JOMO 5, Short 7

$201 \equiv 1 \pmod{1000}$

$201^2 \equiv 401 \pmod{1000}$

$201^3 \equiv 601 \pmod{1000}$

$201^4 \equiv 801 \pmod{1000}$

$201^5 \equiv 1 \pmod{1000}$

$402 \equiv 2 \pmod{5} \Rightarrow 201^{402} \equiv 201^2 \equiv 401 \pmod{1000}$

So the last $3$ digits of $201^{402}$ are $401$

We want to find $201^{402}\pmod{1000}$ so first we shall apply Euler's Theorem . It states that $a^{\phi(n)}\equiv1\pmod n$ where $\phi(n)$ is Euler's Totient function . It is the sum of all positive integers below $n$ that are coprime to $n$ . Fermat's Little Theorem is a special case for when $n=p$ and $p$ is prime. Thus there are $\phi(p)=p-1$ coprime numbers below.

Anyway using Euler's Product Formula we can compute $\phi(1000)$ . Euler's product formula states that $\phi(n)=n\prod_{p|n}\left(1-\frac1p\right)$ where $p$ is a prime divisor of $n$ . Therefore $\phi(1000)=1000\left(1-\frac12\right)\left(1-\frac15\right)=400$ thus $201^{402}\equiv201^{400}\times201^2\equiv201^2\pmod{1000}$ and simply computing $201^2=40401$ means that the answer is $\boxed{401}$ .

Write $201^{402}$ as $(200+1)^{402}$ . Using the binomial theorem, we get that

$(200+1)^{402}=\binom{402}{0}200^{402}+\binom{402}{1}200^{401}1^1+\cdots+\binom{402}{401}200^11^{401}+\binom{402}{402}1^{402}$

Notice that any term involving $200$ to the power of anything to the power of at least two has last three digits all $0$ so we only need the two terms have that powers of $200$ to the first and zeroth. They are $\binom{402}{401}200^11^{401}+\binom{402}{402}1^{402}=402\cdot200+1=80401$ and thus the desired last three digits are $401$ .