number program

/*
Definition from Aniruddha Guddi:
It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.
Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.
 */


/**************************
 *   author: umang bhatt  *
 * bhatt.umang7@gmail.com *
 **************************/

#include<stdio.h>

#define TRUE 0
#define FALSE 1 


int contains_same_digits(int no1, int no2)
{
    int arr1[10] = {0};

    int arr1_top = -1;
    int arr2[10] = {0};

    int arr2_top = -1;
    int temp = no1;

    int i, j, k;
    int returnval;

    int flag = FALSE;

    //store numbers in array
    while (temp > 0)
    {

        arr1[++arr1_top] = temp % 10;
        temp = temp / 10;
    }

    temp = no2;
    while (temp > 0)
    {

        arr2[++arr2_top] = temp % 10;
        temp = temp / 10;
    }





    for (i = 0; i <= arr1_top; i++)
    {

        flag = FALSE;
        for (j = 0; j <= arr2_top; j++)
        {

            if (arr1[i] == arr2[j])
            {
                flag = TRUE;

                // remove that element from both arrays
                for (k = i; k < arr1_top; k++)
                {

                    arr1[k] = arr1[k + 1];
                }
                arr1_top--;

                for (k = j; k < arr2_top; k++)
                {

                    arr2[k] = arr2[k + 1];
                }
                arr2_top--;


                if (j != 0)
                    j--;
            }
        }

        if (flag == TRUE)
        {
            i = -1;
        }
    }

    if (arr1_top == -1 && arr2_top == -1)
    {
        returnval = TRUE;
    }

    else
    {
        returnval = FALSE;
    }
    return returnval;
}

int main()
{
    char c[5];
    int i, j;

    int flag;
    int flag1;

    for (j = 2; j <= 6; j++)
    {

        i = 1;
        flag = FALSE;
        while (flag == FALSE)
        {

            flag = contains_same_digits(i, (i * j));
            if (flag == TRUE)
            {

                printf("\n %d * %d = %d where digits of %d and %d are same  ", i, j, (i * j), i, (i * j));

                flag = TRUE;
            }
            i++;
        }
    }
    printf("\n");

    return 0;
}

C program : largest palindrome made by multiplication of two 3 digit numbers

/*
Definition from Aniruddha Guddi:
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.
 */


/**************************
 *   author: umang bhatt  *
 * bhatt.umang7@gmail.com *
 **************************/




#include<stdio.h>

#define TRUE 0
#define FALSE 1 

int is_palindrome(int no)
{
    int temp = no;

    int ans = 0;
    int r;
    while (temp > 0)
    {

        ans = ((ans * 10) + (temp % 10));

        temp = temp / 10;
    }
    if (ans == no)
    {

        r = TRUE;
    }
    else
    {
        r = FALSE;
    }

    return r;
}

int main()
{
    int f_no = -1;

    int l_no = -1;
    int last_palindrom = -1;

    int flag = FALSE;
    int i, j;

    for (i = 100; i < 999; i++)
    {

        for (j = 100; j < 999; j++)
        {

            if (is_palindrome(i * j) == TRUE)
            {
                if ((i * j) > last_palindrom)
                {

                    last_palindrom = i * j;
                    f_no = i;

                    l_no = j;
                }
            }
        }
    }
    printf("%d * %d = %d \n", f_no, l_no, last_palindrom);

    return 0;
}