A prime number is a natural number which is greater than 1 and has no positive divisor other than 1 and itself, such as 2, 3, 5, 7, 11, 13, and so on. The user is given two integer numbers, lower value, and upper value. The task is to write the Python program for printing all the prime numbers between the given interval (or range). To print all the prime numbers between the given interval, the user has to follow the following steps: - Step
1: Loop through all the elements in the given range.
- Step 2: Check for each number if it has any factor between 1 and itself.
- Step 3: If yes, then the number is not prime, and it will move to the next number.
- Step 4: If no, it is the prime number, and the program will print it and check for the next number.
- Step 5: The loop will break when it is reached to the upper value.
Example:
The Python Code to Print the Prime Number between the given Interval. Output: Please, Enter the Lowest Range Value: 14
Please, Enter the Upper Range Value: 97
The Prime Numbers in the range are:
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97
ConclusionIn this tutorial, we have shown how to write the code to print the prime numbers between the given interval of numbers.
Prime numberA prime number is an integer greater than 1 whose only factors are 1 and itself. A factor is an integer that can be divided evenly into another number. LogicTo print all the prime numbers up to N, we start one loop from 2 to N and then inside the loop we check current number or “num” is prime or not. To check if it is prime or not we again need one nested loop. It is not an efficient way to check prime
number but it is simpler to understand the basic of looping in Python. See also: Check whether a number is prime number or not Program# Take input from user
upto = int(input("Find prime numbers upto : "))
print("\nAll prime numbers upto", upto, "are : ")
for num in range(2, upto + 1):
i = 2
for i in range(2, num):
if(num % i == 0):
i = num
break;
# If the number is prime then print it.
if(i != num):
print(num, end=" ")OutputFind prime numbers upto : 100
All prime numbers upto 100 are :
3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
View Discussion Improve Article Save Article ReadDiscussView Discussion Improve Article Save Article Given a number N, the task is to print the prime numbers from 1 to N.
Examples: Input: N = 10
Output: 2, 3, 5, 7
Input: N = 5
Output: 2, 3, 5 Algorithm:
- First, take the number N as input.
- Then use a for loop to iterate the numbers from 1 to N
- Then check for each number to be a prime number. If it is a prime number, print it.
Approach 1: Now, according to formal definition, a number ‘n’ is prime if it is not divisible by any number other than 1 and n. In other words a number is prime if it is not divisible by any number from 2 to n-1.
Below is the
implementation of the above approach: C++#include <bits/stdc++.h>
using namespace std;
bool isPrime(int n)
{
if (n == 1 || n == 0)
return false;
for (int i = 2; i < n; i++) {
if (n % i == 0)
return false;
}
return true;
}
int main()
{
int N = 100;
for (int i = 1; i <= N; i++) {
if (isPrime(i))
cout << i << " ";
}
return 0;
}
C#include <stdbool.h>
#include <stdio.h>
bool isPrime(int n)
{
if (n == 1 || n == 0)
return false;
for (int i = 2; i < n; i++) {
if
(n % i == 0)
return false;
}
return true;
}
int main()
{
int N = 100;
for (int i = 1; i <= N; i++) {
if (isPrime(i))
printf("%d ", i);
}
return 0;
}
Javaclass GFG
{
static
boolean isPrime(int n){
if(n==1||n==0)return false;
for(int i=2; i<n; i++){
if(n%i==0)return false;
}
return true;
}
public static void main (String[] args)
{
int N = 100;
for(int
i=1; i<=N; i++){
if(isPrime(i)) {
System.out.print(i + " ");
}
}
}
}
Python3def isPrime(n):
if(n==1 or n==0):
return False
for i in range(2,n):
if(n%i==0):
return False
return True
N = 100;
for i in range(1,N+1):
if(isPrime(i)):
print(i,end=" ")
C#using System;
class GFG
{
static bool isPrime(int n){
if(n==1||n==0) return
false;
for(int i=2; i<n; i++) {
if(n%i==0) return false;
}
return true;
}
public static void Main (String[] args)
{
int N = 100;
for(int i=1; i<=N; i++) {
if(isPrime(i)) {
Console.Write(i + " ");
}
}
}
}
Javascript<script>
function isPrime( n)
{
if(n == 1 || n == 0) return false;
for(var i = 2; i < n; i++)
{
if(n % i == 0) return false;
}
return true;
}
var N = 100;
for(var i = 1; i <= N; i++)
{
if(isPrime(i)) {
console.log( i );
}
}
</script>
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 Time Complexity: O(N^2), Auxiliary Space: O(1) Approach 2: For checking if a number is prime or not do we really need to iterate through all the number from 2 to n-1? We already know that a number ‘n’ cannot be divided by any number greater than ‘n/2’. So, according to
this logic we only need to iterate through 2 to n/2 since number greater than n/2 cannot divide n.
C++#include <bits/stdc++.h>
using namespace std;
bool isPrime(int n){
if(n==1||n==0) return false;
for(int i=2; i<=n/2; i++) {
if(n%i==0) return false;
}
return true;
}
int
main()
{
int N = 100;
for(int i=1; i<=N; i++){
if(isPrime(i)) {
cout << i << " ";
}
}
return 0;
}
Javaclass GFG
{
static boolean isPrime(int n){
if(n==1||n==0) return
false;
for(int i=2; i<=n/2; i++){
if(n%i==0)return false;
}
return true;
}
public static void main (String[] args)
{
int N = 100;
for(int i=1; i<=N; i++){
if(isPrime(i)) {
System.out.print(i + " ");
}
}
}
}
Python3def isPrime(n):
if(n==1 or n==0):
return False
for i in range(2,(n//2)+1):
if(n%i==0):
return
False
return True
N = 100;
for i in range(1,N+1):
if(isPrime(i)):
print(i,end=" ")
C#using System;
class GFG
{
static bool isPrime(int n){
if(n==1||n==0)return false;
for(int
i=2; i<=n/2; i++){
if(n%i==0)return false;
}
return true;
}
public static void Main (String[] args)
{
int N = 100;
for(int i=1; i<=N; i++){
if(isPrime(i)) {
Console.Write(i + " ");
}
}
}
}
Javascript<script>
function isPrime(n)
{
if(n == 1 || n == 0) return false;
for(let i = 2; i <= n / 2; i++)
{
if(n % i == 0) return false;
}
return true;
}
let N = 100;
for(let i = 1; i <= N; i++)
{
if(isPrime(i))
{
document.write(i + " ");
}
}
</script>
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 Time Complexity: O(N2), Auxiliary Space: O(1),since no extra space has been taken. Approach 3: If a number ‘n’ is not divided by any number less than or equals to the square root of n then, it will not be divided by any other number greater than the square root of n. So, we only need to check up to the square root of n.
C++#include <bits/stdc++.h>
using namespace std;
bool isPrime(int n){
if(n==1||n==0)return false;
for(int i=2; i*i<=n; i++){
if(n%i==0)return false;
}
return true;
}
int main()
{
int N = 100;
for(int i=1; i<=N; i++){
if(isPrime(i)) {
cout << i << " ";
}
}
return 0;
}
Javaclass GFG
{
static boolean isPrime(int n){
if(n==1||n==0)return false;
for(int i=2; i*i<=n; i++){
if(n%i==0)return
false;
}
return true;
}
public static void main (String[] args)
{
int N = 100;
for(int i=1; i<=N; i++){
if(isPrime(i)) {
System.out.print(i + " ");
}
}
}
}
Python3def isPrime(n):
if(n==1
or n==0):
return False
for i in range(2,int(n**(1/2))+1):
if(n%i==0):
return False
return True
N = 100;
for i in
range(1,N+1):
if(isPrime(i)):
print(i,end=" ")
C#using System;
class GFG
{
static bool isPrime(int n){
if(n==1||n==0)return false;
for(int i=2; i*i<=n; i++){
if(n%i==0)return
false;
}
return true;
}
public static void Main (String[] args)
{
int N = 100;
for(int i=1; i<=N; i++){
if(isPrime(i)) {
Console.Write(i + " ");
}
}
}
}
Javascript<script>
const isPrime = (n) => {
if(n === 1||n === 0)return
false;
for(let i = 2; i <= Math.floor(Math.sqrt(n)); i++)
{
if(n % i == 0)return false;
}
return true;
}
let N = 100;
for(let i=1; i<=N; i++)
{
if(isPrime(i)) {
document.write(i);
}
}
</script>
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 Time
Complexity: O(N^(3/2)), Auxiliary Space: O(1) You can further optimize the time complexity to O(n*log(log(n))). CheckSieve of Eratosthenes.
How do you print prime numbers from 1 to n in python?
Step 1: Loop through all the elements in the given range. Step 2: Check for each number if it has any factor between 1 and itself. Step 3: If yes, then the number is not prime, and it will move to the next number. Step 4: If no, it is the prime number, and the program will print it and check for the next number.
How do you find prime numbers from 1 to n?
C program for prime numbers between 1 to n. #include<stdio.h>. int main(){. int num,i,count,n; printf("Enter max range: ");. scanf("%d",&n);. for(num = 1;num<=n;num++){. count = 0;. for(i=2;i<=num/2;i++){ if(num%i==0){. count++; break;. How do you print prime numbers from 1 to 10 in python?
Program Code. numr=int(input("Enter range:")). print("Prime numbers:",end=' '). for n in range(1,numr):. for i in range(2,n):. if(n%i==0):. break.. print(n,end=' '). How do you print prime numbers in Python?
Python program to print prime numbers using while loop
The break statement is used to come out of the loop as soon we get any positive divisor then no further check is required. At last print(” %d” %num, end = ' ') is used for printing the prime numbers.
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