Thursday, 4 August 2011

Finding the count of factors of a number



  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,

Methods to find factors


   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.



#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*2+diff;
    if(count > max) {
        max = count;
        if(max > 499)