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. 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 -1string 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 notbool 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 theoremvoid 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 codeint 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 theoremimport java.util.*;class GFG{// Returns repeating sequence of a fraction.// If repeating sequence doesn't exits,// then returns -1static 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 notstatic 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 theoremstatic 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 codepublic 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 |
Python3
# Python3 program to demonstrate extended# Midy's theorem# Returns repeating sequence of a fraction.# If repeating sequence doesn't exits,# then returns -1def 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; if rem == 0: return "-1" return res[mp[rem]:] # Checks whether a number is prime or notdef isPrime(n): for i in range(2, n // 2 + 1): if (n % i == 0): return False; return True;# If all conditions are met,# it proves Extended Midy's theoremdef ExtendedMidys(str, n, m): if (not isPrime(n)): print("Denominator is not prime, thus Extended Midy's theorem is not applicable"); return; l = len(str); part1 = 0 part2 = 0; if (l % 2 == 0 and l % m == 0): # Dividing repeated part into m parts part = [0 for _ in range(m)]; sum = 0 res = 0; for i in range(l): var = i // m part[var] = part[var] * 10 + int(str[i]); # Computing sum of parts. for i in range(m): sum = sum + part[i]; print(part[i], end = " "); # Checking for Extended Midy print() res = pow(10, m) - 1; if (sum % res == 0): print("Extended Midy's theorem holds!"); else: print("Extended Midy's theorem doesn't hold!"); elif (l % 2 != 0): print("The repeating decimal is of odd length thus Extended Midy's theorem is not applicable"); elif (l % m != 0): print("The repeating decimal can not be divided into m digits"); # Driver codenumr = 1denr = 17m = 4;res = fractionToDecimal(numr, denr);if (res == "-1"): print("The fraction does not have repeating decimal");else: print("Repeating decimal =", res); ExtendedMidys(res, denr, m);# This code is contributed by phasing17 |
C#
// C# program to demonstrate extended // Midy's theoremusing System;using System.Collections;using System.Collections.Generic;class GFG{// Returns repeating sequence of a fraction.// If repeating sequence doesn't exits,// then returns -1static 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 notstatic 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 theoremstatic 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 codepublic 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. |
Javascript
// JavaScript program to demonstrate extended// Midy's theorem// Returns repeating sequence of a fraction.// If repeating sequence doesn't exits,// then returns -1function 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 notfunction 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 Extended Midy's theoremfunction ExtendedMidys(str, n, m){ if (!isPrime(n)) { console.log( "Denominator is not prime, thus Extended Midy's theorem is not applicable"); return; } let l = str.length; let part1 = 0, part2 = 0; if (l % 2 == 0 && l % m == 0) { // Dividing repeated part into m parts let part = new Array(m).fill(0); let sum = 0, res = 0; for (var i = 0; i < l; i++) { var var_ = Math.floor(i / m); part[var_] = part[var_] * 10 + parseInt(str[i]); } // Computing sum of parts. for (var i = 0; i < m; i++) { sum = sum + part[i]; process.stdout.write(part[i] + " "); } // Checking for Extended Midy console.log(); res = Math.pow(10, m) - 1; if (sum % res == 0) console.log("Extended Midy's theorem holds!"); else console.log( "Extended Midy's theorem doesn't hold!"); } else if (l % 2 != 0) { console.log( "The repeating decimal is of odd length thus Extended Midy's theorem is not applicable"); } else if (l % m != 0) { console.log( "The repeating decimal can not be divided into m digits"); }}// Driver codelet numr = 1, denr = 17, m = 4;let res = fractionToDecimal(numr, denr);if (res == "-1") console.log( "The fraction does not have repeating decimal");else { console.log("Repeating decimal = " + res); ExtendedMidys(res, denr, m);}// This code is contributed by phasing17 |
Output:
Repeating decimal = 0588235294117647 588 2352 9411 7647 Extended Midy's theorem holds!
