WhiZTiM:
. . . .Happy Coding ....
. . .As for the prime number generation, you may want to try Sieve of Eratosthenes... or other Primality tests...

...my favorite, "...most prime numbers from 3 and above obeys ....(6k + 1) or (6k - 1), where k is a natural number"... THat should speed up implementation to some extent....

WhiZTiM:

@olyjosh.
I am really impressed! ...Would love to chat up with you Sir.

olyjosh:


Thanks for reminding me of Sieve of Eratosthenes. That will work fine for the example test case but not for natural number that is as big as 600851475143, it wont work in java since the largest lenght of array or Set(and Set implementation) you can have is 2^31 (int precision) and Sieve of Eratosthenes reqires boolean flagging over such array.

kudaisi:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

WhiZTiM:
Comparing the Runtime of olyjosh's simple and elegant solution, (Java) with mine, (Python)
You can see some abominable things happening... Python 3x faster than Java. ....HOw?
Memorization. ....

http://ideone.com/XNVT5m ....Java (greedy) ...0.07seconds 320MB
http://ideone.com/VSTUQe .....Python (Dynamic Programming) ...0.02seconds 8MB

Java will regain its glory if DP/Memorization is used.

olyjosh:
public class LargestPrimeFactor {
static boolean isPrime(long n){
if(n==1)return false;
if(n==2)return true;
if(n%2==0) return false;
int max = (int)Math.sqrt(n);
for (int i = 2; i <= max; i++) {
if(n%i==0)return false;
}
return true;
}

static Queue<Long> factors(long n){
Queue fac = new ArrayDeque();
//ArrayList<Long> factors = new ArrayList<>();
long h=Math.round(n/2);
for (long i = 2; i <= h; i++) {
if(n%i==0){fac.add(i);}
}
return fac;
}

public static void main(String[] args) {
long lastPrime=0;
long n =600851475143L;
Queue<Long> factors = factors(n);
for (Iterator<Long> iterator = factors.iterator(); iterator.hasNext() {
long next = iterator.next();
if(isPrime(next))lastPrime=next;
}
System.out.println("largest prime factor is: "+lastPrime);
}
}


largest prime factor is: 6857
This solution got me scared cause it runs for about 2minnutes on my 4gb RAM, 1.65GHz processor

public class Primes
{
    static boolean isPrime(long n)
    {
        if ( n % 2 == 0 )return false;
        long sqrt = (long) Math.sqrt(n);
        sqrt= sqrt%2 == 0 ? sqrt-1 : sqrt; 
        for ( int i = 3; i < sqrt; i += 2 ) 
        {
            if ( n % i == 0 )return false;
        }
        return true;
    }
    
    public static void main(String[] args)
{

    long t = 600851475143L;
    long d = 2;
    while (1==1)
    {
        long tmp = 600851475143L / d;
        if ( t % tmp == 0 &&  isPrime(tmp) )
        {
            System.out.println("= " + tmp);
            break;
        }
        d++;
    }
}
}

olyjosh:
@WhiZTiM check this out
Runs for 10seconds on 4gb RAM, 1.65GHz Duo Core processor

public class Primes
{
    static boolean isPrime(long n)
    {
        if ( n % 2 == 0 )return false;
        long sqrt = (long) Math.sqrt(n);
        sqrt= sqrt%2 == 0 ? sqrt-1 : sqrt; 
        for ( int i = 3; i < sqrt; i += 2 ) 
        {
            if ( n % i == 0 )return false;
        }
        return true;
    }
    
    public static void main(String[] args)
{

    long t = 600851475143L;
    long d = 2;
    while (1==1)
    {
        long tmp = 600851475143L / d;
        if ( t % tmp == 0 &&  isPrime(tmp) )
        {
            System.out.println("= " + tmp);
            break;
        }
        d++;
    }
}
}

Jregz:

in php
function getSum($u){
for ($uu = 0 ; $uu < 1000 ; $uu++){
if ($uu % 3 === 0 || $uu % 5 === 0 ){
$u += $uu ;
}
}
return $u
}
echo getSum(0) ; // 233168

olyjosh:


Welcome to the show

kudaisi:
The sum of the squares of the first ten natural numbers is,

12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

/**
 *
 * @author olyjosh
 */
public class DifferenceSumsSquare {
    
    public static void main(String[] args) {
        final int n=100;
        int sqOfSum = (n*(n+1)*(2*n+1))/6;
        int sumOfsq = (n*(1+n))/2; sumOfsq*=sumOfsq;
        System.out.println("Difference is "+(sumOfsq-sqOfSum));
    }
}

kudaisi:
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10 001st prime number?

public class _10001stPrime {
    
    static boolean isPrime(long n)
    {
        if ( n % 2 == 0 )return false;
        long sqrt = (long) Math.sqrt(n);
        sqrt= sqrt%2 == 0 ? sqrt-1 : sqrt; 
        for ( int i = 3; i < sqrt; i += 2 ) 
            if ( n%i == 0 )return false;
        return true;
    }
    
    public static void main(String[] args) {
        int i =1;
        int n =3;
        while(i<10001){
            if(isPrime(n))i++;
            n+=2;
        }
        System.out.println("10001st prime number: "+n);
    }
}

olyjosh:


Thanks for reminding me of Sieve of Eratosthenes. That will work fine for the example test case but not for natural number that is as big as 600851475143, it wont work in java since the largest lenght of array or Set(and Set implementation) you can have is 2^31 (int precision) and Sieve of Eratosthenes reqires boolean flagging over such array.

kudaisi:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

 
var sum = 0;
(function ()  {
for (var i = 0; i < 1000; i += 1) {
if (i % 5 === 0 || i % 3 === 0) {
sum += i;
} 
} 
return sum;
} () ) ;
// 233168

kudaisi:
The sum of the squares of the first ten natural numbers is,

12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

 
(function ()  {
var sum = 0;
for (var i = 0; i <= 100; i += 1) {
sum += i;
} 
return Math.pow(sum, 2);
} ()) -  (function ()  {
var sum = 0;
for (var i = 0; i <= 100; i += 1)  {
sum += Math.pow(i, 2);
} 
return sum;
} () ) 
// 25164150

kudaisi:
The sum of the squares of the first ten natural numbers is,

12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

blueyedgeek:

 
(function ()  {
var sum = 0;
for (var i = 0; i <= 100; i += 1) {
sum += i;
} 
return Math.pow(sum, 2);
} ()) -  (function ()  {
var sum = 0;
for (var i = 0; i <= 100; i += 1)  {
sum += Math.pow(i, 2);
} 
return sum;
} () ) 
// 25164150

kudaisi:
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

Borwe:


What does this question mean LOL

def pythagoras(limit):
  for i in range(1, limit):
    for j in range(1, limit):
      for k in range(1, limit):

        if i**2 + j** 2 == k**2:
          if i + j + k == 1000:
            print i, j, k
            print i * j * k
            return True

pythagoras(500)

kudaisi:
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

kudaisi:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

(see image for inserts)
For example, (see image for inserts)

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

andreT:
python

def pythagoras(limit):
  for i in range(1, limit):
    for j in range(1, limit):
      for k in range(1, limit):

        if i**2 + j** 2 == k**2:
          if i + j + k == 1000:
            print i, j, k
            print i * j * k
            return True

pythagoras(500)

kudaisi:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

(see image for inserts)
For example, (see image for inserts)

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

/**
 *
 * @author olyjosh
 */
public class PythagoreanTriplet {
    
    public static void main(String[] args) {
        int lim=1000/3;
        for (int i = 2; i <=lim; i++) {
            for (int j = i+1; j <= lim; j++) {
                int c = j*j+i*i,
                        b=j*j-i*i,
                        a=2*i*j;
                if(1000==a+b+c)System.out.println("product: "+(a*b*c));
            }
        }
    }
   
}

/**
 *
 * @author olyjosh
 */
public class PythagoreanTriplet {
    
    public static void main(String[] args) {
        int lim=1000/3;
        for (int i = 2; i <=lim; i++) {
            for (int j = i+1; j <= lim; j++) {
                int c = j*j+i*i,
                        b=j*j-i*i,
                        a=2*i*j;
                if(1000==a+b+c)System.out.println("product: "+(a*b*c));
            }
        }
    }
   
}