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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 check java

Write a program to read n numbers. The first number specified as input will be n. Next, the program should read n integer numbers.
The program should check for each number if it is prime as well as if its reverse is prime.
Display all such numbers in ascending order.
Consider below example for input and output:
Input:
7
11
12
23
19
7
113
101
Output:
7
11
101
113
My code
public class Prime {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
int temp;
int[] a = new int [x];
int[] r = new int [x];
int[]c = new int[a.length+r.length];
int[] rev = new int [x];
for(int i=0;i<x;i++){
a[i] = sc.nextInt();
rev[i]=a[i];
}
for(int i = 0; i < a.length; i++) {
while(rev[i] != 0) {
r[i] = r[i] * 10;
r[i] = r[i] + rev[i]%10;
rev[i] = rev[i]/10;
}
}
for(int i = 0; i < a.length; i++) {
boolean isPrime = true;
for (int j = 2; j < i; j++) {
if((a[i]%j==0) || (r[i]%j==0)) {
isPrime = false;
break;
}
}
if(isPrime)
System.out.println(a[i]);
System.out.println(r[i]);
}
}
}
Somewhere I am stuck I don't know how to eliminate repeated no, how to merge the array at last and also it is printing 1 and 2 as prime no when I give input and 2

You need to use TreeSet - which will contain only distinct elements and give result in sorted form. You can refer to following code-
Set<Integer> set = new TreeSet<>();
for(int i = 0; i < a.length; i++) {
boolean isPrime = true;
if(isPrime(a[i]) && isPrime(r[i]))
set.add(a[i]);
}
Iterator it = set.iterator();
while(it.hasNext())
System.out.print(it.next() + " ");
Also create a function for checking prime numbers -
private static boolean isPrime(int num) {
if(num==1) return false;
for(int i = 2; i <= num/2; ++i)
{
if(num % i == 0)
{
return false;
}
}
return true;
}

You can try below code. Hope it helps you,
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Set;
import java.util.TreeSet;
public class PrimeNumberTest {
public static void main(String[] args) {
List<Integer> list = new ArrayList<>(Arrays.asList(7, 11, 12, 23, 19, 7 ,113, 101));
//To remove duplicates
Set<Integer> set = new TreeSet<>(list);
System.out.println(getPrimeNumbers(set).toString().replaceAll(",", "").replace("]", "").replace("[", ""));
}
//Method to get unique ordered set of prime numbers
private static Set<Integer> getPrimeNumbers(Set<Integer> set) {
Set<Integer> resultList=new TreeSet<>();
set.forEach(ele->{
//check for prime
if(isPrime(ele)){
//if prime number check for reverse and if true, add to result
if(isPrime(reverserNumb(ele)))
resultList.add(ele);
}
});
return resultList;
}
private static boolean isPrime(int num){
if(num<2)
return false;
// Check for even numbers
if (num % 2 == 0) {
return num == 2;
}
// Check for odd numbers
for (int i = 3; i*i <= num; i += 2) {
if (num % i == 0) {
return false;
}
}
return true;
}
private static int reverserNumb(int num) {
return Integer.valueOf(new StringBuilder(String.valueOf(num)).reverse().toString());
}
}

Here is the code for prime test using √n approach
static boolean isPrime(int n){
//corner case
if (n <= 1) return false;
if (n <= 3) return true;
//middle 5 number
if (n % 2 == 0 || n % 3 == 0) return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
Use can use set for remove duplicate element

Try if this code works, hope it helps
Code:
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
public class PrimeNumbers {
static ArrayList<Integer> prime = new ArrayList<>();
public static void main(String[] args) {
Scanner user_input = new Scanner(System.in);
System.out.print("Enter Number of Integers: ");
int number_count = user_input.nextInt();
int[] numbers = new int[number_count];
for (int i = 0; i < numbers.length; i++) {
System.out.print("Enter Integer: ");
numbers[i] = user_input.nextInt();
}
System.out.print("Values Entered: ");
for (int i = 0; i < numbers.length; i++) {
System.out.print(numbers[i] + " ");
checkPrime(numbers[i],false); //false don't reverse
checkPrime(numbers[i],true); //true reverse value to check if it is also prime
}
System.out.print("\nList of prime numbers: ");
Collections.sort(prime);
for(int n : prime){
System.out.print(n + " ");
}
if(prime.isEmpty()){
System.out.print("no prime numbers on list\n");
}
}
//check for duplicates
static void insertValueToPrime(int n){
for(int p : prime){
if(n == p){
return;
}
}
prime.add(n);
}
static void checkPrime(int n,boolean isReverse) {
int i, m = 0, flag = 0,realn = n;
if(isReverse){
n = reverseNumber(n);
}
m = n / 2;
if (n == 0 || n == 1) {
//no a prime number
} else {
for (i = 2; i <= m; i++) {
if (n % i == 0) {
// not a prime number
flag = 1;
break;
}
}
if (flag == 0) {
insertValueToPrime(realn);
}
}
}
static int reverseNumber(int n){
String reverse = "",str=""+n;
for(int i = str.length() - 1; i >= 0; i--)
{
reverse = reverse + str.charAt(i);
}
return Integer.parseInt(reverse);
}
}

trying to find the sum of even fibonacci numbers to 4 million

I am trying to find the sum of the even Fibonacci numbers up untill 4 million.
I found the numbers but i can't get them add up... in the if(n % 2 ==0) loop
8
34
144
610
2584
10946
46368
196418
832040
3524578
public static void number2()
{
int number = 40;
int a, b, c;
int numLim = 0;
a = 1;
b = 2;
while(numLim < 4000000)
{
c = a + b;
a = b;
b = c;
numLim = b;
if(numLim > 4000000)
{
break;
}
int sum = 0;
if(numLim % 2 == 0)
{
System.out.println(numLim);
sum = sum + numLim;
System.out.println("sum :" +sum);
}
}
}

You must define sum outside the while loop, or it will become 0 each iteration.
int sum = 0;
...
while ...
Remember not to set sum to 0 each iteration.

public class Euler2 {
public static void main(String[] args) {
int fibonacci;
int num = 0;
int num2 = 1;
int loop;
int sum = 0;
System.out.println(num2);
for (loop = 0; loop <= 32; loop++) {
fibonacci = num + num2;
num = num2;
num2 = fibonacci;
System.out.println("Fibonacci number : " + fibonacci);
sum += fibonacci;
System.out.println("This is the sum " +sum);
}
}
}
So I solved it like this, it's a little more efficient and the math works but Euler hates me, hope this helps.

public class Euler2 {
public static void main(String[] args) {
int fibonacci;
int num = 0;
int num2 = 1;
int loop;
int sum = 0;
System.out.println(num2);
for (loop = 0; loop <= 31; loop++) {
fibonacci = num + num2;
num = num2;
num2 = fibonacci;
System.out.println("Fibonacci number : " + fibonacci);
if (fibonacci%2 == 0) {
sum += fibonacci;
System.out.println(sum);
}
}
}
Sorry, this code works.

Tried doing the above in Java and here is my solution that works
public static void main(String[] args) {
int first = 1;
int second = 2;
int sum = 0;
int sumOfEvenValuedTerms = second;
for (int i = 0; i < 30; i++) {
sum = first + second;
if (sum <= 4000000) {
if (sum % 2 == 0) {
sumOfEvenValuedTerms += sum;
}
first = second;
second = sum;
}
}
System.out.println(sumOfEvenValuedTerms);
}
Output is 4613732

public static int getSumOfEvenNumbers(int n) {
int prev = 0;
int i =1;
int sum = 0;
while (i<n){
int nextNumber = i + prev;
if(nextNumber %2 ==0) {
System.out.println(nextNumber);
sum +=nextNumber;
}
prev = i;
i = nextNumber;
}
return sum;
}

public class evenFib {
public static void main(String[] args) {
double a = 1, b = 2, c = 0, sum = 0;
for (double i = 0; i <= 1000; i++) {
c = a + b;
a = b;
b = c;
if (c % 2 == 0 && sum < 4000000) {
sum = sum + c;
}
}
System.out.println(sum + 2);
}
}

Validate credit card number using luhn algorithm [duplicate]

I tried to check the validation of credit card using Luhn algorithm, which works as the following steps:
Double every second digit from right to left. If doubling of a digit results in a two-digit number, add up the two digits to get a single-digit number.
2 * 2 = 4
2 * 2 = 4
4 * 2 = 8
1 * 2 = 2
6 * 2 = 12 (1 + 2 = 3)
5 * 2 = 10 (1 + 0 = 1)
8 * 2 = 16 (1 + 6 = 7)
4 * 2 = 8
Now add all single-digit numbers from Step 1.
4 + 4 + 8 + 2 + 3 + 1 + 7 + 8 = 37
Add all digits in the odd places from right to left in the card number.
6 + 6 + 0 + 8 + 0 + 7 + 8 + 3 = 38
Sum the results from Step 2 and Step 3.
37 + 38 = 75
If the result from Step 4 is divisible by 10, the card number is valid; otherwise, it is invalid. For example, the number 4388576018402626 is invalid, but the number 4388576018410707 is valid.
Simply, my program always displays valid for everything that I input. Even if it's a valid number and the result of sumOfOddPlace and sumOfDoubleEvenPlace methods are equal to zero. Any help is appreciated.
import java.util.Scanner;
public class CreditCardValidation {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int count = 0;
long array[] = new long [16];
do
{
count = 0;
array = new long [16];
System.out.print("Enter your Credit Card Number : ");
long number = in.nextLong();
for (int i = 0; number != 0; i++) {
array[i] = number % 10;
number = number / 10;
count++;
}
}
while(count < 13);
if ((array[count - 1] == 4) || (array[count - 1] == 5) || (array[count - 1] == 3 && array[count - 2] == 7)){
if (isValid(array) == true) {
System.out.println("\n The Credit Card Number is Valid. ");
} else {
System.out.println("\n The Credit Card Number is Invalid. ");
}
} else{
System.out.println("\n The Credit Card Number is Invalid. ");
}
}
public static boolean isValid(long[] array) {
int total = sumOfDoubleEvenPlace(array) + sumOfOddPlace(array);
if ((total % 10 == 0)) {
for (int i=0; i< array.length; i++){
System.out.println(array[i]);}
return true;
} else {
for (int i=0; i< array.length; i++){
System.out.println(array[i]);}
return false;
}
}
public static int getDigit(int number) {
if (number <= 9) {
return number;
} else {
int firstDigit = number % 10;
int secondDigit = (int) (number / 10);
return firstDigit + secondDigit;
}
}
public static int sumOfOddPlace(long[] array) {
int result = 0;
for (int i=0; i< array.length; i++)
{
while (array[i] > 0) {
result += (int) (array[i] % 10);
array[i] = array[i] / 100;
}}
System.out.println("\n The sum of odd place is " + result);
return result;
}
public static int sumOfDoubleEvenPlace(long[] array) {
int result = 0;
long temp = 0;
for (int i=0; i< array.length; i++){
while (array[i] > 0) {
temp = array[i] % 100;
result += getDigit((int) (temp / 10) * 2);
array[i] = array[i] / 100;
}
}
System.out.println("\n The sum of double even place is " + result);
return result;
}
}

You can freely import the following code:
public class Luhn
{
public static boolean Check(String ccNumber)
{
int sum = 0;
boolean alternate = false;
for (int i = ccNumber.length() - 1; i >= 0; i--)
{
int n = Integer.parseInt(ccNumber.substring(i, i + 1));
if (alternate)
{
n *= 2;
if (n > 9)
{
n = (n % 10) + 1;
}
}
sum += n;
alternate = !alternate;
}
return (sum % 10 == 0);
}
}
Link reference: https://github.com/jduke32/gnuc-credit-card-checker/blob/master/CCCheckerPro/src/com/gnuc/java/ccc/Luhn.java

Google and Wikipedia are your friends. Instead of long-array I would use int-array. On Wikipedia following java code is published (together with detailed explanation of Luhn algorithm):
public static boolean check(int[] digits) {
int sum = 0;
int length = digits.length;
for (int i = 0; i < length; i++) {
// get digits in reverse order
int digit = digits[length - i - 1];
// every 2nd number multiply with 2
if (i % 2 == 1) {
digit *= 2;
}
sum += digit > 9 ? digit - 9 : digit;
}
return sum % 10 == 0;
}
You should work on your input processing code. I suggest you to study following solution:
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
boolean repeat;
List<Integer> digits = new ArrayList<Integer>();
do {
repeat = false;
System.out.print("Enter your Credit Card Number : ");
String input = in.next();
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
if (c < '0' || c > '9') {
repeat = true;
digits.clear();
break;
} else {
digits.add(Integer.valueOf(c - '0'));
}
}
} while (repeat);
int[] array = new int[digits.size()];
for (int i = 0; i < array.length; i++) {
array[i] = Integer.valueOf(digits.get(i));
}
boolean valid = check(array);
System.out.println("Valid: " + valid);
}

I took a stab at this with Java 8:
public static boolean luhn(String cc) {
final boolean[] dbl = {false};
return cc
.chars()
.map(c -> Character.digit((char) c, 10))
.map(i -> ((dbl[0] = !dbl[0])) ? (((i*2)>9) ? (i*2)-9 : i*2) : i)
.sum() % 10 == 0;
}
Add the line
.replaceAll("\\s+", "")
Before
.chars()
If you want to handle whitespace.
Seems to produce identical results to
return LuhnCheckDigit.LUHN_CHECK_DIGIT.isValid(cc);
From Apache's commons-validator.

There are two ways to split up your int into List<Integer>
Use %10 as you are using and store it into a List
Convert to a String and then take the numeric values
Here are a couple of quick examples
public static void main(String[] args) throws Exception {
final int num = 12345;
final List<Integer> nums1 = splitInt(num);
final List<Integer> nums2 = splitString(num);
System.out.println(nums1);
System.out.println(nums2);
}
private static List<Integer> splitInt(int num) {
final List<Integer> ints = new ArrayList<>();
while (num > 0) {
ints.add(0, num % 10);
num /= 10;
}
return ints;
}
private static List<Integer> splitString(int num) {
final List<Integer> ints = new ArrayList<>();
for (final char c : Integer.toString(num).toCharArray()) {
ints.add(Character.getNumericValue(c));
}
return ints;
}

I'll use 5 digit card numbers for simplicity. Let's say your card number is 12345; if I read the code correctly, you store in array the individual digits:
array[] = {1, 2, 3, 4, 5}
Since you already have the digits, in sumOfOddPlace you should do something like
public static int sumOfOddPlace(long[] array) {
int result = 0;
for (int i = 1; i < array.length; i += 2) {
result += array[i];
}
return result;
}
And in sumOfDoubleEvenPlace:
public static int sumOfDoubleEvenPlace(long[] array) {
int result = 0;
for (int i = 0; i < array.length; i += 2) {
result += getDigit(2 * array[i]);
}
return result;
}

this is the luhn algorithm implementation which I use for only 16 digit Credit Card Number
if(ccnum.length()==16){
char[] c = ccnum.toCharArray();
int[] cint = new int[16];
for(int i=0;i<16;i++){
if(i%2==1){
cint[i] = Integer.parseInt(String.valueOf(c[i]))*2;
if(cint[i] >9)
cint[i]=1+cint[i]%10;
}
else
cint[i] = Integer.parseInt(String.valueOf(c[i]));
}
int sum=0;
for(int i=0;i<16;i++){
sum+=cint[i];
}
if(sum%10==0)
result.setText("Card is Valid");
else
result.setText("Card is Invalid");
}else
result.setText("Card is Invalid");
If you want to make it use on any number replace all 16 with your input number length.
It will work for Visa number given in the question.(I tested it)

Here's my implementation of the Luhn Formula.
/**
* Runs the Luhn Equation on a user inputed CCN, which in turn
* determines if it is a valid card number.
* #param c A user inputed CCN.
* #param cn The check number for the card.
* #return If the card is valid based on the Luhn Equation.
*/
public boolean luhn (String c, char cn)
{
String card = c;
String checkString = "" + cn;
int check = Integer.valueOf(checkString);
//Drop the last digit.
card = card.substring(0, ( card.length() - 1 ) );
//Reverse the digits.
String cardrev = new StringBuilder(card).reverse().toString();
//Store it in an int array.
char[] cardArray = cardrev.toCharArray();
int[] cardWorking = new int[cardArray.length];
int addedNumbers = 0;
for (int i = 0; i < cardArray.length; i++)
{
cardWorking[i] = Character.getNumericValue( cardArray[i] );
}
//Double odd positioned digits (which are really even in our case, since index starts at 0).
for (int j = 0; j < cardWorking.length; j++)
{
if ( (j % 2) == 0)
{
cardWorking[j] = cardWorking[j] * 2;
}
}
//Subtract 9 from digits larger than 9.
for (int k = 0; k < cardWorking.length; k++)
{
if (cardWorking[k] > 9)
{
cardWorking[k] = cardWorking[k] - 9;
}
}
//Add all the numbers together.
for (int l = 0; l < cardWorking.length; l++)
{
addedNumbers += cardWorking[l];
}
//Finally, check if the number we got from adding all the other numbers
//when divided by ten has a remainder equal to the check number.
if (addedNumbers % 10 == check)
{
return true;
}
else
{
return false;
}
}
I pass in the card as c which I get from a Scanner and store in card, and for cn I pass in checkNumber = card.charAt( (card.length() - 1) );.

Okay, this can be solved with a type conversions to string and some Java 8
stuff. Don't forget numbers and the characters representing numbers are not the same. '1' != 1
public static int[] longToIntArray(long cardNumber){
return Long.toString(cardNumber).chars()
.map(x -> x - '0') //converts char to int
.toArray(); //converts to int array
}
You can now use this method to perform the luhn algorithm:
public static int luhnCardValidator(int cardNumbers[]) {
int sum = 0, nxtDigit;
for (int i = 0; i<cardNumbers.length; i++) {
if (i % 2 == 0)
nxtDigit = (nxtDigit > 4) ? (nxtDigit * 2 - 10) + 1 : nxtDigit * 2;
sum += nxtDigit;
}
return (sum % 10);
}

private static int luhnAlgorithm(String number){
int n=0;
for(int i = 0; i<number.length(); i++){
int x = Integer.parseInt(""+number.charAt(i));
n += (x*Math.pow(2, i%2))%10;
if (x>=5 && i%2==1) n++;
}
return n%10;
}

public class Creditcard {
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
String cardno = sc.nextLine();
if(checkType(cardno).equals("U")) //checking for unknown type
System.out.println("UNKNOWN");
else
checkValid(cardno); //validation
}
private static String checkType(String S)
{
int AM=Integer.parseInt(S.substring(0,2));
int D=Integer.parseInt(S.substring(0,4)),d=0;
for(int i=S.length()-1;i>=0;i--)
{
if(S.charAt(i)==' ')
continue;
else
d++;
}
if((AM==34 || AM==37) && d==15)
System.out.println("AMEX");
else if(D==6011 && d==16)
System.out.println("Discover");
else if(AM>=51 && AM<=55 && d==16)
System.out.println("MasterCard");
else if(((S.charAt(0)-'0')==4)&&(d==13 || d==16))
System.out.println("Visa");
else
return "U";
return "";
}
private static void checkValid(String S) // S--> cardno
{
int i,d=0,sum=0,card[]=new int[S.length()];
for(i=S.length()-1;i>=0;i--)
{
if(S.charAt(i)==' ')
continue;
else
card[d++]=S.charAt(i)-'0';
}
for(i=0;i<d;i++)
{
if(i%2!=0)
{
card[i]=card[i]*2;
if(card[i]>9)
sum+=digSum(card[i]);
else
sum+=card[i];
}
else
sum+=card[i];
}
if(sum%10==0)
System.out.println("Valid");
else
System.out.println("Invalid");
}
public static int digSum(int n)
{
int sum=0;
while(n>0)
{
sum+=n%10;
n/=10;
}
return sum;
}
}

Here is the implementation of Luhn algorithm.
public class LuhnAlgorithm {
/**
* Returns true if given card number is valid
*
* #param cardNum Card number
* #return true if card number is valid else false
*/
private static boolean checkLuhn(String cardNum) {
int cardlength = cardNum.length();
int evenSum = 0, oddSum = 0, sum;
for (int i = cardlength - 1; i >= 0; i--) {
System.out.println(cardNum.charAt(i));
int digit = Character.getNumericValue(cardNum.charAt(i));
if (i % 2 == 0) {
int multiplyByTwo = digit * 2;
if (multiplyByTwo > 9) {
/* Add two digits to handle cases that make two digits after doubling */
String mul = String.valueOf(multiplyByTwo);
multiplyByTwo = Character.getNumericValue(mul.charAt(0)) + Character.getNumericValue(mul.charAt(1));
}
evenSum += multiplyByTwo;
} else {
oddSum += digit;
}
}
sum = evenSum + oddSum;
if (sum % 10 == 0) {
System.out.println("valid card");
return true;
} else {
System.out.println("invalid card");
return false;
}
}
public static void main(String[] args) {
String cardNum = "4071690065031703";
System.out.println(checkLuhn(cardNum));
}
}

public class LuhnAlgorithm {
/**
* Returns true if given card number is valid
*
* #param cardNum Card number
* #return true if card number is valid else false
*/
private static boolean checkLuhn(String cardNum) {
int cardlength = cardNum.length();
int evenSum = 0, oddSum = 0, sum;
for (int i = cardlength - 1; i >= 0; i--) {
System.out.println(cardNum.charAt(i));
int digit = Character.getNumericValue(cardNum.charAt(i));
if (i % 2 == 0) {
int multiplyByTwo = digit * 2;
if (multiplyByTwo > 9) {
/* Add two digits to handle cases that make two digits after doubling */
String mul = String.valueOf(multiplyByTwo);
multiplyByTwo = Character.getNumericValue(mul.charAt(0)) + Character.getNumericValue(mul.charAt(1));
}
evenSum += multiplyByTwo;
} else {
oddSum += digit;
}
}
sum = evenSum + oddSum;
if (sum % 10 == 0) {
System.out.println("valid card");
return true;
} else {
System.out.println("invalid card");
return false;
}
}
public static void main(String[] args) {
String cardNum = "8112189875";
System.out.println(checkLuhn(cardNum));
}
}
Hope it may works.

const options = {
method: 'GET',
headers: {Accept: 'application/json', 'X-Api-Key': '[APIkey]'}
};
fetch('https://api.epaytools.com/Tools/luhn?number=[CardNumber]&metaData=true', options)
.then(response => response.json())
.then(response => console.log(response))
.catch(err => console.error(err));

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);
}