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Find all the missing numbers in the below conventional division problem

+1 vote
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posted Nov 13, 2014 by Bhavish D A

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1 Solution

+1 vote
 
Best answer

One case is 49 dividing 59241 remainder is 000 and quotient is 1209

See the following program

#include <stdio.h>

main()
{
 int i, j, k, l, m, number;

 for (i=1; i<10; i++)
   for (j=0; j<10; j++)
   {
    for (k=1; k<10; k++)
     for (l=1; l<10; l++)
       for (m=1; m<10; m++)
       {
          number = (10*i + j) * (1000*k+100*l+0+m);
          if (number <100000)
          {
            if ((number%1000 >199) && (number%1000 <300))
            {  
               if (((10*i + j)*(10*k + l)) != (number/100))
               {
                 if (((number/1000) - ((10*i + j)*k)) > 9) 
                 {
                   if (((10*i + j) * l) < 100)
                   printf ("%d %d * %d %d 0 %d = %d\n", i, j, k, l, m, number);
                 }
               }
            }
          }
       }

   }
}

Catch is the middle digit should be 2, second digit in the quotient is zero and one more point...

But very good puzzle keep it up...

solution Nov 13, 2014 by Salil Agrawal
But your answer is not proper for the given problem.
In the second stage, i have given only 2 spaces means it will accept 2 digits.
In your case it is accepting 3 digits.
Got it let me modify the program and that should narrow down further.
Updated it see the original solution with program

Answer is
19 dividing 59241 remainder is 000 and quotient is 1209
Sir, this answer is totally wrong for my scenario...
How can we take 19*1=19, we can divide that by 19*3=57 for 59.
U have done a typo 19 should be 49 as output of the program is 49
Oh sorry, my bad corrected the answer. Thanks for pointing out :)



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