# Midy’s theorem

Last Updated : 24 Mar, 2023

According to Midy’s theorem, if the period of a repeating decimal for , where p is prime and is a reduced fraction, has an even number of digits, then dividing the repeating portion into halves and adding gives a string of 9s.

Examples :

a = 1 and p = 7
1/7 = 0.14285714285..
So 1/7 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.
a = 2 and p = 11
2/11 = 0.18181818181..
Here, repeating decimal is 18. Now this is even in number therefore 1+8 = 9 which again shows the validity of Midy’s theorem.

Given numerator and denominator, the task is to find if the resultant floating point number follows Midy’s theorem or not.

Approach :
Let us simulate the process of converting fraction to decimal. Let us look at the part where we have already figured out the integer part which is floor(numerator/denominator). Now we are left with ( remainder = numerator%denominator ) / denominator.
If you remember the process of converting to decimal, at each step we do the following :

1. Multiply the remainder by 10.
2. Append remainder / denominator to result.
3. Remainder = remainder % denominator.

At any moment, if remainder becomes 0, we are done.
However, when there is a recurring sequence, remainder never becomes 0. For example if you look at 1/3, the remainder never becomes 0.

Below is one important observation :
If we start with remainder â€˜remâ€™ and if the remainder repeats at any point of time, the digits between the two occurrence of â€˜remâ€™ keep repeating.
So the idea is to store seen remainders in a map. Whenever a remainder repeats, we return the sequence before the next occurrence.

Below is the implementation of Midy’s theorem :

## C++

 // C++ implementation as a // proof of the Midy's theorem #include 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. */    map<int, int> mp;     mp.clear();           // 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 Midy's theorem void Midys(string str, int n) {     int l = str.length();     int part1 = 0, part2 = 0;     if (!isPrime(n))         {          cout << "Denominator is not prime, "             << "thus Midy's theorem is not applicable";     }     else if (l % 2 == 0)      {         for (int i = 0; i < l / 2; i++)          {             part1 = part1 * 10 + (str[i] - '0');             part2 = part2 * 10 + (str[l / 2 + i] - '0');         }         cout << part1 << " + " << part2 << " = "              << (part1 + part2) << endl;         cout << "Midy's theorem holds!";     }     else     {         cout << "The repeating decimal is of odd length "             << "thus Midy's theorem is not applicable";     } }   // Driver code int main() {     int numr = 2, denr = 11;     string res = fractionToDecimal(numr, denr);     if (res == "-1")         cout << "The fraction does not have repeating decimal";     else {         cout << "Repeating decimal = " << res << endl;         Midys(res, denr);     }     return 0; }

## Java

 // Java implementation as a // proof of the 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 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 Midy's theorem static void Midys(String str, int n) {     int l = str.length();     int part1 = 0, part2 = 0;           if (!isPrime(n))         {          System.out.print("Denominator is not prime, " +                           "thus Midy's theorem is not " +                          "applicable");     }     else if (l % 2 == 0)      {         for(int i = 0; i < l / 2; i++)          {             part1 = part1 * 10 + (str.charAt(i) - '0');             part2 = part2 * 10 + (str.charAt(l / 2 + i) - '0');         }         System.out.println(part1 + " + " + part2 +                            " = " + (part1 + part2));         System.out.print("Midy's theorem holds!");     }     else     {         System.out.print("The repeating decimal is " +                           "of odd length thus Midy's "+                          "theorem is not applicable");     } }   // Driver code public static void main(String []args) {     int numr = 2, denr = 11;     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);         Midys(res, denr);     } } }   // This code is contributed by rutvik_56

## Python3

 # Python3 implementation as a # proof of the Midy's theorem   # Returns repeating sequence of a fraction. # If repeating sequence doesn't exits, # then returns -1 def fractionToDecimal(numerator, denominator):     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. '''    mp = dict()       # Find first remainder     rem = numerator % denominator       # Keep finding remainder until either remainder     # becomes 0 or repeats     while ((rem != 0) and (rem not in mp)):           # Store this remainder         mp[rem] = len(res)           # Multiply remainder with 10         rem = rem * 10          # Append rem / denr to result         res_part = (rem // denominator)         res += str(res_part)           # Update remainder         rem = rem % denominator       return ["-1", res[mp[rem]:]][rem != 0]     # Checks whether a number is prime or not def isPrime(n):     for i in range(2, 1 + n // 2):         if (n % i == 0):             return False    return True    # If all conditions are met, # it proves Midy's theorem def Midys(str, n):       l = len(str)     part1 = 0    part2 = 0    if (not isPrime(n)):         print("Denominator is not prime, thus Midy's theorem is not applicable")       elif (l % 2 == 0):           for i in range(l // 2):               part1 = part1 * 10 + int(str[i])             part2 = part2 * 10 + int(str[(l // 2) + i])           print(part1, "+", part2, "=", (part1 + part2))         print("Midy's theorem holds!")       else:           print(             "The repeating decimal is of odd length thus Midy's theorem is not applicable")     # Driver code numr = 2denr = 11res = fractionToDecimal(numr, denr) if (res == "-1"):     print("The fraction does not have repeating decimal")   else:     print("Repeating decimal =", res)     Midys(res, denr)     # This code is contributed by phasing17.

## C#

 // C# implementation as a // proof of the 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 Midy's theorem static void Midys(String str, int n) {     int l = str.Length;     int part1 = 0, part2 = 0;        if (!isPrime(n))         {          Console.Write("Denominator is not prime, " +                           "thus Midy's theorem is not " +                          "applicable");     }     else if (l % 2 == 0)      {         for(int i = 0; i < l / 2; i++)          {             part1 = part1 * 10 + (str[i] - '0');             part2 = part2 * 10 + (str[l / 2 + i] - '0');         }         Console.WriteLine(part1 + " + " + part2 +                            " = " + (part1 + part2));         Console.Write("Midy's theorem holds!");     }     else     {         Console.Write("The repeating decimal is " +                           "of odd length thus Midy's "+                          "theorem is not applicable");     } }   // Driver code public static void Main(string []args) {     int numr = 2, denr = 11;     string res = fractionToDecimal(numr, denr);           if (res == "-1")         Console.Write("The fraction does not " +                          "have repeating decimal");     else    {         Console.WriteLine("Repeating decimal = " + res);         Midys(res, denr);     } } }   // This code is contributed by pratham76.

## Javascript

 // JavaScript implementation as a // proof of the Midy's theorem     // Returns repeating sequence of a fraction. // If repeating sequence doesn't exits, // then returns -1 function fractionToDecimal(numerator, denominator) {     let 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. */    let mp = {};           // Find first remainder     let rem = numerator % denominator;       // Keep finding remainder until either remainder     // becomes 0 or repeats     while ((rem != 0) && (!mp.hasOwnProperty(rem)))      {         // Store this remainder         mp[rem] = res.length;           // Multiply remainder with 10         rem = rem * 10;           // Append rem / denr to result         let res_part = Math.floor(rem / denominator);         res += (res_part.toString());           // Update remainder         rem = rem % denominator;     }     return (rem == 0) ? "-1" : res.substr(mp[rem]); }   // Checks whether a number is prime or not function isPrime(n) {     for (var i = 2; i <= n / 2; i++)              if (n % i == 0)             return false;    return true; }   // If all conditions are met, // it proves Midy's theorem function Midys(str, n) {     var l = str.length;     var part1 = 0, part2 = 0;     if (!isPrime(n))         {          console.log("Denominator is not prime, thus Midy's theorem is not applicable");     }     else if (l % 2 == 0)      {         for (var i = 0; i < l / 2; i++)          {             part1 = part1 * 10 + parseInt(str[i]);             part2 = part2 * 10 + parseInt(str[(Math.floor(l / 2) + i)]);         }         console.log(part1 + " + " + part2 + " = " + (part1 + part2));         console.log("Midy's theorem holds!");     }     else     {         console.log("The repeating decimal is of odd length thus Midy's theorem is not applicable");     } }   // Driver code let numr = 2; let denr = 11; let res = fractionToDecimal(numr, denr); if (res == "-1")     console.log("The fraction does not have repeating decimal");   else {     console.log("Repeating decimal = " + res);     Midys(res, denr); }   // This code is contributed by phasing17.

Output :

Repeating decimal = 18
1 + 8 = 9
Midy's theorem holds!

Time Complexity: O(n*log(n))

Auxiliary Space: O(n)
More about Midy’s theorem can be found on
http://www.kurims.kyoto-u.ac.jp/EMIS/journals/INTEGERS/papers/h2/h2.pdf
http://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1047&context=mathfacpub

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