Prove that $76$ raised to any integer power and then divided by $100$ (ignoring the remainder) is always divisible by $57$ Related (but different) question: Is it true that $76^n=76\pmod{100}$ for all $n>0$?
If you raise $76$ to an integer power ($n\ge2$) and ignore the last two digits, it appears that you always have an integer that is divisible by $57$, because the prime factorisation always seems to involve $3$ and $19$:
$76^2$ = 5776, and 57 = 3 x 19
$76^3$ = 438976, and 4389 = 3 x 7 x 11 x 19
$76^4$ = 33362176, and 333621 = 3 x 3 x  19 x 1951
$76^5$ = 2535525376, and 25355253 = 3 x 7 x 11 x 19 x 53 x 109
$76^6$ = 192699928576, and 1926999285 = 3 x 5 x 19 x 71 x 95231
$76^7$ = 14645194571776, and 146451945717= 3 x 3 x 7 x 11 x 19 x 1951 x 5701
 A: Another way to look at it:
$\frac {76^n - 76}{100} = \frac {76(76^{n-1} - 1)}{100} = \frac {19(76^{n-1}-1)}{25} = 19* \frac {(76 -1)(76^{n-2} + 76^{n-3} + ...... + 76 + 1)}{25} =$
$19*\frac {75(76^{n-2} + 76^{n-3} + ...... + 76 + 1)}{25} = 3*19*(76^{n-1} + 76^{n-2} + ...... + 76 + 1)$
So that's that.
But there so some interesting observations.  If $n = m + 1$ is odd and $m$ is even then:  $(76^{n-1=m} - 1) = (76^{2} - 1)(76^{m-2}+76^{m-4} + ... + 1) = (76-1)(76 + 1)(76^{m-2}+76^{m-4} + ... + 1)$ so $7=7*11|7^{n-1} - 1$ so for ever odd $n$, $57*77$ will divide it.
Likewise if $n = 3m + 1$ for some $m$ then $(76^{3m} - 1) = (76-1)(76^2 + 76 + 1)(76^{3m -3} + 76^{3m-6}.... + 1) = 7*(3*1951)*(76^{3m -3} + 76^{3m-6}.... + 1)$.
So for $n = 4,7,10, etc.$ you will get $3*19*3*1951$ will divide the values.
And we can find these patterns all day long.
$n = km + 1$ will always mean $76^{m-1}+ 76^{m-2} + ... + 76+ 1$ will divide the result.
For $n = 4m + 1$ we well always have $\frac{(76^2 + 1)(76^2 - 1)}{25}=3*53*109*7*11$ will always divide the results.
And so on.  
A: Split the numbers in two parts, separating the last two digits from the rest. 
We can easily calculate:
$$76^2 = 5776 = 57\cdot100 + 76$$
Clearly the hundreds part is divisible by $57$. But note that the low part is again $76$.
Let's assume that we can write the powers like this (with some integer $k_n$):
$$76^n = 57 \cdot k_n \cdot100 + 76 $$
then
$$76^{n+1} = 57 \cdot 76 \cdot k_n \cdot100 + 5776  $$
$$ =  (57 \cdot 76 \cdot k_n + 57) \cdot 100 + 76 $$
$$ = 57 \cdot (76 \cdot k_n + 1) \cdot 100 + 76  $$
And the hundreds part is again divisible by 57. There's also the $76$ in the low part, so $76^{n+1}$ is also in the assumed form ($k_{n+1} = 76 \cdot k_n + 1$). Since the assumption held for $n=2$ ($k_2 = 1$), it also holds for any higher power.
For $n=0$ or $n=1$, the hundreds part is zero, and therefore trivially divisible by anything.
A: Once we accept that $76^n$ always ends in $76$, your question becomes: is it always true that 
$$
   \frac{76^n - 76}{100}
$$
is always divisible by $57$ (that is, by both $3$ and $19$)?
It's enough to focus on the numerator $76^n - 76$, since the denominator $100$ has no factors of $3$ or $19$ in it. The numerator


*

*is obviously divisible by $19$, since $76 = 4 \cdot 19$, and

*becomes $1^n - 1$ when we reduce it modulo $3$,


so it's divisible by $57$.
A: Using the formula for the sum of a geometric series:
$$
\begin{align}
\frac{76^n-76}{76-1}
&=76\,\frac{76^{n-1}-1}{76-1}\\
&=76\left(76^{n-2}+76^{n-3}+\cdots+1\right)
\end{align}
$$
Therefore, since $75\cdot76=5700$,
$$
\begin{align}
\frac{76^n-76}{100}
&=57\left(76^{n-2}+76^{n-3}+\cdots+1\right)
\end{align}
$$
