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Find First element in AP which is multiple of given Prime in C++
Concept
With respect of given first term (A) and common difference (d) of an Arithmetic Progression, and a prime number (P), our task is to determine the position of the first element in the given AP which is treated as a multiple of the given prime number P.
Input
A = 3, d = 4, P = 5
Output
3
Explanation
The fourth term of the given AP is a multiple of prime number 5.
First Term = 3
Second Term = 3+4 = 7
Third Term = 3+2*4 = 11
Fourth Term = 3+3*4 = 15
Method
Assume the term be AN. As a result of this,
AN = (A + (N-1)*d)
So, it is given that AN is a multiple of P. As aresult of this,
A + (N-1)*d = l*P
Here, l is a constant.
So assume A be (A % P) and d be (d % P). Now, we have (N-1)*d = (l*P – A).
With the help of adding and subtracting P on RHS, we obtain −
(N-1)*d = P(l-1) + (P-A),
In this case, P-A is treated as a non-negative number
(because A is replaced by A%P which is smaller than P) At last taking mod on both sides −
((N-1)*d)%P = (P-A)%P or, ((N-1)d)%P = P-A
Assume calculate a Y < P, so that (d*Y)%P = 1. Now, this Y is termed as the inverse modulo of d with respect to P.
Finally answer N is −
((Y*(P-A)) % P) + 1.
Example
#include <bits/stdc++.h>
using namespace std;
// Shows iterative Function to calculate
// (x1^y1)%p1 in O(log y1) */
int power(int x1, int y1, int p1){
// Used to initialize result
int res1 = 1;
// Used to update x if it is more than or
// equal to p
x1 = x1 % p1;
while (y1 > 0) {
// It has been seen that if y1 is odd, multiply x1 with
result
if (y1 & 1)
res1 = (res1 * x1) % p1;
// y1 must be even now
y1 = y1 >> 1; // y1 = y1/2
x1 = (x1 * x1) % p1;
}
return res1;
}
// Shows function to find nearest element in common
int NearestElement1(int A, int d, int P){
// Shows base conditions
if (A == 0)
return 0;
else if (d == 0)
return -1;
else {
int Y = power(d, P - 2, P);
return (Y * (P - A)) % P;
}
}
// Driver code
int main(){
int A = 3, d = 4, P = 5;
// Used to module both A and d
A %= P;
d %= P;
// Shows function call
cout << NearestElement1(A, d, P);
return 0;
}Output
3
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