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.
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.
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?
$713=23\cdot 31$
$439 \pmod {23}=2$ 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, $z=25+31y=25+31(6+31w)=713w+211 \equiv 211 {\pmod {713}}$
Here`s the algorithm,basically it is Modular exponentiation.
function modular_pow(base, exponent, modulus) result := 1 while exponent > 0 if (exponent mod 2 == 1): result := (result * base) mod modulus exponent := exponent >> 1 base = (base * base) mod modulus return result
Also here is the working code in c++ which can work for upto 10^4 test cases,in 0.7 seconds Also the ideone link http://ideone.com/eyLiOP
#include using namespace std; #define mod 1000000007 long long int modfun(long long int a,long long int b) { long long int result = 1; while (b > 0) { if (b & 1) { a=a%mod; result = (result * a)%mod; result=result%mod; } b=b>>1; a=a%mod; a = (a*a)%mod; a=a%mod; } return result; } int main() { int t; cin>>t; while(t--) { long long int a,b; cin>>a>>b; if(a==0) cout<<0<<"\n"; else if(b==0) cout<<1<<"\n"; else if(b==1) cout<
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 $713$.
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 $439^{233}=211 \pmod{713}$;