Given the first term (A) and common difference (D) of an Arithmetic Progression, and a prime number (P). The task is to find the position of the first element in the given AP which is a multiple of the given prime number P.
Examples:
Input: A = 4, D = 9, P = 11
Output: 2
Explanation :
The third term of the given AP is
a multiple of prime number 11.
First Term = 4
Second Term = 4+9 = 13
Third Term = 4+2*9 = 22
Input: A = 5, D = 6, P = 7
Output: 5
Explanation :
The sixth term of the given AP is
a multiple of prime number 7.
First Term = 5
Second Term = 5+6 = 11
Third Term = 5+2*6 = 17
Fourth Term = 5+3*6 = 23
Fifth Term = 5+4*6 = 29
Sixth Term = 5+5*5 = 35
Approach:
Let the term be AN. Therefore,
AN = (A + (N-1)*D)
Now, it is given that AN is a multiple of P. Then,
A + (N-1)*D = k*P Where, k is a constant.
Now let A be (A % P) and D be (D % P). So, we have (N-1)*D = (k*P – A).
Adding and subtracting P on RHS, we get:
(N-1)*D = P(k-1) + (P-A), Where P-A is a non-negative number (since A is replaced by A%P which is less than P)
Finally taking mod on both sides:
((N-1)*D)%P = (P-A)%P or, ((N-1)D)%P = P-A
Lets find a X < P, such that (D*X)%P = 1. This X is known as the inverse modulo of D with respect to P.
Thus answer N is:
((X*(P-A)) % P) + 1.
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>using namespace std;// Iterative Function to calculate // (x^y)%p in O(log y) */int power(int x, int y, int p){ // Initialize result int res = 1; // Update x if it is more than or // equal to p x = x % p; while (y > 0) { // If y is odd, multiply x with result if (y & 1) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res;}// function to find nearest element in commonint NearestElement(int A, int D, int P){ // base conditions if (A == 0) return 0; else if (D == 0) return -1; else { int X = power(D, P - 2, P); return (X * (P - A)) % P; }}// Driver codeint main(){ int A = 4, D = 9, P = 11; // module both A and D A %= P; D %= P; // function call cout << NearestElement(A, D, P); return 0;} |
C
#include <stdio.h>// Iterative Function to calculate // (x^y)%p in O(log y) */int power(int x, int y, int p){ // Initialize result int res = 1; // Update x if it is more than or // equal to p x = x % p; while (y > 0) { // If y is odd, multiply x with result if (y & 1) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res;}// function to find nearest element in commonint NearestElement(int A, int D, int P){ // base conditions if (A == 0) return 0; else if (D == 0) return -1; else { int X = power(D, P - 2, P); return (X * (P - A)) % P; }}// Driver codeint main(){ int A = 4, D = 9, P = 11; // module both A and D A %= P; D %= P; // function call printf("%d",NearestElement(A, D, P)); return 0;}// This code is contributed by kothavvsaakash. |
Java
// Java Program to Find First// element in AP which is // multiple of given primeclass GFG{// Iterative Function to // calculate (x^y)%p in// O(log y) */static int power(int x, int y, int p){ // Initialize result int res = 1; // Update x if it is // more than or equal to p x = x % p; while (y > 0) { // If y is odd, multiply // x with result if ((y & 1) != 0) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res;}// function to find nearest// element in commonstatic int NearestElement(int A, int D, int P){ // base conditions if (A == 0) return 0; else if (D == 0) return -1; else { int X = power(D, P - 2, P); return (X * (P - A)) % P; }}// Driver codepublic static void main(String args[]){ int A = 4, D = 9, P = 11; // module both A and D A %= P; D %= P; // function call System.out.println(NearestElement(A, D, P));}}// This code is contributed // by Arnab Kundu |
Python 3
# Python 3 Program to Find First # element in AP which is # multiple of given prime# Iterative Function to calculate # (x^y)%p in O(log y) def power(x, y, p) : # Initialize result res = 1 # Update x if it is more than or # equal to p x = x % p while y > 0 : # If y is odd, multiply x with result if y & 1 : res = (res * x) % p # y must be even now # y = y/2 y = y >> 1 x = (x * x) % p return res# function to find nearest element in commondef NearestElement(A, D, P) : # base conditions if A == 0 : return 0 elif D == 0 : return -1 else : X = power(D, P - 2, P) return (X * (P - A)) % P # Driver Codeif __name__ == "__main__" : A, D, P = 4, 9, 11 # module both A and D A %= P D %= P # function call print(NearestElement(A, D, P)) # This code is contributed by ANKITRAI1 |
C#
// C# Program to Find First// element in AP which is // multiple of given primeusing System;class GFG{// Iterative Function to // calculate (x^y)%p in// O(log y) */static int power(int x, int y, int p){ // Initialize result int res = 1; // Update x if it is // more than or equal to p x = x % p; while (y > 0) { // If y is odd, multiply // x with result if ((y & 1) != 0) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res;}// function to find nearest// element in commonstatic int NearestElement(int A, int D, int P){ // base conditions if (A == 0) return 0; else if (D == 0) return -1; else { int X = power(D, P - 2, P); return (X * (P - A)) % P; }}// Driver codepublic static void Main(){ int A = 4, D = 9, P = 11; // module both A and D A %= P; D %= P; // function call Console.WriteLine(NearestElement(A, D, P));}}// This code is contributed // by chandan_jnu. |
PHP
<?php// Iterative Function to calculate // (x^y)%p in O(log y) */function power($x, $y, $p){ // Initialize result $res = 1; // Update x if it is more // than or equal to p $x = $x % $p; while ($y > 0) { // If y is odd, multiply // x with result if ($y & 1) $res = ($res * $x) %$p; // y must be even now $y = $y >> 1; // y = y/2 $x = ($x * $x) %$p; } return $res;}// function to find nearest// element in commonfunction NearestElement($A, $D, $P){ // base conditions if ($A == 0) return 0; else if ($D == 0) return -1; else { $X = power($D, $P - 2, $P); return ($X * ($P - $A)) %$P; }}// Driver code$A = 4; $D = 9; $P = 11;// module both A and D$A %= $P;$D %= $P;// function callecho NearestElement($A, $D, $P);// This code is contributed // by chandan_jnu.?> |
Javascript
<script>// Javascript Program to Find First// element in AP which is // multiple of given prime// Iterative Function to // calculate (x^y)%p in// O(log y) */function power(x, y, p){ // Initialize result let res = 1; // Update x if it is // more than or equal to p x = x % p; while (y > 0) { // If y is odd, multiply // x with result if ((y & 1) != 0) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res;} // function to find nearest// element in commonfunction NearestElement(A, D, P){ // base conditions if (A == 0) return 0; else if (D == 0) return -1; else { let X = power(D, P - 2, P); return (X * (P - A)) % P; }}// driver program let A = 4, D = 9, P = 11; // module both A and D A %= P; D %= P; // function call document.write(NearestElement(A, D, P)); </script> |
Output:
2
Time Complexity: O(log y)
Auxiliary Space: O(1)
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