What is $57^{46}$ divided by 17? $$\frac{57^{46}}{17}$$
What is best and quickest way to evaluate an approximate answer to this by hand?
 A: Why not simply do the computation? The most time-consuming part is to compute the $46$th power of $57$, but that can be done quite fast using repeated squaring in 6 multiplications. That results in a number with 81 digits. Dividing by 17 using long division is faster, and it takes about 80 easy steps.
When you are done, you can check your result with 
Google.
A: With mental arithmetic alone I would try the following. Undoubtedly you have needed the value of
$$
\tan\frac\pi6=\frac1{\sqrt3} \approx0.57\ldots
$$
enough many times to have it memorized. Therefore $57\approx 100/\sqrt3$. This means that
$57^{46}\approx 10^{92}/3^{23}$. Because $\sqrt{10}\approx3$ we also have $\log_{10}3\approx 0.5$ Thereforer $3^{23}\approx10^{11.5}$. So $57^{46}$ is probably somewhere between $10^{80}$ and $10^{81}$. Dividing this by 17 gives a result most likely between $10^{79}$ and $10^{80}$. If such a ballpark figure is enough, then we're done. If you want a significant digit, I need to work a bit harder, and use more accurate estimates.
A: Euler's theorem gives you the remainder.
A: You can do it by Fermats little theorem as well.
A: It really depends what you mean by "approximate answer". Since you tagged arithmetic...
Notice 57 = 3 . 19, you can say that $57^{46} / 17 = 3 \cdot 57^{45} \cdot \frac{19}{17}$. So if you know what $57^{45}$ is, you can approximate to within 2% if you multiply that by 3.3. 
Now, compute by hand $57^5 = 601692057$, which is less than 0.3% off from $6\cdot 10^8$. Observe that $57^{45} = (57^5)^9$. Using the binomial theorem you see that you can approximate that by $(6\cdot 10^8)^9 = 10 077 696 \cdot 10^{72}$ to within 3%. Rounding off the lower digits won't matter much in the error, so you have $57^{45} \sim 10^{79}$ to 3%. 
So putting it all together you have that to within 5%, 
$$ 57^{46} \sim 3.3 \cdot 10^{79} $$
A: HINT $\rm \mod\ 17:\ \ 57\equiv 6,\ \ 6^{46}\equiv 1/6^2,\ \ 1/6\equiv 3\ \ $ which immediately yields the result
A: By Euler's theorem $57^{16}\equiv1 \pmod{17}$.
It could be done using this.
A: The result is exactly:
$
34657277165715299429841134098673896005946880584360167911317357675111364398875720+\frac{9}{17}
$
