$\begin{array} { c c c } 25 ^ 1 & \equiv 25 & \pmod{1000} \\ 25 ^ 2 & \equiv 625 & \pmod{1000} \\ 25 ^ 3 & \equiv 625 & \pmod{1000} \\ 25 ^ 4 & \equiv 625 & \pmod{1000} \\ \vdots & \vdots & \vdots \\ \end{array}$

We know that the last 3 digits of $25 ^ n$ will always be a constant 625 for large enough integer values of $n$ .

What is the smallest positive integer value $a$ such that the last 5 digits of $\left( 5 ^ a \right) ^n$ will always be a constant for large enough integer values of $n?$

4
6
8
10

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For large enough $n$ , we need that $10^5 | (5^{a(n+1)}-5^{an})$ , or $10^5 | 5^{an}(5^a-1)$ . For large enough $n$ , this happens if and only if $2^5 | (5^a-1)$ . The multiplicative order of $5$ mod $32$ is easily found to be $\fbox{8}$ .