abc_bca_cab_are_in_gp.c

/*

    abc_bca_cab_are_in_gp.c

    Task from Subir Singha:
    https://www.facebook.com/groups/1941400139499670/user/100002028618086/
    https://www.facebook.com/groups/1941400139499670/permalink/2392741967698816/

    'abc' is a 3-digit number such that
    'abc', 'bca' &'cab' are in G.P. Find all such numbers.

    G.P. is geometric progression.
    https://en.wikipedia.org/wiki/Geometric_progression

    x    y     z

    x   r*x (r^2)*x

    y/x = r and z/y = r

    bca/abc = r and cab/bca = r

    Condition is:
    bca/abc = cab/bca = r

    Since 'abc', 'bca' and 'cab' are 3-digit numbers, a, b an c must not be 0.


    You can find all my C programs at Dragan Milicev's pastebin:

    https://pastebin.com/u/dmilicev

*/

#include <stdio.h>

int main(void)
{
    int a, b, c, abc, bca, cab, solved=0, number_of_solutions=0;

    printf("\n  abc  bca  cab \n");

    for(a=1; a<10; a++)
        for(b=1; b<10; b++)
            for(c=1; c<10; c++)
            {
                abc = a*100 + b*10 + c;
                bca = b*100 + c*10 + a;
                cab = c*100 + a*10 + b;

                if( (float)bca/abc == (float)cab/bca )
                {
                    solved = 1;
                    number_of_solutions++;
                    printf("\n %4d %4d %4d , r = %d%d%d/%d%d%d = %.2f , r = %d%d%d/%d%d%d = %.2f \n",
                        abc, bca, cab, b,c,a, a,b,c, (float)bca/abc, c,a,b, b,c,a, (float)cab/bca );
                }
            }

    if( !solved )
        printf("\n There is no such digits. \n");
    else
        printf("\n There are %d such numbers. \n",number_of_solutions);


    return 0;

} // main()