Thursday, 4 August 2011

Finding the count of factors of a number

Question

This is more of a mathematical and a GMAT/CAT question rather than a programming one. But found this question very interesting and found it from Project Euler.

Before seeing the solution, please try out yourselves.

Let me reiterate the question,

“Find a number which has more than 500 factors. The number is part of a triangle series or natural series of additions”

Triangular number is given a name since it can fill every point in a triangle and will add up like this,

1+2+3+4…+n = n(n+1)/2

The above is sometimes called arithmetic series sum as well.

If you need some mathematical background about finding factors, see the below video,

I really loved the second brute force method given in the above video since its simple and easier to visualize and program.

I got into lots of problem since I didn’t make up the triangular number in my mind after I got that clear, zoom I got the answer. Number of trails would be at least 20 times. :)

First tried manually using calculators. But it didn’t work out since I didn’t see that I need to calculate only for triangular number.

Do the second method of factorizing start from 1 and go up by triangular series and count the number of factors.

The Method is,

1.  find sqrt(number)

2. If the sqrt is a perfect square, then set diff to –1

3. run a loop from 1 to sqrt(number) and count how many numbers are dividing the triangular number selected

4. count*2+diff gives the total factors.

Code

#include <math.h>

void main(int argc, char* argv[])
{
int data=0; //pow(35.00,13.00);
int next=1;
int max = 0;

while(1) {
int count=0;
int diff = 0;
data = data+next;
int mid = (int) floor(sqrt((double)data));
if(mid * mid == data)
diff = -1;

for(int j=1;j<=mid;j++) {
if(data % j == 0) {
++count;
}
}
count = count*2+diff;
if(count > max) {
max = count;
cout<<data<<"="<<count<<endl;
if(max > 499)
break;
}
next++;
}
}