Approach
Let N be a nonnegative integer and m be the number of digits in the decimal representation of N. Let d_{(m1)}d_{(m  2)}.......d_{1}d_{0} be the decimal representation of N.
So, N = d_{(m  1)} * 10^{(m1)} + d_{(m  2)} * 10^{(m 2)} +.......+d_{1} * 10^{1} + d_{0} * 10^{0}.
taking mod with 11,
N%11 = (d_{(m  1)} * 10^{(m1)} + d_{(m  2)} * 10^{(m 2)} +.......+d_{1} * 10 + d_{0})%11
1 mod 11 = (1 + 11)mod11 = 10mod11
Using above equation,
So, N%11 = d_{(m  1)} * 1^{(m  1)} + d_{(m  2)} * 1^{(m  2)} + ...........  d_{1} + d_{0}
We can use the above equation to find X.
Given, P * Q = R
taking mod with 11,
(P * Q) mod 11 = R mod 11
(P mod 11) * (Q mod 11) = (R mod 11)
The lefthand side (P mod 11) * (Q mod 11) is a constant, and on the righthand side, X is unknown.
R%11 = r_{(m  1)} * 1^{(m  1)} + r_{(m  2)} * 1^{(m  2)} +..........+ X * 1^{i} + ...........  r_{1} + r_{0}
So if we shift all the terms of C to the lefthand side except X * 1^{i}, we can get the value of X. If X is at an odd place, we will multiply it with 1.
Code
#include <bits/stdc++.h>
using namespace std;
int main(){
string P, Q, R;
cin >> P >> Q >> R;
//size of P
int m = P.size();
//to store P mod 11
int Pmod11 = 0;
//keep the track of sign of current digit
int sign = 1;
//calculate the value of P mod 11 using the equation 1
for(int i = m  1; i >= 0; i){
Pmod11 = Pmod11 + sign * (P[i]  '0');
Pmod11 = Pmod11%11;
//change the sign
sign = sign;
}
//size of Q
m = Q.size();
//to store Q mod 11
int Qmod11 = 0;
//keep the track of sign of current digit
sign = 1;
//calculate the value of P mod 11 using the equation 1
for(int i = m  1; i >= 0; i){
Qmod11 = Qmod11 + sign * (Q[i]  '0');
Qmod11 = Qmod11%11;
//change the sign
sign = sign;
}
//size of R
m = R.size();
//to store (Q mod 11  X * sign of X)
int Rmod11 = 0;
//keep the track of sign of current digit
sign = 1;
//keep the track of sign of X
int signofX = 1;
for(int i = m  1; i >= 0; i){
if(R[i] != 'X'){
Rmod11 = Rmod11 + sign * (R[i]  '0');
Rmod11 = Rmod11%11;
}
else{
//store the sign of X
signofX = sign;
}
//change the sign
sign = sign;
}
//find value of X by shifting all the terms of R to LHS expect X
int X = (Pmod11 * Qmod11  Rmod11 + 11)%11;
//signofX is negative
if(signofX == 1)
X = (X + 11)%11;
cout << X << "\n";
}
Input
89568956
788966631
70666917457X07236
Output
5
Complexity
Time Complexity
The time complexity is O(log10P + log10Q + log10R).
Space Complexity
The space complexity is O(1).
FAQs

How 1 mod 11 == 10 mod 11??
According to modulo arithmetic,
(x  y)%mod = (x%mod  y%mod + mod)%mod
The extra mod is added to keep the result positive.
by using this formula,
1%11 = (0  1)%11
= (0%11  1%11 + 11)%11
= (11  1)%11
= 10%11
So, 1%11 = 10

Why the time complexity is O(log10P + log10Q + log10R)??
The number of digits in the base 10 representation of an integer N is log10N. So the length of the given strings P, Q, and R is log10P, log10Q, and log10R, respectively. We are iterating each string once, so the time complexity is O(log10P + log10Q + log10R).
Key Takeaways
In this article, we solved a number theory problem using modulo arithmetic. If you are practicing competitive programming or preparing for coding interviews, then number theory is one of the most important topics. Check out this library to get a better hold on it.
Check out this problem  First Missing Positive
To learn more about such data structures and algorithms, Coding Ninjas Studio is a onestop destination. This platform will help you to improve your coding techniques and give you an overview of interview experiences in various productbased companies by providing Online Mock Test Series and many more benefits.
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