Expert: Joseph Moore - 8/31/2009

Question
Write a program that examines all the numbers from 1 to 999, displaying all those for which the sum of cubes of the digits equals the number itself. For example, 153, 371.

1^3+5^3+3^3=153
3^3+7^3+1^3=371 ans so on....

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Sir,
 I am not getting the solution of this question, please make the correction so that I get the answer. Given below the wrong solution :



#include<stdio.h>
main()
{
     int i,rem,z;
     long sum=0;
     printf("\nEnter the mximum number :");
     scanf("%d",&i);
     
        while(i>0)
        {
           rem=i%10;
           sum=sum*10+rem;
           i=i/10;
           z=sum*sum*sum;
           printf("Answer: %d",z);
        }
getch();
}

Answer
Hi, Abhimanyu.

You clearly understand some of the basic concepts needed to solve this, such as the / and % operators, as evidenced by your attempted solution.  There is a bit more to it than what you have, so I've gone ahead and written out a program that will do this for you.  I've tried to explain everything in great detail in the comments in the code, so please, compile and run the code I give you, read over the comments, and if you have any questions at all about why it works or how it's doing what it does, do not hesitate to ask.

#include <stdio.h>

void main()
{
   // Storage for the maximum number and the loop counter.
   int maxNum = 0;
   int i;

   // Request the maximum number from the user
   printf("Enter the maximum number: ");
   scanf("%d",    &maxNum);

   // Loop until we've exhausted all of the numbers
   for (i = 1; i <= maxNum; ++i)
   {
       // The curDigit variable stores the current digit we're processing (IE, in the number 93,
       // there are two digits, 9 and 3.  curDigit will hold the digit we're concerned with at
       // this moment.  The curDigit variable is initialized to hold the 10's digit, so in 93,
       // it's initialized to 3.
       int curDigit = i % 10;

       // The sum variable stores the sum of the cubes.  It is initialized to hold the cube of
       // the first digit.  So in 93, sum is initialized to 27 (3 * 3 * 3).
       int sum = curDigit * curDigit * curDigit;

       // The curTest variable stores the current digit test.  This variable is used to determine
       // if we need to continue testing the number (IE, have we run out of digits to test?) and
       // it's used to determine what the current digit is.  See below.  It's initialized to 10,
       // because the digit in the 10's place has already been included with the two initializers
       // above.
       int curTest = 10;

       // Continue testing while there are more digits to test.
       while (i / curTest)
       {
           // Get the next digit.  We must account for all of the digits we've already processed,
           // which is what the "/ curTest" section does.  The "% 10" simply grabs the digit we
           // are interested in.
           curDigit = (i / curTest) % 10;

           // Now add the current digit's cube to the sum.
           sum += curDigit * curDigit * curDigit;

           // We need to move to the next digit, which, in a base-10 numbering system like ours,
           // means multiplying by 10.
           curTest = curTest * 10;
       }

       // If the sum of the cubes of the digits is equal to the number itself...
       if (sum == i)
           // ... then print the number to screen.
           printf("Sum of Digit Cubes == Number: %d\n", i);
   }
}