Member Avatar for Hiroshe

Can anyone help me with doing problem 12 on project euler? The question is asking which triangle number has more than 500 factors. The first few triangle numbers are: 1, 3, 6, 10, 15, 21, 28 ... Hope you understand the pattern, cause a can't explain it in words vary well.

Anyways I written the code in C, and it's just too slow. Here's an english version I wrote for easyness:

Generate the triangle number under (500^2)

loop until the triangle number's factors is more than 500
  generate next triangle number

return triangle number

to find how many factors a number has:
  divide number by prime number going from lowest to higher until it divides evenly
  if the result is prime, tally the prime number and the result
  if the result is not prime, tally the prime number and the result becomes the number and loop back to the "divide number by pr..." stage
  add 1 and multiply everything in the tally that is not zero
  the product is the number of factors

to generate prime numbers:
  start with 2 and 3
  int n = 1, prime = 0
  while prime is not prime
    increase n if called on an odd time
    loop back with prime = 6n-1 on every odd time and
    loop back with prime = 6n+1 on every even time
  return prime

and finally to check if a number is prime:
  if number is under 2, its not prime
  //this is a prime sieve
  if number is 2, its prime
  if number is divisible by 2, its not prime
  if number is 3, its prime
  if number is divisible by 3, its not prime
  ...
  int count = what ever last sieve was
  while count < sqrt number
    if number mod count = 0
      return 0 (not prime)
  return 1 (is prime)

I'm going to attach the c version becouse its 341 lines long!

I think the time is being waisted mostly at the isprime step, or the countdiv (factoring) step.

Any suggestions to speed this puppy up? Thanks alot.

/* Euler problem 12:
   The sequence of triangle numbers is generated by adding the natural numbers.
   So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first
   ten terms would be:

   1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

   Let us list the factors of the first seven triangle numbers:

     1: 1
     3: 1,3
     6: 1,2,3,6
    10: 1,2,5,10
    15: 1,3,5,15
    21: 1,3,7,21
    28: 1,2,4,7,14,28

   We can see that 28 is the first triangle number to have over five divisors.
   What is the value of the first triangle number to have over five hundred
   divisors?

   When I say first and second dimention of an array, I meen which 1D dimention
   I am calling. array[x][1] would be the first, array[x][2] would be the
   second. */

#include <stdio.h>
#include <math.h>

int trianglediv(int nofdiv);
int triunder(int max, int *count);
int countdiv(int number);
void findprimefactor(int number, int output[][2]);
void fillprime(int numofpfactor, int output[][2]);
int nextprime(int *n, int *c);
int isprime(int number);
void tally(int number, int output[][2]);
void init(int size, int array[][2]);

int main() {
  printf("The first triangle number to have over 500 divisors is: %d\n",
         trianglediv(500));

  return 0;
}

int trianglediv(int nofdiv) {
/* This function returns first triangle number to have over 'nofdiv'  divisors.
   It works by generating triangle numbers until it finds one with over
   'nofdiv'. We can start generating the triangle numbers at the triangle number
   under 'nofdiv'^2. */

  int count = 0;
  int triangle = triunder(pow(nofdiv, 2), &count); /* Fast forward to a number
                                                     which can possible hold
                                                     nofdiv factors. */

  while(countdiv(triangle) <= nofdiv) { /* while there arnt enough divisors */
    triangle += count;                  /* Get next triangle number */
    count++;
  }

  return triangle;
}

int triunder(int max, int *count) {
/* This function returns the triangle number under 'max'. It also sets 'count'
   to where it needs to be to generate futher triangle numbers. */

  int nexttri = 0;  /* Hold's next possible triangle number */
  int curtri = 0;   /* Holds current triangle number, this is returned if nextri
                       fails its test. */

  while(nexttri <= max) {
    curtri = nexttri;     /* nexttri is under max, so its safe to save */
    nexttri += *count;    /* Generate the next triangle number */
    (*count)++;
  }
  (*count)--;

  return curtri;
}

int countdiv(int number) {
/* This function returns the number of divisors. It finds the prime divisors,
   adds 1 to each, and finds the product of all of them. It returns the product.
   */

  int count = 0;
  int product = 1;
  int primefactor[number][2];       /* Make a variable length array. The
                                         second dimention holds prime numbers,
                                         and the fist dimention holds how many
                                         of that prime number there is. */

  init(number, primefactor);          /* Replace everything in the array with
                                         0's */
  fillprime(number, primefactor);        /* Fill the second dimention. */
  findprimefactor(number, primefactor);  /* Fill the first dimention */

  while(count < number) {
    if(primefactor[count][0] != 0)  /* If there is a number of that prime */
      product *= (primefactor[count][0]+1); /* add 1 and multiply */
    count++;
  }
  return product;
}

void findprimefactor(int number, int output[][2]) {
/* This function will find all of the prime factors. It works by dividing number
   by primes until it evenly divides. That prime is a prime factor, and if the
   result is also prime, its is a prime factor. If the result is not a prime
   factor, it becomes the new 'number'. */
   
  int n = 0, c = 0;
  int p = nextprime(&n, &c);
   
  while(1) {
    if(number % p == 0) { /* number evenly divides into p */
      if(isprime(number / p)) { /* if result is prime */
        tally(p, output);         /* Add that 'p' to its place on first dim */
        tally((number / p), output); /* Add result to its place on first dim */
        return;
      } 
      else if(number / p == 1) { /* if result is 1 */ 
        tally(number, output); /* Add that 'number' to its place on first dim */
        return;
      }
      else {          /* if result is not prime */
        tally(p, output); /* Tally the prime number */
        number /= p;    /* The result is the new number */
        n = 0;
        c = 0;
      }
    }
  p = nextprime(&n, &c);
  }
}

void fillprime(int numofpfactor, int output[][2]) {
/* This function fills the second dimention of an array with primes. It uses the
   modified version of problem 2 attemp 2's 'nextprime' and 'isprime' used in
   problem 7. Why do all these's questions rely on prime numbers? */
   
  int count = 0;
  int c = 0, n = 0;
   
  while(count < numofpfactor) {
    output[count][1] = nextprime(&n, &c);
    count++;
  }
}

int nextprime(int *n, int *c) {
/* This function will return the next prime number. Simply set n and c at 0 and
   call the function whenever you need the next prime number. This is bassed off
   of the one in problem 3, except will work for more than 12 calls. I thought
   that 6n-1 and 6n+1 would work for all n's, but aparently not. So I just put
   an isprime() to trap it. */

  int prime = 0;
  
  while(!isprime(prime)) {
    if(*n == 0 && *c == 0) {  /* First call, return 2 */
      (*c)++;
       prime = 2;
    }
    else if(*n == 0 && *c == 1) {  /* Second call, return 3 */
      (*c)--;
      (*n)++;
      prime = 3;
    }
    else if(*c == 0) {   /* Even call, return 6*n-1 */
      (*c)++;
      prime = (6*(*n))-1;
    }
    else if(*c == 1) {   /* Odd call, return 6*n+1 */
      (*c)--;
      prime = (6*((*n)++))+1;
    }
  }
  return prime;
}

int isprime(int number) {
/* This function will return 1 if number is prime. It works by dividing 'number'
   by every possible number it could divide evenly into.  If it dosent divide
   into anything upto its square root, the number wont divide into anything.
   Also dont check even numbers. If its divisible by 2, it would be divisible by
   all even numbers. In problem 12 I added a sieve. Basicly, if the number is
   divisible by a prime number, its not prime. */

  int max = (int)sqrt(number) + 2;
  int count = 23; /* Start at 23 so we can increase by 2 for odd numbers. */

  if (number <= 1)  /* if the number is 1 or under, its not prime */
    return 0;
  if (number == 2)  /* if the number is 2, its prime */
    return 1;
  if (number % 2 == 0) /* if the number is divisible by 2 its not prime. */
    return 0;
  /* Extra sieve starts here */
  if(number % 3 == 0) {
    if(number == 3)
      return 1;
    return 0;
  }
  if(number % 5 == 0) {
    if(number == 5)
      return 1;
    return 0;
  }
  if(number % 7 == 0) {
     if(number == 7)
       return 1;
     return 0;
   }
   if(number % 11 == 0) {
     if(number == 11)
       return 1;
    return 0;
  }
  if(number % 13 == 0) {
    if(number == 13)
      return 1;
    return 0;
  }
  if(number % 17 == 0) {
    if(number == 17)
      return 1;
    return 0;
  }
  if(number % 19 == 0) {
    if(number == 19)
      return 1;
    return 0;
  }
  if(number % 23 == 0) {
    if(number == 23)
      return 1;
    return 0;
  }
  if(number % 29 == 0) {
    if(number == 29)
      return 1;
    return 0;
  }
  if(number % 31 == 0) {
    if(number == 31)
      return 1;
    return 0;
   }
   if(number % 37 == 0) {
     if(number == 37)
       return 1;
     return 0;
   }
  if(number % 41 == 0) {
    if(number == 41)
      return 1;
    return 0;
 }
  if(number % 43 == 0) {
    if(number == 43)
      return 1;
    return 0;
  }
  if(number % 47 == 0) {
    if(number == 47)
      return 1;
    return 0;
  }
  if(number % 53 == 0) {
    if(number == 53)
      return 1;
    return 0;
  }
  if(number % 59 == 0) {
    if(number == 59)
      return 1;
    return 0;
  }
  if(number % 61 == 0) {
    if(number == 61)
      return 1;
    return 0;
  }
  if(number % 67 == 0) {
    if(number == 67)
      return 1;
    return 0;
  }
  if(number % 71 == 0) {
    if(number == 71)
      return 1;
    return 0;
  }
  if(number % 73 == 0) {
    if(number == 73)
      return 1;
    return 0;
  }
  if(number % 79 == 0) {
    if(number == 79)
      return 1;
    return 0;
  }

  while (count < max) {
    if (number % count == 0) /* % means modulo, basicly just the remainder. */
      return 0;
  count += 2; /* Count = next odd number */
  }
  return 1;
}

void tally(int number, int output[][2]) {
/* This function will find a number on the second dimention, and tally (add 1)
   to its equivalent place on the first dimention. There is no security. */
   
  int count = 0;

  while(1) {
    if(output[count][1] == number) {
      output[count][0]++;
      return;
    }
    count++;
  }
}

void init(int size, int array[][2]) {
/* This function replaces everthing in an [x][2] array with 0's */

  int count = 0;
  
  while(count != size) {
    array[count][0] = 0;
    array[count][1] = 0;
    count++;
  }
}
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  • Latest Post Latest Post by SVR
Member Avatar for Hiroshe

Hmm. Just had an idea (about 5 minutes after posting the above). Maybe the thing that slowing it down is the part where it reads the array. Look at function countdiv() I guess I could return the highest prime factor so it knows not to loop past it. Any other idea's?

Member Avatar for Hiroshe

Still room for inprovement, but I got it working a reasonalble speed.. I think I kinda solved this thread on my own, but if there are any other suggestion's, I'm still interested in making my code faster.

Member Avatar for BestJewSinceJC

I actually don't understand the pattern. Enlighten me.

Member Avatar for Hiroshe

I actually don't understand the pattern. Enlighten me.

Err.. try understanding this (not tested):

int count = 1;
int triangle = 1;
while(1) {
  printf("%d\n", triangle);
  count++;
  triangle += count;
}

You add 1 more than you did the last time starting a 1, 1+2, 3+3, 6+4... See I really can't explain it good:$

Member Avatar for O71v13r

Perhaps this explanation can help you out BestJewSinceJC:

The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Member Avatar for SVR

Still room for inprovement, but I got it working a reasonalble speed.. I think I kinda solved this thread on my own, but if there are any other suggestion's, I'm still interested in making my code faster.

The first improvement I found was that you don't need to regenerate all the primes every time. Split the array into separate arrays. One global array of primes and one for factor counts.

This improved from ??? to 10 mins or so.
I don't know how long it originally took, I killed it after a while.

To make it work change the fillprimes function to just fill the next x primes ...

void fillprime( int maxPrime ) {
/* This function fills the second dimention of an array with primes. It uses the
modified version of problem 2 attemp 2's 'nextprime' and 'isprime' used in
   problem 7. Why do all these's questions rely on prime numbers? */
   
   static int c = 0, n = 0;
   
   do {
      primes[primeCount] = nextprime(&n, &c);
   }while(primes[primeCount++] <= maxPrime);
   
}

You see I changed the test to maxPrime and test the prime value instead of the count.
Now everywhere you loop over the array(s) you only loop until maxPrime is reached.

Next is findprimefactor. You call tally which iterated through the entire array again. Now as you are iterating over the prime array you have the index of the factor to increment.

void findprimefactor(int number, int *output) {
/* This function will find all of the prime factors. It works by dividing number
by primes until it evenly divides. That prime is a prime factor, and if the
result is also prime, its is a prime factor. If the result is not a prime
   factor, it becomes the new 'number'. */
   
   int n    = 0;
   int p    = 0;
   int n1   = 0;
   
   while(n<primeCount)
   {
      p = primes[n];
      if(number == p) {             /* if result is 1 */ 
         output[n]++;               /* Add that 'number' to its place on first dim */
         return;
      }

      if( (number % p) ==0) {       /* number evenly divides into p  */
         output[n]++;               /* Add that 'p' to its place on first dim */
         number /= p;
         n1 = findPrime(number);

         if(n1 != -1) {             /* if result is prime */
            output[n1]++;           /* Add result to its place on first dim */
            return;
         } 
         /* if result is not prime */
         /* The result is the new number */
         n = 0;
      }
      else
         n++;
   }

   // should not get here except on 1
   if(number>0)
   {
      printf("\nRemainder - %d",number);
   }
   return;
}

When it checks the result of the division it could just continue as if it isn't prime. That would be the same as the search done in isprime. But a quick binary search across the existing primes works even better.
The binary search code came outta my head so it probably has holes. But it worked for the few values I tested.

Now the time was reduced to around 5 mins.

I tweaked a few standard things without much speed up.

Then the biggie ... to get the factors to come out right you have to gen primes all the way up to 'number'. But for most things you only need primes up to sqrt(number) +1.

The only time you really need all is if number is prime. Hmmm... all primes only have to factors. No need to factor or count primes at all!

int countdiv( int number) {
/* This function returns the number of divisors. It finds the prime divisors,
adds 1 to each, and finds the product of all of them. It returns the product.
*/
   
   int count = 0;
   int product = 1;
   int maxPrime;
   

   if(isprime(number) )
   {
      return 2;
   }

   maxPrime= (int) sqrt(number)+1;
   
   fillprime(maxPrime);                     /* generate next x primes */
   
   init(primeCount, primefactor);          /* clear count array array with 0's */
   findprimefactor(number, primefactor);   /* count each prime factor */
   
   
   while(primes[count] <= maxPrime) {
      if(primefactor[count] != 0)           /* If there is a number of that prime */
         product *= (primefactor[count]+1);  /* add 1 and multiply */
      count++;
   }
   
   return product;
}

Boom! It takes less than a second to calculate the first 500 term triangle.

I've attached the changed file.

/* Euler problem 12:
The sequence of triangle numbers is generated by adding the natural numbers.
So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first
ten terms would be:

  1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
  
    Let us list the factors of the first seven triangle numbers:
    
      1:  1
      3:  1,3
      6:  1,2,3,6
      10: 1,2,5,10
      15: 1,3,5,15
      21: 1,3,7,21
      28: 1,2,4,7,14,28
      
  We can see that 28 is the first triangle number to have over five divisors.
  What is the value of the first triangle number to have over five hundred
  divisors?
        
*/

#include <stdio.h>
#include <math.h>
#include <malloc.h>
#include <time.h>


/* global */

#define MAX_PRIMES 76576500 //20000 //11765 76576500

int *primes       = 0;
int *primefactor  = 0;

int primeCount    = 0;


/* defs */

int trianglediv(int nofdiv);
int triunder(int max, int *count);
int countdiv(int number);
void findprimefactor(int number, int *output);
void fillprime(int maxPrime);
int nextprime(int *n, int *c);
int isprime(int number);
void tally(int *output, int n);
void init(int size, int *array);

/* main */
int main() {
   int cnt = 500;
   
   primes      = (int *) malloc( sizeof(int)*MAX_PRIMES);
   primefactor = (int *) malloc( sizeof(int)*MAX_PRIMES);
   
   printf("\nThe first triangle number to have at least %d divisors is: %d\n",
      cnt,trianglediv(cnt));
   
   free(primes);
   free(primefactor);
   
   
   getchar();
   
   return 0;
   
}


int trianglediv(int nofdiv) {
/* This function returns first triangle number to have over 'nofdiv'  divisors.
It works by generating triangle numbers until it finds one with over
'nofdiv'. We can start generating the triangle numbers at the triangle number
   under 'nofdiv'^2. */
   
   int hrs  = 0;
   int mins = 0;
   int secs = 0;
   int n    = 0;
   int terms = 0;
   int maxTerms = 0;
   int tics;
   int count = 1;
  
   
                     /* Fast forward to a number which can possible hold nofdiv factors. */
   int triangle = 1; //triunder((int) pow(nofdiv, 2), &count);
   count++;
   
   
   /* Make a variable length array. The
   second dimention holds prime numbers,
   and the fist dimention holds how many
   of that prime number there is. */
   
   while((terms = countdiv(triangle)) < nofdiv) { /* while there arent enough divisors */
      n++;
      if( terms >maxTerms )
      {
         maxTerms = terms;
         tics = clock()/1000;
         hrs = tics/3600;
         tics -= hrs * 3600; 
         mins = (tics/60);
         tics -= mins * 60;
         secs = tics;
         
         printf("\n%02d:%02d:%02d - %8d tested (max terms %5d for %9d) primes %8d\n",hrs,mins,secs, n, maxTerms, triangle,primeCount);
      }
      else if( (n&0xF) == 0)
      {
         printf(".");
      }

      triangle += count;                  /* Get next triangle number */
      count++;
   }
   n++;
   
   maxTerms = terms;
   tics = clock()/1000;
   hrs = tics/3600;
   tics -= hrs * 3600; 
   mins = (tics/60);
   tics -= mins * 60;
   secs = tics;
   
   printf("\nDONE!\n%02d:%02d:%02d - %8d tested (max terms %5d for %9d) primes %8d\n",hrs,mins,secs, n, maxTerms, triangle,primeCount);
   
   
   //  printf("Final %d -> %d\n", triangle, terms);
   return triangle;
}

int triunder(int max, int *count) {
/* This function returns the next triangle number after 'max'. It also sets 'count'
   to where it needs to be to generate futher triangle numbers. */
   
   int nexttri = 0;        /* Hold's next possible triangle number */
   
   do {
      nexttri += *count;   /* Generate the next triangle number */
      (*count)++;
   }while(nexttri < max);
   
   return nexttri;
}



int countdiv( int number) {
/* This function returns the number of divisors. It finds the prime divisors,
adds 1 to each, and finds the product of all of them. It returns the product.
*/
   
   int count = 0;
   int product = 1;
   int maxPrime;
   

   if(isprime(number) )
   {
      return 2;
   }

   maxPrime = (int) sqrt(number)+1;
   
   fillprime(maxPrime);                     /* generate next x primes */
   
   init(primeCount, primefactor);          /* clear count array array with 0's */
   findprimefactor(number, primefactor);   /* count each prime factor */
   
   
   while(primes[count] <= maxPrime) {
      if(primefactor[count] != 0)           /* If there is a number of that prime */
         product *= (primefactor[count]+1);  /* add 1 and multiply */
      count++;
   }
   
   return product;
}

int findPrime1(int p)
{
   int n = 0;
   while(primes[n] < p)
   {
      n++;
   }
   if(primes[n] == p)
      return n;
   return -1;
}

int findPrime(int p)
{
   
   int n = primeCount>>1;
   int d = n>>1;
   
   while(d)
   {
      if(primes[n] < p)
      {
         n = n + (d);
      }
      else if(primes[n] > p)
      {
         n = n - (d);
      }
      else
         return n;
      
      d>>= 1;
   }
   
   return -1;
   
}

void findprimefactor(int number, int *output) {
/* This function will find all of the prime factors. It works by dividing number
by primes until it evenly divides. That prime is a prime factor, and if the
result is also prime, its is a prime factor. If the result is not a prime
   factor, it becomes the new 'number'. */
   
   int n    = 0;
   int p    = 0;
   int n1   = 0;
   
   while(n<primeCount)
   {
      p = primes[n];
      if(number == p) {             /* if result is 1 */ 
         output[n]++;               /* Add that 'number' to its place on first dim */
         return;
      }

      if( (number % p) ==0) {       /* number evenly divides into p  */
         output[n]++;               /* Add that 'p' to its place on first dim */
         number /= p;
         n1 = findPrime(number);

         if(n1 != -1) {             /* if result is prime */
            output[n1]++;           /* Add result to its place on first dim */
            return;
         } 
         /* if result is not prime */
         /* The result is the new number */
         n = 0;
      }
      else
         n++;
   }

   // should not get here except on 1
   if(number>0)
   {
      printf("\nRemainder - %d",number);
   }
   return;
}

void fillprime( int maxPrime ) {
/* This function fills the second dimention of an array with primes. It uses the
modified version of problem 2 attemp 2's 'nextprime' and 'isprime' used in
   problem 7. Why do all these's questions rely on prime numbers? */
   
   static int c = 0, n = 0;
   
   do {
      primes[primeCount] = nextprime(&n, &c);
   }while(primes[primeCount++] <= maxPrime);
   
}


int nextprime(int *n, int *c) {
/* This function will return the next prime number. Simply set n and c at 0 and
call the function whenever you need the next prime number. This is bassed off
of the one in problem 3, except will work for more than 12 calls. I thought
that 6n-1 and 6n+1 would work for all n's, but aparently not. So I just put
   an isprime() to trap it. */
   
   int prime = 0;
   
   do {
      if(*n == 0 && *c == 0) {  /* First call, return 2 */
         (*c)++;
         prime = 2;
      }
      else if(*n == 0 && *c == 1) {  /* Second call, return 3 */
         (*c)--;
         (*n)++;
         prime = 3;
      }
      else if(*c == 0) {   /* Even call, return 6*n-1 */
         (*c)++;
         prime = (6*(*n))-1;
      }
      else if(*c == 1) {   /* Odd call, return 6*n+1 */
         (*c)--;
         prime = (6*((*n)++))+1;
      }
   }while(!isprime(prime));
   return prime;
}


int isprime1(int number) {
   
   int maxCnt = (int)sqrt(number) + 2;
   int count = 0;
   
   if(number < 25)
      return 1;
   
   while (primes[count] < maxCnt) {
      if (number % primes[count] == 0) /* % means modulo, basicly just the remainder. */
         return 0;
      count ++; /* Count = next prime number */
   }
   return 1;
   
}

int isprime(int number) {
/* This function will return 1 if number is prime. It works by dividing 'number'
by every possible number it could divide evenly into.  If it dosent divide
into anything upto its square root, the number wont divide into anything.
Also dont check even numbers. If its divisible by 2, it would be divisible by
all even numbers. In problem 12 I added a sieve. Basicly, if the number is
   divisible by a prime number, its not prime. */
   
   int max = (int)sqrt(number) + 2;
   int count = 79; // 23; /* Start at 23 so we can increase by 2 for odd numbers. */
   
   if (number <= 1)  /* if the number is 1 or under, its not prime */
      return 0;
   if (number == 2)  /* if the number is 2, its prime */
      return 1;
   if (number % 2 == 0) /* if the number is divisible by 2 its not prime. */
      return 0;
   /* Extra sieve starts here */
   if(number % 3 == 0) {
      if(number == 3)
         return 1;
      return 0;
   }
   if(number % 5 == 0) {
      if(number == 5)
         return 1;
      return 0;
   }
   if(number % 7 == 0) {
      if(number == 7)
         return 1;
      return 0;
   }
   if(number % 11 == 0) {
      if(number == 11)
         return 1;
      return 0;
   }
   if(number % 13 == 0) {
      if(number == 13)
         return 1;
      return 0;
   }
   if(number % 17 == 0) {
      if(number == 17)
         return 1;
      return 0;
   }
   if(number % 19 == 0) {
      if(number == 19)
         return 1;
      return 0;
   }
   if(number % 23 == 0) {
      if(number == 23)
         return 1;
      return 0;
   }
   if(number % 29 == 0) {
      if(number == 29)
         return 1;
      return 0;
   }
   if(number % 31 == 0) {
      if(number == 31)
         return 1;
      return 0;
   }
   if(number % 37 == 0) {
      if(number == 37)
         return 1;
      return 0;
   }
   if(number % 41 == 0) {
      if(number == 41)
         return 1;
      return 0;
   }
   if(number % 43 == 0) {
Edited by SVR because: n/a
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