Code with prime numbers - primes

Question

I am trying to solve one problem from on-line judging system. I have a solution which works, but not efficient enough. Here is the problem:
Which the least number n can we imagine in product n = a∙b like k ways? Products a∙b and b∙a is one of the way, where all numbers is natural (1≤ k ≤50).
Input One number k.
Output One number n.
My code did not pass four tests. It is too slow for k=31, 37, 47. I have been thinking on this problem 2 days,but no improvement. Here is my code, please share, if you have any ideas.
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int prime[10000];
long x,j,i,flag,k,length,p,checker,count,number;
int main()
{
prime[0]=2;
scanf("%ld",&k);
//I find prime numbers between 1 and 1000. 1000 can be changed, just for testing
for (i=3;i<=1000;i=i+2)
{
flag=0;
for (j=2;j<=sqrt(i);j++)
{
if(i%j==0)
{
flag=1;
break;
}
}
if(flag==0)
{
x++;
prime[x]=i;
}
}
length=x;
//this loop is too big I know, again for testing. I suspect, there must be a way to make some changes to this for loop
for (i=1;i<10000000000;i++)
{
number=i;
p=1;
for(x=0;x<=length;x++)
{
if(prime[x]>sqrt(i))
break;
count=0;
while(number%prime[x]==0)
{
number=number/prime[x];
count++;
}
p=p*(count+1);
//I find prime factors of numbers and their powers, then calculate number of divisors
}
//printf("%d\n",p);
//number of ways is just number of divisors/2 or floor (divisors/2)+1
if(p%2==0)
checker=p/2;
else
checker=floor(p/2)+1;
if(checker==k)
{
printf("%ld\n",i);
break;
}
}
return 0;
}

Answer 1

If I understand the problem correctly it's asking you which is the least number n with exactly 2k divisors (should I consider 1 and n?)
in fact if a number has a divisor a, then n / a = b is an integer and n = a* b (counting only one time a and b, so you should divide by two the number of divisors)
edit
Doing that is time consuming indeed. So this is the idea;
for a number n in the form n = p1^(a1)*p2^(a2)...pn^(an) (this is the prime factorization of the number) the number of divisor is (a1 + 1)(a2+1)...(an+1)
Hence, if you want to find a number that has k divisor, factorize k. then assign the biggest factor to the smallest prime; eg if k = 2*5*7, then n should be 2^7*3^5*5^2
I know it is not since i didnt take into account that (a, b) is equal to (b, a) but play around it a little and it should work
example
take k = 37. Then double the number - (to consider the symmetry). You get 74.
Now, if you can imagine n as n = n * 1, then you just need to factor 74 (that is 2 * 37);
then give 36 to 2 and 1 to 3, leading n = 2^(36)*3 = 206158430208
if you can't, then you need to add 1 to the number you got previously (in this case, 74 + 1 = 75 = 25*3); this way you get n = 2^24 * 3^2 = 150994944
If it's none of the above, then I am probably wrong...