Three digits

First, we note that finding the last three digits is the same as performing the operation of given number modulo $1000$ . Also, $\gcd(143,1000)=1$ . Next, we use the Euler's Theorem (which is a generalization of Fermat's Little Theorem ) to ease out computations of residue by reducing exponents on the given number. The theorem states that:

$a^{\phi(n)}\equiv 1\pmod{n}~\forall a,n\in\mathbb{Z^+} \land \gcd(a,n)=1$

Note: $\phi(n)$ is the Euler's Totient Function which counts the totatives of $n$ .

$\phi(1000)=\phi(5^3\times 2^3)=1000\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5}\right)=400$

$\therefore\quad143^{400}\equiv 1\pmod{1000}\implies 143^{1000000}=(143^{400})^{2500}\equiv 1\pmod{1000}$

$143^{999999}=143^{1000000}\times 143^{-1}\equiv 1\times 143^{-1}\equiv 143^{-1}\pmod{1000}$

Let $z\equiv 143^{-1}\pmod{1000}$ . Then, we have, $143z\equiv 1\pmod{1000}$ . Using Extended Euclidean Algorithm , we have,

$143z\equiv 1\pmod{1000}\implies z\equiv 7\pmod{1000}$

$\therefore\quad 143^{999999}\equiv 7\pmod{1000}$

Required sum $=0+0+7=\boxed{7}$

Euler Totient Function: $\phi(1000) = 1000 \times \left (1 - \frac 1 2 \right ) \left (1 - \frac 1 5 \right ) = 400$

Consider modulo $1000$

$\begin{aligned} 143^{999999} & \equiv & 143^{10^6 - 1} \\ & \equiv & 143^{10^6 \pmod {400} } \cdot 143^{-1} \\ & \equiv & 143^0 \cdot \frac {1}{143} \\ & \equiv & \frac {1}{143} \\ \end{aligned}$

So we want to to find modulo inverse of $143$ modulo $1000$ , which is easily computed to be $\boxed{7}$ because $1 = 7(143) - 1(1000)$ .

I got the answer using binomial expansion of (140+3)^999999

We can let n be the exponent of 143. Then we must look for a pattern. If the problem asks us to find what is the last digit or the last two digits, it would be much easier. Let n = 1

143^1 = 143.

n = 2, 143^2 = 20449. But we don't need to find out the other digits.

After solving for the last three digits in 143^n, we multiply that number to 143 to get 143^(n+1). If n = 2, the last three digits are 449. n = 3: 449 x 143 = ...207. Just solve for the last three digits to avoid wasting time. Repeat until you find a pattern. After n = 20, the last three digits repeat.

We divide 999999 by 20, then we get a remainder of 19. if n = 19, the last three digits are 007. Hence, the answer is 7.

Of course there are more elegant ways to solve this but for pattern-oriented people such as me this solution works best.

last digit of 143^999999 is same as the last digit of 3^999999......... now last digit of 3^1=3 , 3^2=9 , 3^3=7 , 3^4=1 , 3^5=3 , so after power 4 the repetition start so, now 999999 % 4 = 3 , so answer is 7