Challenge for champions !!! 2

Let $m$ be a natural number and $r$ be the last digit of $m$ .

---

Then $m \equiv r (mod 10) \implies m^{10} \equiv r^{10} mod(10^{2})$

$\implies (m^{10})^{10} \equiv (r^{10})^{10} mod (10^{3})$

i.e. $m^{100} \equiv r^{100} (mod 1000)$

---

$\textbf{Case 1:-}$

If $r = 0$ , $r^{100} \equiv 0 (mod 1000)$ . So last three digits are $000$

---

$\textbf{Case 2:-}$

If $r=1,3,7,9$ , then $r^{2} \equiv 1 (mod 8)$ , so $r^{100} \equiv 1 (mod 8)$ , Also $\emptyset (125) = 100$ By Euler's Theorem, as $(r,5)=1, r^{100} \equiv 1 (mod 125)$ . Hence, $r^{100} \equiv 1(mod 1000)$ . So last 3 digits are $001$

---

$\textbf{Case 3:-}$

If $r=5$ then $r^{2} \equiv 1 (mod 8)$ . Hence $r^{100} \equiv 1(mod 8)$ and $r^{100} \equiv 0 (mod 125)$ . Using Chinese remainder theorem, $r^{100} \equiv 625 (mod 1000)$ . So last 3 digits are $625$

---

$\textbf{Case 4:-}$

If $r = 2,4,6,8$ then $r^{100} \equiv 0 (mod 8)$ . Since $(r,5)=1$ we get $r^{100} \equiv 1 (mod 125)$ . By Chinese remainder theorem, $r^{100} \equiv 376 (mod 1000)$ . So last 3 digits are $376$

---

Adding all the above numbers we get addition as $\boxed {1002}$