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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.

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5 Answers 5

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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?

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  • $\begingroup$ Had we needed to compute a^b%c and a>c. Can we change a to a%c and then do a^b%c? $\endgroup$
    – asn
    Commented Aug 2, 2019 at 15:40
  • $\begingroup$ Is there any answer on ((abc*d.....)^(kk^pp))%mod? $\endgroup$
    – asn
    Commented Aug 2, 2019 at 15:46
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$713=23\cdot 31$

$439 \pmod {23}=2$ 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, $z=25+31y=25+31(6+31w)=713w+211 \equiv 211 {\pmod {713}}$

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  • 1
    $\begingroup$ Thank you for your answer! I'll ask a friend of mine to explain this to me, I'm fairly new to modulus calculation haha. $\endgroup$ Commented Sep 14, 2012 at 10:03
  • $\begingroup$ I believe you made a small mistake and set 439^223 rather than 439^233 on the 3rd line $\endgroup$
    – Mart
    Commented Oct 24, 2021 at 14:18
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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 $713$.

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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 <iostream>
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<<a%mod<<"\n";
           else
           {
               cout<<(modfun(a,b))%mod<<"\n";
           }

        }
    return 0;
}
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You can go fast if you incrementally square: $439^2=211 \pmod{713}$; $211^2=315 \pmod{713}$; $315^2=118 \pmod{713}$; $118^2=377 \pmod{713}$; $377^2=242 \pmod{713}$; $242^2=98 \pmod{713}$; etc.

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

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

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