# Java Program to Find Prime Number between 1 to 100

## Finding Prime number between 1 to 100

Here, in this section we will discuss a program to find prime number between 1 to 100 in java. A prime number is a positive integer having exactly two factors. If 11 is a prime, then it’s only factors are necessarily 1 and 11 itself ## Methods Discussed on page

• Method 1: Basic checking prime by only checking first n
• Method 2: Basic checking prime by only checking first n/2 divisors
• Method 3: Checking prime by only checking first √n divisors
• Method 4: Checking prime by only checking first √n divisors, but also skipping even iterations. ## Method 1

Basic checking prime by only checking first n

### Java Code

Run
```public class Main
{

public static void main(String[] args) {
int a=1,b=100;
for(int i=a;i<=b;i++){
if(checkPrime(i)){
System.out.print(i+" " );
}
}
}
public static boolean checkPrime(int num){

// 0, 1 and negative numbers are not prime
if(num<2){
return false;
}
else{

// no need to run loop till num-1 as for any number x the numbers in
// the range(num/2 + 1, num) won't be divisible anyways.
// Example 36 wont be divisible by anything b/w 19-35
int x= num/2;
for(int i=2;i<x;i++){
if(num%i==0){
return false;
}
}
}
// the number would be prime if we reach here
return true;
}
}
```

### Output

`2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97`

## Algorithm (Method 2)

Basic checking prime by only checking first n/2 divisors

• Run a loop in the iteration of (i) b/w 1 and 100 bounds.
• For each, i check if its prime or not using function checkPrime(i)
• If i is prime print it else move to the next iteration

### Java Code

Run
```public class Main
{
public static void main(String[] args) {

for(int i=1;i<=100;i++){
if(checkPrime(i)){
System.out.print(i+" " );
}
}
}
public static boolean checkPrime(int num){

// 0, 1 and negative numbers are not prime
if(num<2){
return false;
}
else{
// no need to run loop till num-1 as for any number x the numbers in
// the range(num/2 + 1, num) won't be divisible anyways.
// Example 36 wont be divisible by anything b/w 19-35
int x= num/2;
for(int i=2;i<x;i++){
if(num%i==0){
return false;
}
}
}
// the number would be prime if we reach here
return true;
}
}
```

### Output

`2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97`

## Method 3

Checking prime by only checking first √n divisors

The outer logic remains the same. Only the method to check prime changes to make code more optimized. Has better time complexity of O(√N)

• Run a loop in the iteration of (i) b/w 1 and 100 bounds.
• For each, i check if its prime or not using function checkPrime(i)
• If i is prime print it else move to next iteration

### Java Code

Run
```import java.lang.Math;
public class Main
{
public static void main(String[] args)
{

for(int i=1;i<=100;i++)
{
if(checkPrime(i))
{
System.out.print(i+" " );
}
}
}
public static boolean checkPrime(int num)
{
// 0, 1 and negative numbers are not prime
if(num<2)
{
return false;
}
else
{
// A number n is not a prime, if it can be factored into two factors a & b:
// n = a * b
/*Now a and b can't be both greater than the square root of n, since
then the product a * b would be greater than sqrt(n) * sqrt(n) = n.
So in any factorization of n, at least one of the factors must be
smaller than the square root of n, and if we can't find any factors
less than or equal to the square root, n must be a prime.*/
for(int i=2;i<Math.sqrt(num);i++)
{
if(num%i==0)
{
return false;
}
}
}
// the number would be prime if we reach here
return true;
}
}

// This method is obviously  faster as has better time complexity

```

### Output

`2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97`

## Method 4

Checking prime by only checking first √n divisors, but also skipping even iterations.

The outer logic remains the same. Only the method to check prime changes to make code more optimized. Has same time complexity of O(√N).

But makes around half lesser checks

• Run a loop in the iteration of (i) b/w 1 to 100 bounds.
• For each, i check if its prime or not using function checkPrime(i)
• If i is prime print it else move to next iteration

### Java Code

Run
```import java.lang.Math;
public class Main
{
public static void main(String[] args) {

for(int i=1;i<=100;i++){
if(checkPrime(i)){
System.out.print(i+" " );
}
}
}
public static boolean checkPrime(int n){
// 0 and 1 are not prime numbers
// negative numbers are not prime
if(n<=1){
return false;

// special case as 2 is the only even number that is prime
}else if(n==2){
return true;

// Check if n is a multiple of 2 thus all these won't be prime
}else if(n%2==0){
return false;
}
// If not, then just check the odds
for(int i=3;i<Math.sqrt(n);i+=2){
if(n%i==0){
return false;
}
}

return true;
}
}

// This method is obviously faster as makes around half lesser comparision due skipping even iterations in the loop
```

### Output

`2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97`

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## Working with Numbers

### 2 comments on “Java Program to Find Prime Number between 1 to 100”

• Satendra

import java.util.*;class main
{
public static void main(String[]args){
int a, b,i,j;
System.out.println(“first nu”);
Scanner s= new Scanner (System.in);
a=s.nextInt();
System.out.println(“second nu”);
b=s.nextInt();
for(i=a; i<=b; i++){
for(j=2; j<=i ; j++){
if(i%j==0){
break;
}
}
if(i==j){
System.out.println(j);
}
}
}

} 0
• BitCoin

correct solution:
public class Prime
{
public static void main(String arg[])
{
int j,i,count;
System.out.print(“Enter n value : “);
System.out.println(“Prime numbers between 1 to 100 are “);
for(j=2;j<=100;j++)
{
count=0;
for(i=1;i<=j;i++)
{
if(j%i==0)
{
count++;
}
}
if(count==2)
System.out.print(j+" ");
}
}
} 0