How do you calculate the modulo of a high-raised number?

Question

I need some help with this problem:

$$439^{233} \mod 713$$

I can't calculate $439^{223}$ since it's a very big number, there must be a way to do this.

Thanks.

Answer 1

There are often tricks to this if the numbers are nice enough, but even if they're not, here's a way that's not entirely horrible.

You already know what 439 is mod 713. What is $439^2 \mod 713$? What about $439^4$? (Hint: take your answer for $439^2$ after reducing it mod 713, and then square it again.) In the same way, calculate $439^8, 439^{16}, \dots, 439^{128} \mod 713$. Now just note that 233 = 128 + 64 + 32 + 8 + 1. So multiply the appropriate powers of 439 together - again, one calculation at a time, reducing mod 713 each time.

Now you should only have to do 11 calculations, and now all your numbers are 6 digits or less. Rather than impossible, it's now simply tedious. :)

By the way, one thing to notice: 713 = 23 * 31. Perhaps your calculations will be easier if you do them mod 23 and 31, then apply the Chinese remainder theorem?

Answer 2

You can do this step by step by first computing $439^2\equiv 211 {\ \rm mod\ }713$. Then you compute $439^4\equiv 211^2 \equiv 315 {\ \rm mod\ }713$. Continue to square and reduce mod $713$ and build up a list of powers $$439^1, \qquad 439^2, \qquad 439^4 \qquad \dots \qquad 439^{128}\qquad \qquad ({\rm mod\ }713)$$ Finally, write $233$ as a sum of powers of $2$, and compute $$439^{233}=439^{1+8+32+64+128}=439\cdot 439^8\cdot 439^{32}\cdot439^{64}\cdot 439^{128} \qquad ({\rm mod\ }713)$$ Just remember to reduce mod $713$ each time you get a product which is larger than $713$.

Answer 3

You can go fast if you incrementally square: $439^2=211 \pmod{713}$; $211^2=315 \pmod{713}$; $315^2=118 \pmod{713}$; $315^2=377 \pmod{713}$; $377^2=242 \pmod{713}$; $377^2=98 \pmod{713}$; etc.

The last one means $439^{64}=98 \pmod{713}$; In this way you can combine and reach 233 fast.

Ultimately $439^{233}=211 \pmod{713}$;

Answer 4

$713=23\cdot 31$

$439 \pmod {23}=2$ and $\phi(23)=22$ and $233\equiv 13{\pmod {22}}$

So, $439^{223} {\pmod {23}} \equiv 2^{22\cdot 10 + 13}\equiv {(2^{22})}^{10}2^{13}\equiv 2^{13} {\pmod {23}}$ using Euler's Totient Theorem.

$2^6\equiv 18 {\pmod {23}}, 2^7\equiv 36 \equiv 13$

$\implies 2^{13}\equiv 18\cdot 13=234\equiv 4 {\pmod {23}}=4+23x$ for some integer $x$.

$439 \pmod {31}=5$ and $\phi(31)=30$ and $233\equiv 23{\pmod {30}}$

So, $439^{223} {\pmod {31}} \equiv 5^{23} {\pmod {31}}$

$5^3 \equiv 1 {\pmod {31}} \implies 5^{23}\equiv({5^3})^7 5^2\equiv 5^2{\pmod {31}}=25+31y$ for some integer $y$.

So, we need to find $z$ such that $z=25+31y=4+23x$

Expressing as continued fraction, $$\frac{31}{23}=1+\frac{8}{23}=1+\frac{1}{\frac{23}{8}}=1+\frac{1}{2+\frac{7}{8}}$$

$$=1+\frac{1}{2+\frac{1}{\frac{8}{7}}}=1+\frac{1}{2+\frac{1}{1+\frac{1}{7}}}$$

So, the last but one convergent is $$1+\frac{1}{2+\frac{1}{1}}=\frac{4}{3}$$

So, $23\cdot 4- 31\cdot 3=-1$

$25+31y=4+23x\implies 23x=31y+21(31\cdot 3-23\cdot 4)$ $\implies 23(x+84)=31(y+63)$ $$\implies x+84=\frac{31(y+63)}{23}$$

So, $23\mid (y+63)$ as $x+84$ is integer and $(31,23)=1$ i.e., $ 23\mid (y+69-6)\implies 23\mid (y-6) \implies y=6+23w$

So, $z=25+31y=25+31(6+31w)=713w+211 \equiv 211 {\pmod {713}}$