Print the nearest prime number formed by adding prime numbers to N (original) (raw)
Last Updated : 11 Jul, 2025
Given a number N. The task is to print the nearest prime if the number is not prime by making it prime by adding prime numbers sequentially from 2.
Examples:
Input: N = 8
Output: 13
8 is not prime, so add the first prime to it to get 10
10 is not prime, hence add the second prime, i.e., 3 to get 13 which is prime.
Input: N = 45
Output: 47
Naive Approach : In this approach we add every prime number to given number N until we find the desired output.
- First run the loop from 2 to N*N and find a prime number.
- Then add that prime number to variable sum and check then the new sum formed is prime or not.
- If it is a Prime Number then return sum and if not then find another prime number and perform the same task again until sum become a prime number.
Implementation :
C++ `
// C++ code for the naive approach
#include <bits/stdc++.h> using namespace std;
// function to check if a number is prime or not bool isPrime(int n) { if (n <= 1) { return false; } for (int i = 2; i <= n/2; i++) { if (n % i == 0) { return false; } } return true; }
// function to add all prime numbers to a given number until it becomes a prime number int makePrime(int n) { int sum = n;
// to check every number prime or not
for(int i=2 ;i< n*n ;i++){
// the number is number then add it to sum
if(isPrime(i)){
sum+=i;
// check new sum formed is prime or not
if(isPrime(sum)){
// sum is prime then return ans
return sum;
}
}
}
return -1;}
// Driver Code int main() { int N = 8;
// function call
int result = makePrime(N);
cout << result << endl;
return 0;}
// this code is contributed by bhardwajji
Java
// Java code for the naive approach
import java.util.*;
public class Main { // function to check if a number is prime or not static boolean isPrime(int n) { if (n <= 1) { return false; } for (int i = 2; i <= n / 2; i++) { if (n % i == 0) { return false; } } return true; }
// function to add all prime numbers to a given number
// until it becomes a prime number
static int makePrime(int n)
{
int sum = n;
// to check every number prime or not
for (int i = 2; i < n * n; i++) {
// the number is number then add it to sum
if (isPrime(i)) {
sum += i;
// check new sum formed is prime or not
if (isPrime(sum)) {
// sum is prime then return ans
return sum;
}
}
}
return -1;
}
// Driver Code
public static void main(String[] args)
{
int N = 8;
// function call
int result = makePrime(N);
System.out.println(result);
}} // This code is contributed by sarojmcy2e
Python3
function to check if a number is prime or not
def isPrime(n): if n <= 1: return False for i in range(2, int(n/2) + 1): if n % i == 0: return False return True
function to add all prime numbers to a given number until it becomes a prime number
def makePrime(n): sum = n
# to check every number prime or not
for i in range(2, n*n):
# the number is number then add it to sum
if isPrime(i):
sum += i
# check new sum formed is prime or not
if isPrime(sum):
# sum is prime then return ans
return sum
return -1Driver Code
N = 8
function call
result = makePrime(N) print(result)
C#
using System;
class Program { // function to check if a number is prime or not static bool IsPrime(int n) { if (n <= 1) { return false; } for (int i = 2; i <= n / 2; i++) { if (n % i == 0) { return false; } } return true; }
// function to add all prime numbers to a given number
// until it becomes a prime number
static int MakePrime(int n)
{
int sum = n;
// to check every number prime or not
for (int i = 2; i < n * n; i++) {
// the number is prime then add it to sum
if (IsPrime(i)) {
sum += i;
// check new sum formed is prime or not
if (IsPrime(sum)) {
// sum is prime then return ans
return sum;
}
}
}
return -1;
}
static void Main(string[] args)
{
int N = 8;
// function call
int result = MakePrime(N);
Console.WriteLine(result);
}}
JavaScript
// JavaScript code for the naive approach
// function to check if a number is prime or not function isPrime(n) { if (n <= 1) { return false; } for (let i = 2; i <= n/2; i++) { if (n % i == 0) { return false; } } return true; }
// function to add all prime numbers to a given number until it becomes a prime number function makePrime(n) { let sum = n; // to check every number prime or not for(let i=2 ;i< n*n ;i++){
// the number is number then add it to sum
if(isPrime(i)){
sum+=i;
// check new sum formed is prime or not
if(isPrime(sum)){
// sum is prime then return ans
return sum;
}
}}
return -1; }
// Driver Code let N = 8;
// function call let result = makePrime(N); console.log(result);
`
Time Complexity: O((N * N) * N) // run loop from 2 to N*N to find the prime number. and N to check every number is prime or not.
Auxiliary Space: O(1) // no extra space used
Approach Using Sieve of Eratosthenes, mark the prime index by 1 in isprime[] list and store all the prime numbers in a list prime[]. Keep adding prime numbers sequentially to N, till it becomes prime.
Below is the implementation of the above approach:
C++ `
// C++ program to print the // nearest prime number by // sequentially adding the // prime numbers #include<bits/stdc++.h> using namespace std;
// Function to store prime // numbers using prime sieve void prime_sieve(int MAX, vector &isprime, vector &prime) {
// iterate for all
// the numbers
int i = 2;
while (i * i <= MAX)
{
// If prime[p] is not changed,
// then it is a prime
if (isprime[i] == 1)
{
// append the prime
// to the list
prime.push_back(i);
// Update all multiples of p
for (int j = i * 2; j < MAX; j += i)
{
isprime[j] = 0;
}
}
i += 1;
}}
// Function to print // the nearest prime int printNearest(int N) { int MAX = 1e6;
// store all the
// index with 1
vector<int> isprime(MAX, 1);
// 0 and 1 are not prime
isprime[0] = isprime[1] = 0;
// list to store
// prime numbers
vector<int> prime;
// variable to
// add primes
int i = 0;
// call the sieve function
prime_sieve(MAX, isprime, prime);
// Keep on adding prime
// numbers till the nearest
// prime number is achieved
while (!isprime[N])
{
N += prime[i];
i += 1;
}
// return the
// nearest prime
return N ;}
// Driver Code int main() { int N = 8; printf("%d", printNearest(N)); return 0; }
// This code is contributed // by Harshit Saini
Java
// Java program to print the // nearest prime number by // sequentially adding the // prime numbers import java.util.*;
class GFG {
// Function to store prime // numbers using prime sieve static void prime_sieve(int MAX, int []isprime, Vector prime) {
// iterate for all
// the numbers
int i = 2;
while (i * i <= MAX)
{
// If prime[p] is not changed,
// then it is a prime
if (isprime[i] == 1)
{
// append the prime
// to the list
prime.add(i);
// Update all multiples of p
for (int j = i * 2;
j < MAX; j += i)
{
isprime[j] = 0;
}
}
i += 1;
}}
// Function to print // the nearest prime static int printNearest(int N) { int MAX = (int) 1e6;
// store all the
// index with 1 except 0,1 index
int [] isprime = new int[MAX];
for(int i = 2; i < MAX; i++)
isprime[i] = 1;
// list to store
// prime numbers
Vector<Integer> prime = new Vector<Integer>();
// variable to add primes
int i = 0;
// call the sieve function
prime_sieve(MAX, isprime, prime);
// Keep on adding prime
// numbers till the nearest
// prime number is achieved
while (isprime[N] == 0)
{
N += prime.get(i);
i += 1;
}
// return the
// nearest prime
return N ;}
// Driver Code public static void main(String[] args) { int N = 8; System.out.printf("%d", printNearest(N)); } }
// This code is contributed by Rajput-Ji
Python3
Python3 program to print the nearest prime
number by sequentially adding the prime numbers
Function to store prime numbers using prime sieve
def prime_sieve(MAX, isprime, prime):
# iterate for all the numbers
i = 2
while (i * i <= MAX):
# If prime[p] is not changed,
# then it is a prime
if (isprime[i] == 1):
# append the prime to the list
prime.append(i)
# Update all multiples of p
for j in range(i * 2, MAX, i):
isprime[j] = 0
i += 1
Function to print the nearest prime
def printNearest(N):
MAX = 10**6
# store all the index with 1
isprime = [1] * MAX
# 0 and 1 are not prime
isprime[0] = isprime[1] = 0
# list to store prime numbers
prime = []
# variable to add primes
i = 0
# call the sieve function
prime_sieve(MAX, isprime, prime)
# Keep on adding prime numbers
# till the nearest prime number
# is achieved
while not isprime[N]:
N += prime[i]
i += 1
# return the nearest prime
return N Driver Code
N = 8 print(printNearest(N))
C#
// C# program to print the // nearest prime number by // sequentially adding the // prime numbers using System; using System.Collections.Generic;
class GFG {
// Function to store prime // numbers using prime sieve static void prime_sieve(int MAX, int []isprime, List prime) {
// iterate for all the numbers
int i = 2;
while (i * i <= MAX)
{
// If prime[p] is not changed,
// then it is a prime
if (isprime[i] == 1)
{
// append the prime to the list
prime.Add(i);
// Update all multiples of p
for (int j = i * 2;
j < MAX; j += i)
{
isprime[j] = 0;
}
}
i += 1;
}}
// Function to print // the nearest prime static int printNearest(int N) { int MAX = (int) 1e6; int i = 0;
// store all the
// index with 1 except 0,1 index
int [] isprime = new int[MAX];
for(i = 2; i < MAX; i++)
isprime[i] = 1;
// list to store
// prime numbers
List<int> prime = new List<int>();
// variable to add primes
i = 0;
// call the sieve function
prime_sieve(MAX, isprime, prime);
// Keep on adding prime
// numbers till the nearest
// prime number is achieved
while (isprime[N] == 0)
{
N += prime[i];
i += 1;
}
// return the
// nearest prime
return N;}
// Driver Code public static void Main(String[] args) { int N = 8; Console.Write("{0}", printNearest(N)); } }
// This code is contributed by Princi Singh
JavaScript
`
Time Complexity: O(N * log(logN))
Auxiliary Space: O(N)