Trying to remove a mod from an equation. $376$ is a number that for positive integers $n$, $376^n$ will always end with the number $376$. Now knowing that $376^k \mod 1000 = 376$.
How do you prove that the following is true.
$$
376^k \mod 1000 = 376^{k+1} \mod 1000
$$
Is there a way to cancel the mods or can you plug $376^k \mod 100 = 376$ into the equation somehow?
 A: Of course it is true that
$$
a \mod b = a - \left\lfloor \frac{a}{b}\right\rfloor b
$$
but removing mod from your equation will make your life harder, not easier.  You need to do some number theory.
Consider the difference $376^{k+1} - 376^k$, for $k \ge 1$.
This is equal to $376^{k - 1} ( 376^2 - 376 ) $.
Now,
\begin{align*}
376^2 - 376
&= 376(376 - 1) \\
&= 376(375) \\
&= (8 \cdot 47)(125 \cdot 3) \\
&= (8 \cdot 125)(3 \cdot 47) = (1000) (3 \cdot 47)
\end{align*}
which is a multiple of $1000$.  So the difference between the two numbers is a multiple of $1000$;
hence they have the same remainder, which proves equation 1.
A: Proof is easiest  by induction. You have stated your answer  but you just need to word it right.
First off
$$
376^2 \equiv 376 \mod 1000
$$
can be verified directly.
The general statement as you point out is
$$
376^k\equiv 376 \mod 1000
$$
Assume it to be true for $k$. Now
$$
376^{k+1}\equiv 376^k \cdot 376 \equiv 376 \cdot 376 \equiv 376^2 \equiv  376\mod 1000
$$
This proves your statement.
Note: I am essentially rewriting what you state. I hope this is what you are looking for!
A: Hint $\,\ a^2\equiv a,\,\ \color{#c00}{a^n\equiv a}\ \Rightarrow\, a^{n+1}\equiv a\color{#c00}{a^n}\equiv a\color{#c00}a\equiv a$
