Number of solutions to Modular Equations

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) = .... = 2 

Input : 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

CPP


[sourcecode language=”CPP”]
/* 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 isnt 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 code
int main()
{
int A = 26, B = 2;
numberOfPossibleWays(A, B);
A = 21, B = 5;
numberOfPossibleWays(A, B);
return 0;
}
[/sourcecode]

C#


[sourcecode language=”CSharp”]
/* 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 isnt 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
[/sourcecode]

Java


[sourcecode language=”Java”]
// Java Program to print all cube free
// numbers smaller than or equal to n.
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 isnt 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
[/sourcecode]

Python3


[sourcecode language=”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 B
def 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 isnt 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 = 26
B = 2
numberOfPossibleWays(A, B)

A = 21
B = 5
numberOfPossibleWays(A, B)

# Contributed by _omg
[/sourcecode]

PHP


[sourcecode language=”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 isnt 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.
?>
[/sourcecode]


Output:

For A = 26 and B = 2, X can take 6 values
For A = 21 and B = 5, X can take 2 values

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))



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