Trailing Digits #1

In fact all positive integer powers of $25$ end in the digits $25$ . This is easy to prove by induction; say $25^k \equiv 25 \pmod{100}$

(which clearly holds for $k=1$ ). Then $25^{k+1} \equiv 25\times 25 = 625 \equiv 25 \pmod{100}$

which proves the claim.

We can use the Chinese remainder theorem to solve the problem. Let us consider the last three digits of $25^{293864}$ . Note that $1000 = 2^3 \times 5^3 = 8 \times 125$ . Then we have:

$\begin{aligned} 25^{293864} & \equiv 0 \pmod {125} \\ 25^{293864} & \equiv (24+1)^{293864} \equiv 1 \pmod 8 \\ \implies 25^{293864} & \equiv 8 \blue n + 1 & \small \blue{\text{where }n \text{ is an integer.}} \\ 8n + 1 & \equiv 0 \pmod {125} \\ \implies n & \equiv 78 \\ 25^{293864} & \equiv 8(78)+1 \equiv 625 \pmod {1000} \\ \implies 25^{293864} & \equiv \boxed{25} \pmod {100} \end{aligned}$