Given A and B, the task is to find the number of possible values that X can take such that the given modular equation (A mod X) = B holds good. Here, X is also called a solution of the modular equation.
Examples:
Input : A = 26, B = 2
Output : 6
Explanation:
X can be equal to any of {3, 4, 6, 8, 12, 24} as A modulus any of these values equals 2
i. e., (26 mod 3) = (26 mod 4)
= (26 mod 6) = (26 mod 8)
= …. = 2Input : 21 5
Output : 2
Explanation
X can be equal to any of {8, 16} as A modulus any of these values equals 5
i.e. (21 mod 8) = (21 mod 16) = 5
If we carefully analyze the equation A mod X = B its easy to note that if (A = B) then there are infinitely many values greater than A that X can take. In the Case when (A < B), there cannot be any possible value of X for which the modular equation holds. So the only case we are left to investigate is when (A > B).So now we focus on this case in depth.
Now, in this case we can use a well known relation i.e.
Dividend = Divisor * Quotient + Remainder
We are looking for all possible X i.e. Divisors given A i.e Dividend and B i.e., remainder. So,
We can say, A = X * Quotient + B Let Quotient be represented as Y ? A = X * Y + B A - B = X * Y ? To get integral values of Y, we need to take all X such that X divides (A - B) ? X is a divisor of (A - B)
So, the problem reduces to finding the divisors of (A – B) and the number of such divisors is the possible values X can take.
But as we know A mod X would result in values from (0 to X – 1) we must take all such X such that X > B.
Thus, we can conclude by saying that the number of divisors of (A – B) greater than B, are the all possible values X can take to satisfy A mod X = B
C++
/* C++ Program to find number of possible values of X to satisfy A mod X = B */#include <bits/stdc++.h>using namespace std;/* Returns the number of divisors of (A - B) greater than B */int calculateDivisors(int A, int B){ int N = (A - B); int noOfDivisors = 0; for (int i = 1; i <= sqrt(N); i++) { // if N is divisible by i if ((N % i) == 0) { // count only the divisors greater than B if (i > B) noOfDivisors++; // checking if a divisor isnot counted twice if ((N / i) != i && (N / i) > B) noOfDivisors++; } } return noOfDivisors;}/* Utility function to calculate number of all possible values of X for which the modular equation holds true */int numberOfPossibleWaysUtil(int A, int B){ /* if A = B there are infinitely many solutions to equation or we say X can take infinitely many values > A. We return -1 in this case */ if (A == B) return -1; /* if A < B, there are no possible values of X satisfying the equation */ if (A < B) return 0; /* the last case is when A > B, here we calculate the number of divisors of (A - B), which are greater than B */ int noOfDivisors = 0; noOfDivisors = calculateDivisors(A, B); return noOfDivisors;}/* Wrapper function for numberOfPossibleWaysUtil() */void numberOfPossibleWays(int A, int B){ int noOfSolutions = numberOfPossibleWaysUtil(A, B); // if infinitely many solutions available if (noOfSolutions == -1) { cout << "For A = " << A << " and B = " << B << ", X can take Infinitely many values" " greater than " << A << "\n"; } else { cout << "For A = " << A << " and B = " << B << ", X can take " << noOfSolutions << " values\n"; }}// Driver codeint main(){ int A = 26, B = 2; numberOfPossibleWays(A, B); A = 21, B = 5; numberOfPossibleWays(A, B); return 0;} |
Java
/* Java Program to find number of possible values of X to satisfy A mod X = B */import java.lang.*;class GFG{ /* Returns the number of divisors of (A - B) greater than B */ public static int calculateDivisors(int A, int B) { int N = (A - B); int noOfDivisors = 0; for (int i = 1; i <= Math.sqrt(N); i++) { // if N is divisible by i if ((N % i) == 0) { // count only the divisors greater than B if (i > B) noOfDivisors++; // checking if a divisor isnot counted twice if ((N / i) != i && (N / i) > B) noOfDivisors++; } } return noOfDivisors; } /* Utility function to calculate number of all possible values of X for which the modular equation holds true */ public static int numberOfPossibleWaysUtil(int A, int B) { /* if A = B there are infinitely many solutions to equation or we say X can take infinitely many values > A. We return -1 in this case */ if (A == B) return -1; /* if A < B, there are no possible values of X satisfying the equation */ if (A < B) return 0; /* the last case is when A > B, here we calculate the number of divisors of (A - B), which are greater than B */ int noOfDivisors = 0; noOfDivisors = calculateDivisors(A, B); return noOfDivisors; } /* Wrapper function for numberOfPossibleWaysUtil() */ public static void numberOfPossibleWays(int A, int B) { int noOfSolutions = numberOfPossibleWaysUtil(A, B); // if infinitely many solutions available if (noOfSolutions == -1) { System.out.print("For A = " + A + " and B = " + B + ", X can take Infinitely many values" + " greater than " + A + "\n"); } else { System.out.print("For A = " + A + " and B = " + B + ", X can take " + noOfSolutions + " values\n"); } } // Driver program public static void main(String[] args) { int A = 26, B = 2; numberOfPossibleWays(A, B); A = 21; B = 5; numberOfPossibleWays(A, B); }}// Contributed by _omg |
Python3
# Python Program to find number of possible# values of X to satisfy A mod X = B import math# Returns the number of divisors of (A - B)# greater than Bdef calculateDivisors (A, B): N = A - B noOfDivisors = 0 a = math.sqrt(N) for i in range(1, int(a + 1)): # if N is divisible by i if ((N % i == 0)): # count only the divisors greater than B if (i > B): noOfDivisors +=1 # checking if a divisor isnot counted twice if ((N / i) != i and (N / i) > B): noOfDivisors += 1; return noOfDivisors # Utility function to calculate number of all# possible values of X for which the modular# equation holds true def numberOfPossibleWaysUtil (A, B): # if A = B there are infinitely many solutions # to equation or we say X can take infinitely # many values > A. We return -1 in this case if (A == B): return -1 # if A < B, there are no possible values of # X satisfying the equation if (A < B): return 0 # the last case is when A > B, here we calculate # the number of divisors of (A - B), which are # greater than B noOfDivisors = 0 noOfDivisors = calculateDivisors; return noOfDivisors # Wrapper function for numberOfPossibleWaysUtil() def numberOfPossibleWays(A, B): noOfSolutions = numberOfPossibleWaysUtil(A, B) #if infinitely many solutions available if (noOfSolutions == -1): print ("For A = " , A , " and B = " , B , ", X can take Infinitely many values" , " greater than " , A) else: print ("For A = " , A , " and B = " , B , ", X can take " , noOfSolutions , " values")# main()A = 26B = 2numberOfPossibleWays(A, B)A = 21B = 5numberOfPossibleWays(A, B)# Contributed by _omg |
C#
/* C# Program to find number of possible values of X to satisfy A mod X = B */using System;class GFG{ /* Returns the number of divisors of (A - B) greater than B */ static int calculateDivisors(int A, int B) { int N = (A - B); int noOfDivisors = 0; double a = Math.Sqrt(N); for (int i = 1; i <= (int)(a); i++) { // if N is divisible by i if ((N % i) == 0) { // count only the divisors greater than B if (i > B) noOfDivisors++; // checking if a divisor isnot counted twice if ((N / i) != i && (N / i) > B) noOfDivisors++; } } return noOfDivisors; } /* Utility function to calculate number of all possible values of X for which the modular equation holds true */ static int numberOfPossibleWaysUtil(int A, int B) { /* if A = B there are infinitely many solutions to equation or we say X can take infinitely many values > A. We return -1 in this case */ if (A == B) return -1; /* if A < B, there are no possible values of X satisfying the equation */ if (A < B) return 0; /* the last case is when A > B, here we calculate the number of divisors of (A - B), which are greater than B */ int noOfDivisors = 0; noOfDivisors = calculateDivisors(A, B); return noOfDivisors; } /* Wrapper function for numberOfPossibleWaysUtil() */ public static void numberOfPossibleWays(int A, int B) { int noOfSolutions = numberOfPossibleWaysUtil(A, B); // if infinitely many solutions available if (noOfSolutions == -1) { Console.Write ("For A = " + A + " and B = " + B + ", X can take Infinitely many values" + " greater than " + A + "\n"); } else { Console.Write ("For A = " + A + " and B = " + B + ", X can take " + noOfSolutions + " values\n"); } } public static void Main() { int A = 26, B = 2; numberOfPossibleWays(A, B); A = 21; B = 5; numberOfPossibleWays(A, B); }}// Contributed by _omg |
PHP
<?php/* PHP Program to find number of possible values of X to satisfy A mod X = B *//* Returns the number of divisors of (A - B)greater than B */function calculateDivisors($A, $B){ $N = ($A - $B); $noOfDivisors = 0; for ($i = 1; $i <= sqrt($N); $i++) { // if N is divisible by i if (($N % $i) == 0) { // count only the divisors greater than B if ($i > $B) $noOfDivisors++; // checking if a divisor isnot counted twice if (($N / $i) != $i && ($N / $i) > $B) $noOfDivisors++; } } return $noOfDivisors;}/* Utility function to calculate number of all possible values of X for which the modular equation holds true */function numberOfPossibleWaysUtil($A, $B){ /* if A = B there are infinitely many solutions to equation or we say X can take infinitely many values > A. We return -1 in this case */ if ($A == $B) return -1; /* if A < B, there are no possible values of X satisfying the equation */ if ($A < $B) return 0; /* the last case is when A > B, here we calculate the number of divisors of (A - B), which are greater than B */ $noOfDivisors = 0; $noOfDivisors = calculateDivisors($A, $B); return $noOfDivisors;}/* Wrapper function for numberOfPossibleWaysUtil() */function numberOfPossibleWays($A, $B){ $noOfSolutions = numberOfPossibleWaysUtil($A, $B); // if infinitely many solutions available if ($noOfSolutions == -1) { echo "For A = " , $A, " and B = " ,$B, "X can take Infinitely many values greater than " , $A , "\n"; } else { echo "For A = ", $A , " and B = " ,$B, " X can take ",$noOfSolutions, " values\n"; }}// Driver code $A = 26; $B = 2; numberOfPossibleWays($A, $B); $A = 21; $B = 5; numberOfPossibleWays($A, $B); // This code is contributed ajit.?> |
Javascript
<script>// JavaScript Program to find number of possible// values of X to satisfy A mod X = B // Returns the number of divisors of (A - B)// greater than B function calculateDivisors(A, B){ let N = (A - B); let noOfDivisors = 0; for(let i = 1; i <= Math.sqrt(N); i++) { // If N is divisible by i if ((N % i) == 0) { // Count only the divisors // greater than B if (i > B) noOfDivisors++; // Checking if a divisor // isnot counted twice if ((N / i) != i && (N / i) > B) noOfDivisors++; } } return noOfDivisors;} // Utility function to calculate number of all // possible values of X for which the modular // equation holds true function numberOfPossibleWaysUtil(A, B){ // If A = B there are infinitely many solutions // to equation or we say X can take infinitely // many values > A. We return -1 in this case if (A == B) return -1; // If A < B, there are no possible values of // X satisfying the equation if (A < B) return 0; // The last case is when A > B, here // we calculate the number of divisors // of (A - B), which are greater than B let noOfDivisors = 0; noOfDivisors = calculateDivisors(A, B); return noOfDivisors;} // Wrapper function for numberOfPossibleWaysUtil() function numberOfPossibleWays(A, B){ let noOfSolutions = numberOfPossibleWaysUtil(A, B); // If infinitely many solutions available if (noOfSolutions == -1) { document.write("For A = " + A + " and B = " + B + ", X can take Infinitely " + "many values" + " greater than " + A + "<br/>"); } else { document.write("For A = " + A + " and B = " + B + ", X can take " + noOfSolutions + " values " + "<br/>"); }}// Driver codelet A = 26, B = 2;numberOfPossibleWays(A, B);A = 21, B = 5;numberOfPossibleWays(A, B);// This code is contributed by splevel62</script> |
The Time Complexity of the above approach is nothing but the time complexity of finding the number of divisors of (A – B) ie O(?(A – B))
Auxiliary Space: O(1)
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