Let $$2^{4\times5^k}\equiv a \pmod {5^{k+3}},\\2^{4\times5^k}\equiv b \pmod {5^{k+4}},$$ and $0<a<5^{k+3},0<b<5^{k+4},$ prove that $a=b.$$(k>1)$
This is equivalent to this: if $2^{4\times5^k}=x\times5^{k+3}+a,0<a<5^{k+3},$ then $5\mid x.$
ADD: A similar problem: Prove that if $2^{2\times5^k}=x\times5^{k+4}+a,0<a<5^{k+4},$ then $5\mid x.(k>2)$
Using this, if ord$\displaystyle _{(p^k)}a = d$ where k is a natural number and $p$ odd prime,
we can show that ord$_{(p^{k+1})}a = d$ or $pd$
Now, $\displaystyle 2^2\equiv-1\pmod5\implies 2^4\equiv1\pmod5\implies$ord$_52=4$
$\displaystyle\implies$ord$_{(5^2)}2=4$ or $4\cdot5=20$ which $=\phi(25)$
Now, $\displaystyle2^4=16\not\equiv1\pmod{25}\implies$ ord$_{(5^2)}2=20$
So, $2$ is a primitive root of $25$
using this, $2$ is a primitive root of $5^k$ for integer $k\ge1$
$\displaystyle\implies2^{4\cdot5^k}\equiv1\pmod{5^{k+1}}\equiv1+c\cdot5^{k+1}\pmod{5^r}$ for integer $r\ge k+1$
where $c$ is some integer not divisible by $5$ as $\displaystyle2^{4\cdot5^k}\not\equiv1\pmod{5^r}$ where $r\ge k+1$
