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how to find the sum of all the prime numbers between 1 and 50 in java?

i have the following code, but the answer is coming out to be 2.
public class Main {
public static void main(String[] args) {
int sum = 0;
int count = 0;
for(int i = 2; i <= 50; i++) {
for(int j = 1; j <= i; j++) {
System.out.println(j);
if(i % j == 0) {
count++;
}
if(count == 2) {
System.out.println(i);
sum = sum + i;
}
}
}
System.out.println("The sum: " + sum);
}
}

You were close. But here is all you need.
first, set sum to 2. It is the only even prime and doing this allows a better increment below
now increment the test cases starting at 3 and incrementing by 2. Skip even values.
now divide by each test case by numbers from 2 to the sqrt(i). No need to go beyond that.
as soon as you have a successful division, end the inner loop and start the next of the outer loop.
otherwise, if the inner loop completes, add i to sum.
int sum = 2;
outer:
for(int i = 3; i <= 50; i+= 2) {
for(int j = 2; j <= Math.sqrt(i); j++) {
if(i % j == 0) {
continue outer;
}
}
sum += i;
}
System.out.println("The sum: " + sum);
An improvement would be to store the primes as they are found and then only divide by those instead of arbitrary numbers.

Disclaimer: this code was written by Github Copilot with minimal supervision.
class Scratch {
// find the sum of all prime numbers between 2 and n
public static int sumPrimes(int n) {
int sum = 0;
for (int i = 2; i < n; i++) {
if (isPrime(i)) {
sum += i;
}
}
return sum;
}
// return true if n is prime
public static boolean isPrime(int n) {
if (n == 2) {
return true;
}
if (n % 2 == 0) {
return false;
}
for (int i = 3; i * i <= n; i += 2) {
if (n % i == 0) {
return false;
}
}
return true;
}
public static void main(String[] args) {
// print sum of primes between 2 and 50
System.out.println("The sum: " + sumPrimes(50));
}
}
Prints:
The sum: 328

Maybe a minimal version with recursive method based on Aivean response:
class Scratch {
// find the sum of all prime numbers between 2 and n
public static int sumPrimes(int n) {
if (n >= 2) {
return (isPrime(n) ? n : 0) + sumPrimes(n-1);
}
return 0;
}
// return true if n is prime
public static boolean isPrime(int n) {
for (int i = 2; i < n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
public static void main(String[] args) {
// print sum of primes between 2 and 50
System.out.println("The sum: " + sumPrimes(50));
}
}

Prime Number in java

I wrote some code to find prime numbers up to a given number. Could you guys let me know ways to make my code more efficient or better? Or give insight on how I did? In addition, there is a problem in my code where certain numbers repeats twice or three times in a pattern. How do I fix this?
public class PrimeNumber2 {
public static void main(String[] args)
{
int max_prime = 10000;
for(int i = 3; i < max_prime; i+=2)
{
for(int j = 1; j < Math.sqrt(i); j++)
{
if(i % j == 0)
{
System.out.println(i);
}
}
}
}
}

Have a look at:
https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
This is quite a nice way of doing it.
Here is some code for you to look at:
public void runEratosthenesSieve(int upperBound) {
int upperBoundSquareRoot = (int) Math.sqrt(upperBound);
boolean[] isComposite = new boolean[upperBound + 1];
for (int m = 2; m <= upperBoundSquareRoot; m++) {
if (!isComposite[m]) {
System.out.print(m + " ");
for (int k = m * m; k <= upperBound; k += m)
isComposite[k] = true;
}
}
for (int m = upperBoundSquareRoot; m <= upperBound; m++) {
if (!isComposite[m]) {
System.out.print(m + " ");
}
}

Here´s a tuned version of yours.
I did put the check if the number is prime into a seperate method. That´s also a part of the reason why your version did print values multiple times, since if it found out that it has a divisor, then it would print the values. (Also you´r algorythm would print basicly everything despite it beeing prime or not, for example it prints 15 and 27).
The reason why it did print multiple values was, that once you found a divisor it would have printed i, but it would continue looping. If it would have found another divisor, it would print i again(you can notice that it does not only print prime numbers).
Here is the fixed version of your´s
public static void main(String[] args) {
if(isPrime(2)) {
System.out.println(2);
}
int max_prime = 10000;
for(int i = 3; i < max_prime; i+=2)
{
if(isPrime(i)) {
System.out.println(i);
}
}
}
private static boolean isPrime(int n) {
if(n<=1) return false;
if(n == 2) return true;
for(int i = 2;i*i<=n;++i) {
if(n%i == 0) return false;
}
return true;
}

Try this.
public class PrimeNumber2 {
public static void main(String[] args)
{
int max_prime = 10000;
System.out.println(2);
L: for (int i = 3; i < max_prime; i += 2)
{
for (int j = 3, max = (int)Math.sqrt(i); j <= max ; j += 2)
{
if(i % j == 0)
{
continue L;
}
}
System.out.println(i);
}
}
}

Here a way using parallelStream
System.out.println(2);
IntStream.range(1, 10000000)
.map(i -> i * 2 + 1)
.filter(i -> (i & 1) != 0 && IntStream.range(1, (int) (Math.sqrt(i)-1)/2)
.map(j -> j * 2 + 1)
.noneMatching(j -> i % j == 0)
.forEach(System.out::println);
Note: the ranges test the n-th odd number.

public static void main(String[] args) {
int upperBound = 30;
List<Integer> primes = new ArrayList<>();
// loop through the numbers one by one
for (int number = 2; number < upperBound; number++) {
boolean isPrimeNumber = true;
// check to see if the number is prime
for (int j = 2; j < number; j++) {
if (number % j == 0) {
isPrimeNumber = false;
break; // exit the inner for loop
}
}
// print the number if prime
if (isPrimeNumber) {
primes.add(number);
}
}
System.out.println("The number of prime is: " + primes.size() + ", and they are: " + primes.toString());
}
possible to dupplicate
get prime numbers and total prime numbers in range

public boolean isPrimeNumber(int number) {
for (int i=2; i<=number/2; i++) {
if (number % i == 0) {
return false;
}
}
return true;
}

Find the sum of all the primes below two million.My program doesn't work for very big numbers

This is my code for finding the sum of primes.It works good with some low numbers but if it's 2000000(2 million) it never ends.Anybody can help me?
import java.math.BigInteger;
public class Problem010{
public static void main(String[] args) {
BigInteger sum = new BigInteger("2");
//for (int i=3; i<2000000; i++) {
for(int i=3; i<10; i++){
for (int j=2; j<i; j++){
if (i % j == 0)
break;
else if (i == j+1){
sum = sum.add(BigInteger.valueOf(i));
}
}
}
System.out.println("Sum = "+sum);
}
}

your answer is 142913828922 but how?
I just changed your algorithm a little bit:
public static void main(String[] args) {
BigInteger sum = new BigInteger("2");
boolean isPrime = true;
for (int i=3; i<2000000; i++) {
double aa = Math.sqrt((double)i);
for (int j=2; j<=aa; j++){
if (i % j == 0){
isPrime = false;
break;
}
}
if(isPrime){
sum = sum.add(BigInteger.valueOf(i));
}
isPrime = true;
}
System.out.println("Sum = "+sum);
}
instead of going through all the numbers from 2 to i I just go from 2 to sqrt(i) and this improve your code running time a lot :)

#Lrrr, answer is correct. But algorithm can be further optimised. Look at my isPrime algorithm. For 2 million you don't need the BigInteger.
long sum = 2;// new BigInteger("2");
for (int i=3; i<2000000; i++) {
if(isPrime(i)) {
sum = sum + i;//.add(BigInteger.valueOf(i));
}
}
System.out.println("Sum = "+sum);
Here is isPrime method.
static boolean isPrime(int n) {
if (n < 2) {
return false;
}
if (n == 2 || n == 3) {
return true;
}
if ((n & 1) == 0 || n % 3 == 0) {
return false;
}
int sqrtN = (int) Math.sqrt(n) + 1;
for (int i = 6; i <= sqrtN; i += 6) {// loop 6 step
if (n % (i - 1) == 0 || n % (i + 1) == 0) {
return false;
}
}
return true;
}

An efficient solution could be to use Sieve of Eratosthenes to find out which number is prime below 2,000,000 (or any other number), and than post-process and sum them all:
int n = 2000000;
boolean[] isPrime = new boolean[n];
//preprocess - set up the array
for (int i = 2; i<n;i++) isPrime[i] = true;
//run sieve:
for (int i = 2; i < (int) Math.sqrt(n) + 1; i++) {
if (isPrime[i]) {
for (int j = 2; j*i < n; j++) isPrime[i*j] = false;
}
}
//sum primes:
long sum = 0;
for (int i = 2; i < n; i++) {
if (isPrime[i]) sum+=i;
}
System.out.println(sum);
As opposed to checking for each number at a time if it is prime or not (which takes O(sqrt(n)) - and by doing it for all numbers you get O(nsqrt(n)), in here you aggregate knowledge from previous iterations, effectively lowering the complexity to O(nloglog(n)), which is significantly faster for large enough values of n.
This comes at a cost of O(n) additional space.

You could use Sieve of Eratosthenes algorithm, it is more efficient then yours.
1) Store all numbers between 2 and N in array and mark them all as prime numbers.
2) Start from X = 2, and mark all its i*X (2X, 3X..), where i is natural number less then or equal N, multipliers as not prime. Do not mark X.
3) Find the next number greater then X which is not marked and repeat the procedure. If there is no such number, stop.
4) Remaining numbers in your array are prime
Something like this:
public static boolean[] findPrimes (int N) {
boolean[] primes = new boolean[N + 1];
// assume that all numbers are prime within given range
for (int i = 2; i <= N; i++) {
primes[i] = true;
}
// for all numbers in range, starting from 2
for (int i = 2; i*i <= N; i++) {
// mark natural multiples of i as nonprime
if (primes[i]) {
for (int j = i; i*j <= N; j++) {
primes[i*j] = false;
}
}
return primes;
}
5) Iterate over returned primes and sum indexes of TRUE values

I developed my own solution and it completes in 700 milliseconds to find all below 2 million.
I use the iterative method but I just stop looking for the numbers greater than (n/i)+1 where n is the number being checked if it is prime and i is the number in the iterative loop to see if it is a divisor.
public void run () {
long sumOfPrimes = 2;
int maxNumber = 2000000;
int counter = 0;
for (int i = 3; i <= maxNumber; i = i+2) {
if(isPrimeOptimized(i)){
sumOfPrimes = sumOfPrimes + i;
counter ++;
}
}
System.out.println("num of primes is " + counter);
System.out.println("sum of primes is " + sumOfPrimes);
}
private boolean isPrimeOptimized(int n){
int limitToDivide = n;
for(int i=2;i<=limitToDivide && i<n;i++){
if(n%i == 0)
return false;
else
limitToDivide = (n/i) + 1;
}
return true;
}

So far no one has actually implemented the Sieve correctly. Double check the wikipedia page and pay attention to how you are looping through the numbers. Without any optimizations using an array of int (or booleans) it should take only a few seconds in Java.

Perfect Number program java

I am supposed to create a perfect number class using the following pseudocode:
For i from 2 to “very large”,
For j from 2 to √i,
if (j evenly divides i),
accumulate the sum j and i/j
if √i is an integer
subtract √i ... you added it twice
if the sum of divisors == i
Print the number ... it’s perfect!
So here is my version. It runs, but it doesn't do what I want at all. It just runs and produces nothing as an output. Can someone tell me what is wrong with my program? It's bothering me so much.
import java.util.Scanner;
public class PerfectNumber {
public static void main(String[] args) {
double sum = 0
double newsum = 0;
for (int i = 2; i < 1000000; i++) {
for (int j = 2; i<Math.sqrt(i); j++){
if (i%j==0){
sum = j + (i%j);
}
if (Math.sqrt(i)==(int)i){
newsum = sum - Math.sqrt(i);
}
if (sum == 0) {
System.out.println(sum + "is a perfect number");
}
}
}
}
}

Few mistakes according to the algorithm:
sum = j + (i%j); should be changed to sum = j + (i/j);
This piece:
if (Math.sqrt(i)==(int)i){
newsum = sum - Math.sqrt(i);
}
if (sum == 0) {
System.out.println(sum + "is a prime number");
}
Should be under upper "for"
Math.sqrt(i)==(int)i would never be true unless i is 1. If you want to check this that way you should write Math.sqrt(i)==((int) Math.sqrt(i))
There are much more errors, the simplest way to do it is:
double sum = 0;
for (int i = 1; i <= 10000; i++) {
for (int j = 1; j < i; j++) {
if (i % j == 0) {
sum += j;
}
}
if (i == sum) {
System.out.println(sum + " is a prime number");
}
sum = 0;
}

Your code contains several mistakes. Here is the corrected code, commented with the changes.
// newsum isn't needed; declare sum to be int to avoid floating-point errors
int sum = 0;
for (int i = 2; i < 1000000; i++) {
// Start with 1; every natural number has 1 as a factor.
sum = 1;
// Test if j, not i, is less than the square root of i.
for (int j = 2; j <= Math.sqrt(i); j++){
if (i % j == 0){
// Add to sum; don't replace sum. Use i / j instead of i % j.
sum = sum + j + (i / j);
// Move test inside this if; test if j is square root of i
if (j*j == i){
// I used j because we know it's the square root already.
sum = sum - j;
}
}
// Move print outside of inner for loop to prevent multiple
// printings of a number.
// Test if sum equals the number being tested, not 0.
if (sum == i) {
// Space before is
System.out.println(sum + " is a perfect number");
}
}
}
Output:
6 is a perfect number
28 is a perfect number
496 is a perfect number
8128 is a perfect number

public static void main(String[] args){
int min = 2;
int max = 1000000;
int sum = 0;
for (; min <= max; min++,sum = 0) {
for (int e = 1; e < min; e++)
sum += ((min % e) == 0) ? e : 0;
if (sum == min){
System.out.println(sum);
}
}
}

for(n=1;n<=number;n++){ //calculates the sum of the number.
int i=1;
int sum = 0;
while(i<n){
if(n%i==0)
sum+=i;
i++;
}
if(sum==n){ //if the sum is equal to its sum :
System.out.print(n+": ");
for (int j = 1;j<n;j++){
if(n%j==0){
System.out.print(j+" ");
}
}
System.out.println();
}
}

Here is the simplest and easiest form you can write a program for perfect number....this code gives perfect number within 25 ...you can change as you want
import java.util.Scanner;
public class PerfectNumber {
public static void main(String[] args) {
int n,i,j,count=0;
for(i=2;i<=25;i++) {
for(j=1;j<=i;j++) {
if(i%j ==0) /*count increments if a reminder zero*/ {
count++;
}
}
/*since a perfect number is divided only by 1 and itself
if the count is 2 then its a prime number...*/
if(count==2)
System.out.println(i);
count=0;
}
return 0;
}
}

According to the pseudocode you want to move the second and third if test outside of the inner loop
for (int i = 2; i < 1000000; i++) {
double iroot = Math.sqrt(i);
int sum = 1;
for (int j = 2; j <= iroot; j++){
if (i % j == 0){
sum = sum + j + i / j;
}
}
if (iroot == (int) iroot) {
sum = sum - iroot;
}
if (sum == i) {
System.out.println(sum + "is a perfect number");
}
}

Thanks for watched
public boolean testPerfect(int n){
int i=1;
int sum=0;
while(i<n){
if(n%i==0)
{
sum+=i++;
}
else{
i++;}
}
if (sum==n){
return true;
}
return false;
}

Prime Numbers III Problem 18 - I dont get the right answer even though I think I am right

I am new here. I am trying to solve this exercise Problem 18 just for reinforcing my solving skills. I've already coded the answer. The task asks for "How many of the primes below 1,000,000 have the sum of their digits equal to the number of days in a fortnight?" (a fortnight is 14 days). My answers is 16708, but it is wrong. I hope you can help me. I don't know what my error is. I have 2 methods, 1 for generating the primes, and another for counting the digits of each prime.
This is my code:
import java.util.ArrayList;
import java.util.List;
public class Problema18 {
public static void main(String args[]) {
ArrayList<Integer> num = primes();
System.out.println(num);
count(primes());
}
public static ArrayList<Integer> primes() {
List<Integer> primes = new ArrayList<Integer>();
primes.add(2);
for (int i = 3; i <= 1000000; i += 2) {
boolean isPrime = true;
int stoppingPoint = (int) (Math.pow(i, 0.5) + 1);
for (int p : primes) {
if (i % p == 0) {
isPrime = false;
break;
}
if (p > stoppingPoint) { break; }
}
if (isPrime) { primes.add(i); }
}
// System.out.println(primes);
return (ArrayList<Integer>) primes;
//System.out.println(primes.size());
}
public static void count(ArrayList<Integer> num) {
int count = 0;
for (int i = 0; i <= num.size() - 1; i++) {
int number = num.get(i);
String num1 = String.valueOf(number);
int sum = 0;
for (int j = 0; j < num1.length(); j++) {
sum = Integer.parseInt(num1.charAt(j) + "") + sum;
if (sum == 14) { count++; }
}
System.out.println(sum);
}
System.out.println(count);
}
}

You should check whether sum == 14 outside the inner for loop. What happens now is that you also count those primes for which the sum of digits is larger than 14 but the sum of the digits in some prefix of the prime is equal to 14.

This part...
if (sum == 14) {
count++;
}
should be outside the inner for-loop - i.e. you want to do it each time you pass through the i for-loop, but not each time you pass through the j for-loop.
Like this:
public static void count(ArrayList<Integer> num) {
int count = 0;
for (int i = 0; i <= num.size() - 1; i++) {
int number = num.get(i);
String num1 = String.valueOf(number);
int sum = 0;
for (int j = 0; j < num1.length(); j++) {
sum = Integer.parseInt(num1.charAt(j) + "") + sum;
}
System.out.println(sum);
if (sum == 14) {
count++;
}
}
System.out.println(count);
}