How to solve $9^{89}\equiv x\mod{1000}$ for $0\leq x\leq 999$ without calculating $9^{89}$

Question

Is there a method I can use to get an answer without having to resort to using Wolfram Alpha to calculate $9^{89}$ and taking the last three digits?

Answer 1

Using $9^{89}=(10-1)^{89}$

Now using Binomial Theorem $$(10-1)^{89}$$ $$=10^{89}-\binom{89}{88}10^{88}+\cdots-\binom{89}410^4+\binom{89}310^3-\binom{89}210^2+\binom{89}110-1$$

$$\equiv-\binom{89}210^2+\binom{89}110-1\pmod{10^3}$$

$$\equiv-\frac{89\cdot88\cdot100}2+89\cdot10-1\pmod{10^3}$$

$$\equiv-(100-11)(100-12)50+889\pmod{10^3}$$

$$\equiv -50\{100(100-11-12)+11\cdot12\}+889\pmod{10^3}$$

$$\equiv -50\cdot11\cdot12+889\pmod{10^3}\text{ as }50\cdot100\equiv0\pmod{10^3}$$

$$\equiv889-660\equiv289$$