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31)=10^7 \cdot t$ Observe that the last $7$ digits of $33!$ are $0$ so $c=0$ .

Work $t \pmod {100}$ to retrieve the $8$ th and $9$ th digit of $33!$ .

Clearly, $t \equiv 0 \pmod{4}$

The only obstacle left is to work $t \pmod{25}$

$2^{24} \equiv 1024^2 \cdot 2^4 \equiv 16 \pmod{25}$ $\\ 3^15 \equiv 27^5 \equiv 2^5 \equiv 7 \pmod{25}$ $\\ 7^4 \equiv 1 \pmod{25}$ $\\ 11^3 \equiv 11 \cdot -4 \equiv 6 \pmod{25}$ $\\ 13^2 \equiv -6 \pmod{25}$ $\\17 \cdot 19 \equiv -8 \cdot -6 \equiv -2 \pmod{25}$ $\\23 \cdot 29 \cdot 31 \equiv -2 \cdot 4 \cdot 6 \equiv 2 \pmod{25}$

Now multiply all the above congruences to get:

$t \equiv 16 \cdot 7 \cdot 1 \cdot 6 \cdot(-6) \cdot (-2) \cdot 2 \equiv 16 \cdot 7 \cdot 6 \cdot 24 \equiv 16\cdot 6 \cdot 7 \cdot (-1) \equiv (-4) \cdot (-7)\equiv 3 \pmod{25}$

By the Chinese Remainder Theorem, $t \equiv 28 \pmod{100}$

So $a=2, b=8, c=0$

Answer : $\\ \overline{abc}=\boxed{280}$

Using wolfram alpha:

1

33! mod 10^14