Skip to content
Related Articles

Related Articles

Extended Midy’s theorem
  • Last Updated : 25 Jan, 2021

According to Midy’s theorem, if the period of a repeating decimal for a / p      , where p is prime and a / p      is a reduced fraction, has an even number of digits, then dividing the repeating portion into halves and adding gives a string of 9s. For example 1/7 = 0.14285714285.. is a repeating decimal with 142857 being repeated. Now, according to the theorem, it has even number of repeating digits i.e. 142857. Further if we divide this into two halves, we get 142 and 857. Thus, on adding these two, we get 999 which is string of 9s and matches our theorem. 

In Extended Midy’s theorem if we divide the repeating portion of a/p into m digits, then their sum is a multiple of 10m -1.

Suppose a = 1 and p = 17, 
a/p = 1/17 = 0.0588235294117647… 
So, 0588235294117647 is the repeating portion of the decimal expansion of 1/17. Its repeating portion has 16 digits and it can be divided into m digits where m can be 2, 4, 8. 
If we consider m = 4 then 0588235294117647 can be divided into 16/4 = 4 numbers and if we add these 4 numbers then result should be a multiple of 104 – 1 = 9999 i.e, 
0588 + 2352 + 9411 + 7647 = 19998 = 2*9999

C++




// C++ program to demonstrate extended
// Midy's theorem
#include <bits/stdc++.h>
using namespace std;
 
// Returns repeating sequence of a fraction.
// If repeating sequence doesn't exits,
// then returns -1
string fractionToDecimal(int numerator,
                         int denominator)
{
    string res;
 
    /* Create a map to store already seen remainders
    remainder is used as key and its position in
    result is stored as value. Note that we need
    position for cases like 1/6. In this case,
    the recurring sequence doesn't start from first
    remainder. */
    unordered_map<int, int> mp;
 
    // Find first remainder
    int rem = numerator % denominator;
 
    // Keep finding remainder until either remainder
    // becomes 0 or repeats
    while ((rem != 0) && (mp.find(rem) == mp.end())) {
 
        // Store this remainder
        mp[rem] = res.length();
 
        // Multiply remainder with 10
        rem = rem * 10;
 
        // Append rem / denr to result
        int res_part = rem / denominator;
        res += to_string(res_part);
 
        // Update remainder
        rem = rem % denominator;
    }
 
    return (rem == 0) ? "-1" : res.substr(mp[rem]);
}
 
// Checks whether a number is prime or not
bool isPrime(int n)
{
    for (int i = 2; i <= n / 2; i++)
        if (n % i == 0)
            return false;
    return true;
}
 
// If all conditions are met,
// it proves Extended Midy's theorem
void ExtendedMidys(string str, int n, int m)
{
    if (!isPrime(n)) {
        cout << "Denominator is not prime, "
             << "thus Extended Midy's "
             << "theorem is not applicable";
        return;   
    }
 
    int l = str.length();
    int part1 = 0, part2 = 0;
    if (l % 2 == 0 && l % m == 0) {
 
        // Dividing repeated part into m parts
        int part[m] = { 0 }, sum = 0, res = 0;
        for (int i = 0; i < l; i++) {
            int var = i / m;
            part[var] = part[var] * 10 + (str[i] - '0');
        }
 
        // Computing sum of parts.
        for (int i = 0; i < m; i++) {
            sum = sum + part[i];
            cout << part[i] << " ";
        }
         
        // Checking for Extended Midy
        cout << endl;
        res = pow(10, m) - 1;
        if (sum % res == 0)
            cout << "Extended Midy's theorem holds!";       
        else
            cout << "Extended Midy's theorem"
                 << " doesn't hold!";       
    }
    else if (l % 2 != 0) {
        cout << "The repeating decimal is"
             << " of odd length thus Extended "
            << "Midy's theorem is not applicable";
    }
    else if (l % m != 0) {
        cout << "The repeating decimal can "
             << "not be divided into m digits";
    }
}
 
// Driver code
int main()
{
    int numr = 1, denr = 17, m = 4;
    string res = fractionToDecimal(numr, denr);
    if (res == "-1")
        cout << "The fraction does not"
             << " have repeating decimal";
    else {
        cout << "Repeating decimal = " << res << endl;
        ExtendedMidys(res, denr, m);
    }
    return 0;
}


Java




// Java program to demonstrate extended
// Midy's theorem
import java.util.*;
 
class GFG{
 
// Returns repeating sequence of a fraction.
// If repeating sequence doesn't exits,
// then returns -1
static String fractionToDecimal(int numerator,
                                int denominator)
{
    String res = "";
 
    /* Create a map to store already seen remainders
    remainder is used as key and its position in
    result is stored as value. Note that we need
    position for cases like 1/6. In this case,
    the recurring sequence doesn't start from first
    remainder. */
    HashMap<Integer, Integer> mp = new HashMap<>();
 
    // Find first remainder
    int rem = numerator % denominator;
 
    // Keep finding remainder until either remainder
    // becomes 0 or repeats
    while ((rem != 0) && !mp.containsKey(rem))
    {
         
        // Store this remainder
        mp.put(rem, res.length());
 
        // Multiply remainder with 10
        rem = rem * 10;
 
        // Append rem / denr to result
        int res_part = rem / denominator;
        res += res_part + "";
 
        // Update remainder
        rem = rem % denominator;
    }
 
    return (rem == 0) ? "-1" : res.substring(mp.get(rem));
}
 
// Checks whether a number is prime or not
static boolean isPrime(int n)
{
    for(int i = 2; i <= n / 2; i++)
        if (n % i == 0)
            return false;
             
    return true;
}
 
// If all conditions are met,
// it proves Extended Midy's theorem
static void ExtendedMidys(String str, int n, int m)
{
    if (!isPrime(n))
    {
        System.out.print("Denominator is not prime, " +
                         "thus Extended Midy's theorem " +
                         "is not applicable");
        return;   
    }
 
    int l = str.length();
    int part1 = 0, part2 = 0;
     
    if (l % 2 == 0 && l % m == 0)
    {
         
        // Dividing repeated part into m parts
        int []part = new int[m];
        int sum = 0, res = 0;
        for(int i = 0; i < l; i++)
        {
            int var = i / m;
            part[var] = part[var] * 10 +
                   (str.charAt(i) - '0');
        }
 
        // Computing sum of parts.
        for(int i = 0; i < m; i++)
        {
            sum = sum + part[i];
            System.out.print(part[i] + " ");
        }
         
        // Checking for Extended Midy
        System.out.println();
        res = (int)Math.pow(10, m) - 1;
         
        if (sum % res == 0)
            System.out.print("Extended Midy's " +
                             "theorem holds!");       
        else
            System.out.print("Extended Midy's " +
                             "theorem doesn't hold!");       
    }
    else if (l % 2 != 0)
    {
        System.out.print("The repeating decimal is of " +
                         "odd length thus Extended Midy's " +
                         "theorem is not applicable");
    }
    else if (l % m != 0)
    {
        System.out.print("The repeating decimal can " +
                         "not be divided into m digits");
    }
}
 
// Driver code
public static void main(String []args)
{
    int numr = 1, denr = 17, m = 4;
    String res = fractionToDecimal(numr, denr);
     
    if (res == "-1")
        System.out.print("The fraction does not " +
                         "have repeating decimal");
    else
    {
        System.out.println("Repeating decimal = " + res);
        ExtendedMidys(res, denr, m);
    }
}
}
 
// This code is contributed by rutvik_56


C#




// C# program to demonstrate extended
// Midy's theorem
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
 
// Returns repeating sequence of a fraction.
// If repeating sequence doesn't exits,
// then returns -1
static string fractionToDecimal(int numerator,
                                int denominator)
{
    string res = "";
 
    /* Create a map to store already seen remainders
    remainder is used as key and its position in
    result is stored as value. Note that we need
    position for cases like 1/6. In this case,
    the recurring sequence doesn't start from first
    remainder. */
    Dictionary<int,int> mp = new Dictionary<int,int>();
 
    // Find first remainder
    int rem = numerator % denominator;
 
    // Keep finding remainder until either remainder
    // becomes 0 or repeats
    while ((rem != 0) && !mp.ContainsKey(rem))
    {
         
        // Store this remainder
        mp[rem]= res.Length;
 
        // Multiply remainder with 10
        rem = rem * 10;
 
        // Append rem / denr to result
        int res_part = rem / denominator;
        res += res_part + "";
 
        // Update remainder
        rem = rem % denominator;
    }
 
    return (rem == 0) ? "-1" : res.Substring(mp[rem]);
}
 
// Checks whether a number is prime or not
static bool isPrime(int n)
{
    for(int i = 2; i <= n / 2; i++)
        if (n % i == 0)
            return false;
             
    return true;
}
 
// If all conditions are met,
// it proves Extended Midy's theorem
static void ExtendedMidys(string str, int n, int m)
{
    if (!isPrime(n))
    {
        Console.Write("Denominator is not prime, " +
                         "thus Extended Midy's theorem " +
                         "is not applicable");
        return;   
    }
 
    int l = str.Length;
     
    if (l % 2 == 0 && l % m == 0)
    {
         
        // Dividing repeated part into m parts
        int []part = new int[m];
        int sum = 0, res = 0;
        for(int i = 0; i < l; i++)
        {
            int var = i / m;
            part[var] = part[var] * 10 +
                   (str[i] - '0');
        }
 
        // Computing sum of parts.
        for(int i = 0; i < m; i++)
        {
            sum = sum + part[i];
            Console.Write(part[i] + " ");
        }
         
        // Checking for Extended Midy
        Console.WriteLine();
        res = (int)Math.Pow(10, m) - 1;
         
        if (sum % res == 0)
            Console.Write("Extended Midy's " +
                             "theorem holds!");       
        else
            Console.Write("Extended Midy's " +
                             "theorem doesn't hold!");       
    }
    else if (l % 2 != 0)
    {
        Console.Write("The repeating decimal is of " +
                         "odd length thus Extended Midy's " +
                         "theorem is not applicable");
    }
    else if (l % m != 0)
    {
        Console.Write("The repeating decimal can " +
                         "not be divided into m digits");
    }
}
 
// Driver code
public static void Main(string []args)
{
    int numr = 1, denr = 17, m = 4;
    string res = fractionToDecimal(numr, denr);
     
    if (res == "-1")
        Console.Write("The fraction does not " +
                         "have repeating decimal");
    else
    {
        Console.WriteLine("Repeating decimal = " + res);
        ExtendedMidys(res, denr, m);
    }
}
}
 
// This code is contributed by pratham76.


Output: 

Repeating decimal = 0588235294117647
588 2352 9411 7647 
Extended Midy's theorem holds!

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :