Hint : $3^2 = -1 $ (mod $10$).Hence what can you say about $3^{100}$ ?
Edit: By above observation ,$3^{100}=(3^2)^{50} =1$ (mod10).Hence what can you say about $3^{101}$ ?
I don't want to bury you in notation, so I will settle for examples and I hope you will see the implied pattern.
\begin{align}
1234567\pmod 9
&= (1+2+3+4+5+6+7) \pmod 9\\
&= 28 \pmod 9\\
&= (2+8) \pmod 9\\
&= 10 \pmod 9\\
&= (1+0) \pmod 9\\
&= 1\\
\end{align}
\begin{align}
1234567\pmod{99}
&= (1+23+45+67) \pmod{99}\\
&= 136 \pmod{99}\\
&= (1 + 36) \pmod{99}\\
&= 37\\
\end{align}
\begin{align}
1234567\pmod{999}
&= (1+234+567) \pmod{999}\\
&= 802
\end{align}
The verification of that last computation would look like this
\begin{align}
1234567
&= 1 \times 1000000 + 234 \times 1000 + 567 \pmod{999}\\
&= 1 \times (1 + 999999) + 234 \times (1 + 999) + 567 \pmod{999}\\
&= (1 + 999999) + (234+ 234 \times 999) + 567 \pmod{999}\\
&= 1 + 234 + 567 \pmod{999}\\
\end{align}
So
\begin{align}
123,123, \ldots ,123 \pmod{999}
&= 123+123+\ldots + 123 \pmod{999}\\
&= 100 \times 123 \pmod{999}\\
&= 12300 \pmod{999}\\
&= (12 + 300) \pmod{999}\\
&= 312
\end{align}
For the case $10^n + 1$, it looks like this
\begin{align}
1234567\pmod{11}
&= (7-6+5-4+3-2+1) \pmod{11}\\
&= 4 \pmod{11}\\
&= 4\\
\end{align}
\begin{align}
1234567\pmod{101}
&= (67 - 45 + 23 - 1) \pmod{101}\\
&= 44\\
\end{align}
\begin{align}
1234567\pmod{1001}
&= (567-234+1) \pmod{1001}\\
&= 334
\end{align}
The verification of that last computation would look like this
\begin{align}
1234567
&= 1 \times (1001000 - 1001 + 1) + 234 \times (1001-1) + 567 \pmod{1001}\\
&= 1 - 234 + 567 \pmod{1001}\\
&= 1 \times (-1001 + 1) + 234 \times (-1) + 567 \pmod{1001}\\
&= 567 - 234 + 1 \pmod{1001}\\
&= 334\\
\end{align}
Note similarly that
$$10^9 = 1000000000 = 1001000000 - 1001000 + 1001 - 1 = -1 \pmod{1001}$$
Best Answer
Similar to casting out nines to find the remainder when divided by 9, you can cast out 999's to find the remainder when divided by 999. The reason this works is the same: 1000 = 1 modulo 999, so xyz000...000 = xyz modulo 999. But instead of just adding up the digits, you have to add three-digit groups.
Take for instance the number 12345678. Break this into three-digit groups (starting from the right):
12 345 678
Take the sum:
12 + 345 + 678 = 1035
Reduce modulo 999, using the same trick:
1035 = 1 + 035 = 36 mod 999
So 12345678 = 36 mod 999.
In your case, we have
678 967 896 789
repeated 25 times:
25 * (678 + 967 + 896 + 789) = 25 * 3330 = 83250
So 67896789...6789 = 83250 = 83 + 250 = 333 modulo 999.