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
/* 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 code int main()
{ int A = 26, B = 2;
numberOfPossibleWays(A, B);
A = 21, B = 5;
numberOfPossibleWays(A, B);
return 0;
} |
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!