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Finding the smith number between given range

I will get to the point quickly. Basically smith numbers are: Composite number the sum of whose digits is the sum of the digits of its prime factors (excluding 1). (The primes are excluded since they trivially satisfy this condition). One example of a Smith number is the beast number 666=2·3·3·37, since 6+6+6=2+3+3+(3+7)=18.
what i've tried:
In a for loop first i get the sum of the current number's(i) digits
In same loop i try to get the sum of the number's prime factors digits.
I've made another method to check if current number that is going to proccessed in for loop is prime or not,if its prime it will be excluded
But my code is seems to not working can you guys help out?
public static void main(String[] args) {
smithInrange(1, 50);
}
public static void smithInrange(int start_val, int end_val) {
for (int i = start_val; i < end_val; i++) {
if(!isPrime(i)) { //since we banned prime numbers from this process i don't include them
int for_digit_sum = i, digit = 0, digit_sum = 0, for_factor_purpose = i, smith_sum = 0;
int first = 0, second = 0, last = 0;
// System.out.println("current number is" + i);
while (for_digit_sum > 0) { // in this while loop i get the sum of current number's digits
digit = for_digit_sum % 10;
digit_sum += digit;
for_digit_sum /= 10;
}
// System.out.println("digit sum is"+digit_sum);
while (for_factor_purpose % 2 == 0) { // i divide the current number to 2 until it became an odd number
first += 2;
for_factor_purpose /= 2;
}
// System.out.println("the first sum is " + first);
for (int j = 3; j < Math.sqrt(for_factor_purpose); j += 2) {
while (for_factor_purpose % j == 0) { // this while loop is for getting the digit sum of every prime
// factor that j has
int inner_digit = 0, inner_temp = j, inner_digit_sum = 0;
while (inner_temp > 0) {
inner_digit = inner_temp % 10;
second += inner_digit;
inner_temp /= 10;
}
// System.out.println("the second sum is " + second);
for_factor_purpose /= j;
}
}
int last_temp = for_factor_purpose, last_digit = 0, last_digit_sum = 0;
if (for_factor_purpose > 2) {
while (last_temp > 0) {
last_digit = last_temp % 10;
last += last_digit;
last_temp /= 10;
}
// System.out.println("last is " + last);
}
smith_sum = first + second + last;
// System.out.println("smith num is "+ smith_sum);
// System.out.println(smith_sum);
if (smith_sum == digit_sum) {
System.out.println("the num founded is" + i);
}
}
}
}
public static boolean isPrime(int i) {
int sqrt = (int) Math.sqrt(i) + 1;
for (int k = 2; k < sqrt; k++) {
if (i % k == 0) {
// number is perfectly divisible - no prime
return false;
}
}
return true;
}
the output is:
the num founded is4
the num founded is9
the num founded is22
the num founded is25
the num founded is27
the num founded is49
how ever the smith number between this range(1 and 50) are:
4, 22 and 27
edit:I_ve found the problem which is :
Math.sqrt(for_factor_purpose) it seems i should add 1 to it to eliminate square numbers. Thanks to you guys i've see sthe solution on other perspectives.
Keep coding!

Main loop for printing Smith numbers.
for (int i = 3; i < 10000; i++) {
if (isSmith(i)) {
System.out.println(i + " is a Smith number.");
}
}
The test method to determine if the supplied number is a Smith number. The list of primes is only increased if the last prime is smaller in magnitude than the number under test.
static boolean isSmith(int v) {
int sum = 0;
int save = v;
int lastPrime = primes.get(primes.size() - 1);
if (lastPrime < v) {
genPrimes(v);
}
outer:
for (int p : primes) {
while (save > 1) {
if (save % p != 0) {
continue outer;
}
sum += sumOfDigits(p);
save /= p;
}
break;
}
return sum == sumOfDigits(v) && !primes.contains(v);
}
Helper method to sum the digits of a number.
static int sumOfDigits(int i) {
return String.valueOf(i).chars().map(c -> c - '0').sum();
}
And the prime generator. It uses the list as it is created to determine if a given
number is a prime.
static List<Integer> primes = new ArrayList<>(List.of(2, 3));
static void genPrimes(int max) {
int next = primes.get(primes.size() - 1);
outer:
while (next <= max) {
next += 2;
for (int p : primes) {
if (next % p == 0) {
continue outer;
}
if (p * p > next) {
break;
}
}
primes.add(next);
}
}
}

I do not want to spoil the answer finding, but just some simpler code snippets,
making everything simpler, and more readable.
public boolean isSmith(int a) {
if (a < 2) return false;
int factor = findDivisor(a);
if (factor == a) return false;
int sum = digitSum(a);
// loop:
a /= factor;
sum -= digitSum(factor);
...
}
boolean isPrime(int a){
for(int i = 2; i*i <= a; i++) {
if (a % i == 0) {
return false;
}
}
return true;
}
int findDivisor(int a){
for(int i = 2; i*i <= a; i++) {
if (a % i == 0) {
return i;
}
}
return a;
}
int digitSum(int a) {
if (a < 10) {
return a;
}
int digit = a % 10;
int rest = a / 10;
return digit + digitSum(rest);
}
As you see integer division 23 / 10 == 2, and modulo (remainder) %: 23 % 10 == 3 can simplify things.
Instead of isPrime, finding factor(s) is more logical. In fact the best solution is not using findDivisor, but immediately find all factors
int factorsSum = 0;
int factorsCount = 0;
for(int i = 2; i*i <= a; i++) {
while (a % i == 0) {
factorsSum += digitSum(i);
a /= i;
factorsCount++;
}
}
// The remaining factor >= sqrt(original a) must be a prime.
// (It cannot contain smaller factors.)
factorsSum += digitSum(a);
factorsCount++;

Here is the code. If you need further help, please let me know. The code is pretty self explanatory and a decent bit was taken from your code but if you need me to explain it let me know.
In short, I created methods to check if a number is a smith number and then checked each int in the range.
import java.util.*;
public class MyClass {
public static void main(String args[]) {
System.out.println(smithInRange)
}
public int factor;
public boolean smithInRange(int a, int b){
for (int i=Math.min(a,b);i<=Math.max(a,b);i++) if(isSmith(i)) return true;
return false;
}
public boolean isSmith(int a){
if(a<2) return false;
if(isPrime(a)) return false;
int digits=0;
int factors=0;
String x=a+¨" ";
for(int i=0;i<x.length()-1;i++) digits+= Integer.parseInt(x.substring(i,i+1));
ArrayList<Integer> pF = new ArrayList<Integer>();
pF.add(a);
while(!aIsPrime(pF)){
int num = pF.get(pF.size-1)
pF.remove(pF.size()-1);
pF.add(factor);
pF.add(num/factor)
}
for(int i: pF){
if((factors+"").length()==1)factors+= i;
else{
String ss= i+" ";
int nums=0;
for(int j=0;j<ss.length()-1;j++){
nums+=Integer.parseInt(ss.substring(j,j+1));
}
}
}
return (factors==digits);
}
public boolean isPrime(int a){
for(int i=2;i<=(int)Math.sqrt(a),i++){
String s = (double)a/(double)i+"";
if(s.substring(s.length()-2).equals(".0")){
return false;
factor = i;
}
}
return true;
}
public boolean aIsPrime(ArrayList<int> a){
for(int i: a) if (!isPrime(a)) return false;
return true;
}
}

Can't figure out the error Luhn check

Its supose to tell me if a card is valid or invalid using luhn check
4388576018402626 invalid
4388576018410707 valid
but it keeps telling me that everything is invalid :/
Any tips on what to do, or where to look, would be amazing. I have been stuck for a few hours.
It would also help if people tell me any tips on how to find why a code is not working as intended.
im using eclipse and java
public class Task11 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter a credit card number as a long integer: ");
long number = input.nextLong();
if (isValid(number)) {
System.out.println(number + " is valid");
} else {
System.out.println(number + " is invalid");
}
}
public static boolean isValid(long number) {
return (getSize(number) >= 13) && (getSize(number) <= 16)
&& (prefixMatched(number, 4) || prefixMatched(number, 5) || prefixMatched(number, 6) || prefixMatched(number, 37))
&& (sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 == 0;
}
public static int sumOfDoubleEvenPlace(long number) {
int result = 0;
long start = 0;
String digits = Long.toString(number);
if ((digits.length() % 2) == 0) {
start = digits.length() - 1;
} else {
start = digits.length() - 2;
}
while (start != 0) {
result += (int) ((((start % 10) * 2) % 10) + (((start % 10) * 2) / 2));
start = start / 100;
}
return result;
}
public static int getDigit(int number) {
return number % 10 + (number / 10);
}
public static int sumOfOddPlace(long number) {
int result = 0;
while (number != 0) {
result += (int) (number % 10);
number = number / 100;
}
return result;
}
public static boolean prefixMatched(long number, int d) {
return getPrefix(number, getSize(d)) == d;
}
public static int getSize(long d) {
int numberOfDigits = 0;
String sizeString = Long.toString(d);
numberOfDigits = sizeString.length();
return numberOfDigits;
}
public static long getPrefix(long number, int k) {
String size = Long.toString(number);
if (size.length() <= k) {
return number;
} else {
return Long.parseLong(size.substring(0, k));
}
}
}

You should modiffy your isValid() method to write down when it doesn't work, like this:
public static boolean isValid(long number) {
System.err.println();
if(getSize(number) < 13){
System.out.println("Err: Number "+number+" is too short");
return false;
} else if (getSize(number) > 16){
public static boolean isValid(long number) {
System.err.println();
if(getSize(number) < 13){
System.out.println("Err: Number "+number+" is too short");
return false;
} else if (getSize(number) > 16){
System.out.println("Err: Number "+number+" is too long");
return false;
} else if (! (prefixMatched(number, 4) || prefixMatched(number, 5) || prefixMatched(number, 6) || prefixMatched(number, 37)) ){
System.out.println("Err: Number "+number+" prefix doesn't match");
return false;
} else if( (sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 != 0){
System.out.println("Err: Number "+number+" doesn't have sum of odd and evens % 10. ");
return false;
}
return true;
}
My guess for your problem is on the getPrefix() method, you should add some logs here too.
EDIT: so, got more time to help you (don't know if it's still necessary but anyway). Also, I corrected the method I wrote, there were some errors (like, the opposite of getSize(number) >= 13 is getSize(number) < 13)...
First it will be faster to test with a set of data instead of entering the values each time yourself (add the values you want to check):
public static void main(String[] args) {
long[] luhnCheckSet = {
0, // too short
1111111111111111111L, // too long (19)
222222222222222l // prefix doesn't match
4388576018402626l, // should work ?
};
//System.out.print("Enter a credit card number as a long integer: ");
//long number = input.nextLong();
for(long number : luhnCheckSet){
System.out.println("Checking number: "+number);
if (isValid(number)) {
System.out.println(number + " is valid");
} else {
System.out.println(number + " is invalid");
}
System.out.println("-");
}
}
I don't know the details of this, but I think you should work with String all along, and parse to long only if needed (if number is more than 19 characters, it might not parse it long).
Still, going with longs.
I detailed your getPrefix() with more logs AND put the d in parameter in long (it's good habit to be carefull what primitive types you compare):
public static boolean prefixMatched(long number, long d) {
int prefixSize = getSize(d);
long numberPrefix = getPrefix(number, prefixSize);
System.out.println("Testing prefix of size "+prefixSize+" from number: "+number+". Prefix is: "+numberPrefix+", should be:"+d+", are they equals ? "+(numberPrefix == d));
return numberPrefix == d;
}
Still don't know what's wrong with this code, but it looks like it comes from the last test:
I didn't do it but you should make one method from sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 and log both numbers and the sum (like i did in prefixMatched() ). Add logs in both method to be sure it gets the result you want/ works like it should.
Have you used a debugger ? if you can, do it, it can be faster than adding a lot of logs !
Good luck

EDIT:
Here are the working functions and below I provided a shorter, more efficient solution too:
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. ");
}
in.close();
}
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;
}
}
I also found a solution with less lines of logic. I know you're probably searching for an OO approach with functions, building from this could be of some help.
Similar question regarding error in Luhn algorithm logic:
Check Credit Card Validity using Luhn Algorithm
Link to shorter solution:
https://code.google.com/p/gnuc-credit-card-checker/source/browse/trunk/CCCheckerPro/src/com/gnuc/java/ccc/Luhn.java
And here I tested the solution with real CC numbers:
public class CreditCardValidation{
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);
}
public static void main(String[] args){
//String num = "REPLACE WITH VALID NUMBER"; //Valid
String num = REPLACE WITH INVALID NUMBER; //Invalid
num = num.trim();
if(Check(num)){
System.out.println("Valid");
}
else
System.out.println("Invalid");
//Check();
}
}

How can I make ProjectEuler 23 Solution faster

This is what I have. It is solving the problem, but taking forever. Can I divide the last loop 0 to 28123 into half and run them simultaneously somehow to make it faster, and then in the end add the two sums to get the final result? Will "thread" help? What can I do to make the code solve faster?
/*
A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.
As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
*/
public class power {
public static void main(String[] args){
System.out.println(bigSum());
}
public static boolean isPerfect(int a){
boolean perfect = false;
int sum = 0;
for(int i = 1; i<(a/2)+1; i++)
{
if (a%i == 0)
{
sum = sum + i;
}
}
if(a == sum){
perfect = true;
}
return perfect;
}
public static boolean isAbundant(int a){
boolean Abundant = false;
int sum = 0;
for(int i = 1; i<(a/2)+1; i++)
{
if (a%i == 0)
{
sum = sum + i;
}
}
if(a < sum){
Abundant = true;
}
return Abundant;
}
public static boolean isDeficient(int a){
boolean Deficient = false;
int sum = 0;
for(int i = 1; i<(a/2)+1; i++)
{
if (a%i == 0)
{
sum = sum + i;
}
}
if(a > sum){
Deficient = true;
}
return Deficient;
}
public static boolean isSumOfTwoAbundant(int a){
boolean SumOfTwoAbundant = false;
for(int i = 1; i<a; i++){
if(isAbundant(i) && isAbundant(a-i)){
SumOfTwoAbundant = true;
}
}
return SumOfTwoAbundant;
}
public static long bigSum(){
int sum = 0;
for(int i = 0; i<28123; i++){
if(!isSumOfTwoAbundant(i)){
sum = sum + i;
System.out.println("i: " + i + "; " + "Sum: " + sum);
}
}
return sum;
}
}

You're recomputing whether every number < a is abundant with every call of isSumOfTwoAbundant. Try keeping a list of abundant numbers and add to it when you find one. Then you can loop through that list rather than rechecking for abundance for numbers < a. Something like:
public static boolean isSumOfTwoAbundant(int a){
boolean SumOfTwoAbundant = false;
if(isAbundant(a))
{
abundants.add(a);
}
for(int i = 0; i<abundants.length; i++) {
for(int j = 0; j < abundants.length; j++) {
if(a - abundants[i] == abundants[j]){
SumOfTwoAbundant = true;
}
}
}
return SumOfTwoAbundant ;
}
private ArrayList<int> abundants;
There are a lot of other ways to make this better, too, but Project Euler is about learning those through experience.

Another Java solution: (takes less than a second)
static int sum_Of_Divisors(int n){
int limit = n;
int sum = 0;
for(int i=1;i<limit;i++){
if(n%i==0){
if(i!=1){
if(i != n/i) sum += (i + n/i);
else sum += i;
}
else
sum += i;
limit = n/i;
}
}
return sum;
}
static boolean isAbundant(int n){
int sum = sum_Of_Divisors(n);
return sum>n;
}
static boolean sum_of_Two_Abundant(int n, HashSet<Integer> abundant){
for(Integer i:abundant){
if(abundant.contains(n-i)) return true;
}
return false;
}
static long q23(){
long sum = 0;
HashSet<Integer> abundant = new HashSet<Integer>();
for(int i=2;i<=28123;i++){
if(isAbundant(i))
abundant.add(i);
}
for(int i=1;i<=28123;i++)
if(!sum_of_Two_Abundant(i, abundant)) sum+=i;
return sum;
}

How to factor a number and determine whether its a prime number

So i have this problem where when i factor a number, lets say 15, i have to display this: 15=3x5 but instead what i get is 3x5x5 and i have no clue of how to make it that so it only displays 3x5. And then another problem i have is to find whether the number i inputted is a prime number or not. Any way of fixing this? I just need that and the other stuff im gonna edit after that.
public class PrimeFactor
{
public static void main(String[] args)
{
Scanner input= new Scanner(System.in);
int a;
int d;
int remainder=0;
int count=2;
int c=0;
String s;
System.out.println("Enter an integer to be factored:");
a=input.nextInt();
s="";
d=a;
while(a>1)
{
if(a>1)
{
s="";
while(a>1)
{
remainder=a%count;
if (!(remainder>0))
while(remainder==0)
{
remainder=a%count;
if (remainder==0)
{
a=a/count;
c=c+1;
s=s+count+"x";
if (a==1)
s=s+count;
}
else
count++;
}
else
count++;
}
if (a%count==0)
{
System.out.println(d +"=" + s);
System.out.println(d+" is a prime number.");
}
else
System.out.println(d +"=" + s);
}
// TODO code application logic here
}
}
}

This determines if the number is prime or not the quickest way. Another method would be to use a for loop to determine the number of factors for the number and then say it's prime if it has more than two factors.
int num; // is the number being tested for if it's prime.
boolean isPrime = true;
for (int i = 2; i <= Math.sqrt(num); i++) // only have to test until the square root of the number
{
if (num%i == 0) // if the number is divisible by anything from 2 - the square root of the number
{
isPrime = false; // it is not prime
break; // break out of the loop because it's not prime and no more testing needed
}
}
if (isPrime)
{
System.out.println(num + " is a prime number.");
}
else
{
System.out.println(num + " is a composite number.");
}

You are not constructing the factorization string quite right:
When you find that 3 divides a=15 you set s to 3x and set a to the quotient, so a=5
When you find that 5 divides a=5 you append 5x to s, so now s is 3x5x. Then you set a to the quotient, which is 1. Since the quotient is now 1, you append 5 again, so now you get 3x5x5.
What you'll want to do is append only 5 when a=1, not 5x5. You have to change this:
s=s+count+"x";
if (a==1)
s=s+count;
to this:
if (a==1) {
s=s+count;
} else {
s=s+count+"x";
}

How about trying like this:-
for(int i = input-1; i > 0; i--) {
if((input % i) == 0) {
if(i == 1)
System.out.println("Number is a prime");
else
System.out.println("Number is not a prime");
break;
}
}

These are quite straight-forward methods you can use to factor a number and determine if it is a prime number:
public static int oneFactor(int i) {
for (int j = 2; j < i; j++) {
if (i % j == 0)
return j;
}
return -1;
}
public static Integer[] primeFactors(int i) {
List<Integer> factors = new ArrayList<Integer>();
boolean cont = true;
while (cont) {
int f = oneFactor(i);
if (i > 1 && f != -1) {
i /= f;
factors.add(f);
} else
factors.add(i);
if (f == -1)
cont = false;
}
return factors.toArray(new Integer[factors.size()]);
}
public static boolean isPrime(int i) {
if (i == 2 || i == 3)
return true;
if (i < 2 || i % 2 == 0)
return false;
for (int j = 3, end = (int) Math.sqrt(i); j <= end; j += 2) {
if (i % j == 0) {
return false;
}
}
return true;
}
I am sure one can use faster algorithms, but those would be at the cost of simplicity, and it doesn't seem like you need high speed methods.
They all operate on ints, but its easy to change them to work with longs.
If you have any questions, feel free to ask!

You want to write a loop which loops through numbers 1 to (Inputted Number). And if you found a factor, you print it and divide the input by the factor. (And test if it can be divided again by the same number), else then skip to the next number.
Keep doing this until your input divides down to 1.
This program will break the number down to prime factors:
public class Factor {
public static void main() {
//Declares Variables
int factor = 15;
int i = 2;
System.out.println("Listing Factors:\n");
while (factor>1) {
//Tests if 'i' is a factor of 'factor'
if (factor%i == 0) {
//Prints out and divides factor
System.out.println(i);
factor = factor/i;
} else {
//Skips to next Number
i++;
}
}
}
}
Output:
Listing Factors:
3
5

Largest prime factor program takes aaaages - Java

So this is problem 3 from project Euler. For those who don't know, I have to find out the largest prime factor of 600851475143. I have the below code:
import java.lang.Math;
// 600851475143
public class LargestPrimeFactor {
public static void main(String[] stuff) {
long num = getLong("What number do you want to analyse? ");
long[] primes = primeGenerator(num);
long result = 0;
for(int i = 0; i < primes.length; i++) {
boolean modulo2 = num % primes[i] == 0;
if(modulo2) {
result = primes[i];
}
}
System.out.println(result);
}
public static long[] primeGenerator(long limit) {
int aindex = 0;
long[] ps = new long[primeCount(limit)];
for(long i = 2; i < limit + 1; i++) {
if(primeCheck(i)) {
ps[aindex] = i;
aindex++;
}
}
return ps;
}
public static boolean primeCheck(long num) {
boolean r = false;
if(num == 2 || num == 3) {
return true;
}
else if(num == 1) {
return false;
}
for(long i = 2; i < Math.sqrt(num); i++) {
boolean modulo = num % i == 0;
if(modulo) {
r = false;
break;
}
else if(Math.sqrt(num) < i + 1 && !modulo) {
r = true;
break;
}
}
return r;
}
public static int primeCount(long limit) {
int count = 0;
if(limit == 1 || limit == 2) {
return 0;
}
for(long i = 2; i <= limit; i++) {
if(primeCheck(i)) {
count++;
}
}
return count;
}
public static long getLong(String prompt) {
System.out.print(prompt + " ");
long mrlong = input.nextLong();
input.nextLine();
return mrlong;
}
}
But when I test the program with something (a lot) smaller than 600851475143, like 100000000, then the program takes its time - in fact, 100000000 has taken 20 minutes so far and is still going. I've obviously got the wrong approach here (and yes, the program does work, I tried it out with smaller numbers). Can anyone suggest a less exhaustive way?

public static void main(String[] args) {
long number = 600851475143L;
long highestPrime = -1;
for (long i = 2; i <= number; ++i) {
if (number % i == 0) {
highestPrime = i;
number /= i;
--i;
}
}
System.out.println(highestPrime);
}

public class LargestPrimeFactor {
public static boolean isPrime(long num){
int count = 0;
for(long i = 1; i<=num/2 ; i++){
if(num % i==0){
count++;
}
}
if(count==1){
return true;
}
return false;
}
public static String largestPrimeFactor(long num){
String factor = "none";
for(long i = 2; i<= num/2 ; i++){
if(num % i==0 && isPrime(i)){
factor = Long.toString(i);
}
}
return factor;
}
public static void main(String[] args) {
System.out.println(largestPrimeFactor(13195));
}
}

I have done several dozen of the challenges on Project Euler. Some of the questions can be solved with brute force (they recommend not to do this) but others require "out of the box" thinking. You cannot solve that by problem with brute force.
There is lots of help on the web to lead you in the right direction, for example:
http://thetaoishere.blogspot.com.au/2008/05/largest-prime-factor-of-number.html

The number of prime factors a number can have is always less than sqrt of that number so that there is no need to iterate through the number n to find its largest prime factor.
See this code.
public class LargestPrimeFactor {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
long num=sc.nextLong();
if(num>0 && num<=2)
{
System.out.println("largest prime is:-" + num);
System.exit(0);
}
int i=((Double)Math.sqrt(num)).intValue();
int j=3;
int x=0;
//used for looping through the j value which can also be a prime. for e.g in case of 100 we might get 9 as a divisor. we need to make sure divisor is also a prime number.
int z=0;
//same function as j but for divisor
int y=3;
int max=2;
//divisor is divisible
boolean flag=false;
//we found prime factors
boolean found=false;
while(x<=i)
{
y=3;
flag=false;
if(num % j ==0)
{
if(j>max)
{
for(z=0;z<Math.sqrt(j);z++)
{
if(j!=y && j % y==0)
{
flag=true;
}
y+=2;
}
if(!flag)
{
found=true;
max=j;
}
}
}
j+=2;
x++;
}
if(found){
System.out.println("The maximum prime is :- " + max);
}
else
{
System.out.println("The maximum prime is :- " + num);
}
}
}

change
for(long i = 2; i <= limit; i++)
to
// add the one for rounding errors in the sqrt function
new_limit = sqrt(limit) + 1;
// all even numbers are not prime
for(long i = 3; i <= new_limit; i+=2)
{
...
}
Factoring 1,000,000 for example instead of iterating 1,000,000 times
the thing only needs to do around 500 iterations.