C/C++ Program for Number of solutions to Modular Equations
Last Updated :
26 May, 2022
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
#include <bits/stdc++.h>
using namespace std;
int calculateDivisors( int A, int B)
{
int N = (A - B);
int noOfDivisors = 0;
for ( int i = 1; i <= sqrt (N); i++) {
if ((N % i) == 0) {
if (i > B)
noOfDivisors++;
if ((N / i) != i && (N / i) > B)
noOfDivisors++;
}
}
return noOfDivisors;
}
int numberOfPossibleWaysUtil( int A, int B)
{
if (A == B)
return -1;
if (A < B)
return 0;
int noOfDivisors = 0;
noOfDivisors = calculateDivisors(A, B);
return noOfDivisors;
}
void numberOfPossibleWays( int A, int B)
{
int noOfSolutions = numberOfPossibleWaysUtil(A, B);
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";
}
}
int main()
{
int A = 26, B = 2;
numberOfPossibleWays(A, B);
A = 21, B = 5;
numberOfPossibleWays(A, B);
return 0;
}
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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)) Please refer complete article on Number of solutions to Modular Equations for more details!
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