Question

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Counting trailing zeros of numbers resulted from factorial

It is given an integer “p”. I have to find a number “n” for which “n factorial” has “p” numbers of zero at the end. Here is the solution i thought, but i am not sure if it is a solution for this problem. Do i have to make a function to calculate the factorial and another function to get the zeros?

int p;
int count5=0;
int i;
int copy_i;

printf("Enter p: ");
scanf("%d",&p);

for(i=1; ;i++)
{
    copy_i=i;

    while(copy_i/5)
    {
        if(copy_i%5==0)
        {
            count5++;
            copy_i=copy_i/5;
        }
        else
        {
            break;
        }
    }

    if(count5==p)
    {
        printf("The minimum number n is: %d.",i);
        break;
    }
    else if(count5>p) 
    {
        printf("No match for n! with %d zero.",p);
        break;
    }

}

Answer 1

This sounds like a Project Euler problem, so I won’t give an explicit solution. Here’s two hints though:

• You do not have to actually calculate the factorial to find out how many zeroes it has at the end.
• If a number is divisible by 2^N and by 5^N, the number has at least N zeroes at the end.