- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

# 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