Mod1000 won't help!

Let us write $600!$ on format like below:

$600!\\=\color{#D61F06}{\left(5\times{10}\times{15}\times{20}\ldots\times{595}\times{600}\right)}\times\\{\left(1\times{2}\times{3}\times{4}\right)}\times{\left(6\times{7}\times{8}\times{9}\right)}\times{\ldots\times{\left(596\times{597}\times{598}\times{599}\right)}}\\=\color{#D61F06}{\left({5}^{120}\times{120!}\right)}\times\\{\left(1\times{2}\times{3}\times{4}\right)}\times{\left(6\times{7}\times{8}\times{9}\right)}\times{\ldots\times{\left(596\times{597}\times{598}\times{599}\right)}}\\=\color{#D61F06}{\left({10}^{120}\times{120!}\right)}\times{2^{-120}}\times\\{\left(1\times{2}\times{3}\times{4}\right)}\times{\left(6\times{7}\times{8}\times{9}\right)}\times{\ldots\times{\left(596\times{597}\times{598}\times{599}\right)}}$ .

Let us observe all product of each black item in the bracket. We can see that all last three digits of the product is $24$ . So, we can say that the last three digits of

$\begin{array}{c}&\dfrac{600!}{{10}^{120}}&\equiv\color{#D61F06}{\left(120!\right)}\times{2^{-120}}\underbrace{\left(\times{24}\times{24}\times{24}\ldots\times{24}\right)}_\text{120 times}\\&\equiv\color{#D61F06}{\left(120!\right)}\underbrace{\left(\times{12}\times{12}\times{12}\ldots\times{12}\right)}_\text{120 times}\equiv\color{#D61F06}{\left(120!\right)}\times{{12}^{120}}\pmod{1000}\end{array}$

Using same method above, we will determine the last three digits non zero of $\color{#D61F06}{\left(120!\right)}$ .

$\begin{array}{c}&\dfrac{120!}{{10}^{24}}&\equiv\color{#D61F06}{\left(24!\right)}\times{{12}^{24}}\pmod{1000}\end{array}$

And the last for $~\color{#D61F06}{24!}$

$\dfrac{24!}{{10}^{4}}\equiv\color{#D61F06}{\left(4!\right)}\times{21}\times{22}\times{23}\times{24}\times{{12}^{4}}\equiv576\times{{12}^{4}}\pmod{1000}$ .

Finally, we have

$\begin{array}{c}&\dfrac{600!}{{10}^{120}}&\equiv{12}^{(120+24+4)}\times{576}\equiv{12}^{(148)}\times{576}\pmod{1000}\end{array}$ .

For simplicity, we will work with modulo $8$ and $125$ to use Euler's theorem. Now we have:

$4^{148}\equiv\begin{cases}{16}^{74}\equiv0\pmod{8}\\\left(4^{\phi(125)}\right)\times{4^{48}}\equiv86\pmod{125}\end{cases}$

and $~~3^{148}\equiv\begin{cases}9^{74}\equiv1\pmod{8}\\\left(3^{\phi(125)}\right)\times{3^{48}}\equiv111\pmod{125}\end{cases}$

Note that $\phi(125)=\left(1-\dfrac{1}{5}\right)\times{125}=100$

By using Chinese Reminder Theorem we shall get that $4^{148}\equiv336\pmod{1000}~~$ and $~~3^{148}\equiv361\pmod{1000}$

Hence,

$\begin{array}{c}&\dfrac{600!}{{10}^{120}}&\equiv{4}^{148}\times{3^{148}}\times{576}\equiv336\times{361}\times{576}\equiv69866496 \\&\equiv\boxed{\boxed{\huge{496}}}\pmod{1000}\end{array}$ .

Therefore, the final answer is $4+9+6=19$

You can check this for other similar problem

Here's my java solution. I had to use BigInteger instead of double because factorial of 600 simply exceeds the max limit of double.

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BigInteger fact = new BigInteger("1");
for(int n = 1; n <= 600; n++)
    fact = fact.multiply(BigInteger.valueOf(n));
String ans = fact.toString().replace("0", "");
int sum = 0;
for(int n = 1; n <= 3; n++)
    sum += Character.getNumericValue(ans.charAt(ans.length() - n));
System.out.println(sum);