Three digit number problem

Posted on Apr 17, 2020

Ask the user to enter a three digit number and check if the sum of 'number and number obtained by reversing the digits’ is a three digit number or not. Also check if any of the digits in the result matches the digits in the original number.

Let us understand the problem statement with the help of an example:
Let num = 452,
So, sum of 'number and number obtained by reversing the digits’ = 452 + 254 = 706
As you can see, sum is a three digit number and there is no common digit between the original number and sum.

Let us consider another example,
Let, num = 428
So, sum of 'number and number obtained by reversing the digits’ = 428 + 824 = 1252
As you can see, sum is a four digit number and there is a common digit '2' between the original number and sum.

Program 

#include<stdio.h>
#include<conio.h>
void main()
{
	int number, o, t, h, sum, revnum, flag = 0;

    printf("\nEnter a three digit number\t:");
    scanf("%d",&number);

    o = number % 10;
    h = number / 100;
    t = (number / 10) % 10;

    revnum = (o * 100 + t * 10 + h);

    printf("\nThe reverse number is\t:%d",revnum);

    sum = number + revnum;

    printf("\nThe sum of the number and its reverse is\t:%d",sum);
    
    if(sum <= 999)
    {
		printf("\nThe sum is a three digit number");
		//ones place
		if(((sum % 10) == o) || ((sum % 10) == h) || ((sum % 10) == t))
		{
		    flag = 1;
		}
		//hundreds place
		else if(((sum / 100) == o) || ((sum / 100) == h) || ((sum / 100) == t))
		{
		    flag = 1;
		}
		//tens place
		else if((((sum / 10) % 10) == o) || (((sum / 10) % 10) == h) ||  (((sum / 10) % 10) == t))
		{
		    flag = 1;
		}
    }
    if(sum > 999)
    {
		printf("\nThe sum is a four digit number");
		//ones place
	    if(((sum % 10) == o) || ((sum % 10) == h) || ((sum % 10) == t))
		{
		    flag = 1;
		}
		//thousand digit
		else if(((sum / 1000) == o) || ((sum / 1000) == h) || ((sum / 1000) == t))
		{
		    flag = 1;
		}
		//tens place
		else if((((sum / 10) % 10) == o) || (((sum / 10) % 10) == h) ||  (((sum / 10) % 10) == t))
		{
		    flag = 1;
		}
		//hundreds digit
		else if((((sum / 100) % 10) == o) || (((sum / 100) % 10) == h) ||  (((sum / 100) % 10) == t ))
		{
		    flag = 1;
		}
    }

    if(flag == 1)
    {
		printf("\nDigits in the result matches the digits in the original number");
    }
    else
    {
		printf("\nDigits in the result does not matches the digits in the original number");
    }
    
    getch();
}

Output 

********** Run1 ********** 

Enter a three digit number      :452

The reverse number is   :254
The sum of the number and its reverse is        :706
The sum is a three digit number
Digits in the result does not matches the digits in the original number


********** Run2 ********** 

Enter a three digit number      :143

The reverse number is   :341
The sum of the number and its reverse is        :484
The sum is a three digit number
Digits in the result matches the digits in the original number


********** Run3 ********** 

Enter a three digit number      :458

The reverse number is   :854
The sum of the number and its reverse is        :1312
The sum is a four digit number
Digits in the result does not matches the digits in the original number


********** Run4 ********** 

Enter a three digit number      :428

The reverse number is   :824
The sum of the number and its reverse is        :1252
The sum is a four digit number
Digits in the result matches the digits in the original number

Explanation

A number is taken as an input from the user and stored in the variable named say 'number'.

First step is to extract the individual digits of the number (ones, tens, and hundreds) and store in the variable named 'o', 't' and 'h' respectively. This is done using the expressions:
o = number % 10;
h = number / 100;
t = (number / 10) % 10;

To calculate the reverse of the number, following expression is used.
revnum = (o * 100 + t * 10 + h);

Next step is to calculate the sum of the number and the reversed number.
sum = number + revnum;

Now we have to check whether the 'sum' is of 3 digits or 4 digits. This is done by comparing it with number '999'.
If sum <= 999, then it is of 3 digits.
If sum > 999, then sum is of 4 digits.

Now, we have to check whether any of the digits in 'sum' and the actual number matches or not. For this, we have to extract the individual digits of 'sum' and compare them with the digits of actual number ('o', 't', 'h') extracted previously. If 'sum' is 3 digit number, then we will calculate the ones, tens and hundreds place of the 'sum'. And if the 'sum' is of 4 digit, then we will calculate ones, tens, hundreds, and thousands place of the 'sum'. This is done as follows:

If sum <= 999 (3 digit sum) then,
for ones place,
if(((sum % 10) == o) || ((sum % 10) == h) || ((sum % 10) == t))
(sum % 10) gives the ones place of 'sum'

for hundreds place,
else if(((sum / 100) == o) || ((sum / 100) == h) || ((sum / 100) == t))
(sum / 100) gives the hundreds place of 'sum'

for tens place,
else if((((sum / 10) % 10) == o) || (((sum / 10) % 10) == h) || (((sum / 10) % 10) == t))
((sum / 10) % 10) gives the tens place of 'sum'


If sum > 999 (4 digit sum) then,
for ones place
if(((sum % 10) == o) || ((sum % 10) == h) || ((sum % 10) == t))
(sum % 10) gives the ones place of 'sum'

for thousand digit
else if(((sum / 1000) == o) || ((sum / 1000) == h) || ((sum / 1000) == t))
(sum / 1000) gives the thousands place of 'sum'

for tens place
else if((((sum / 10) % 10) == o) || (((sum / 10) % 10) == h) || (((sum / 10) % 10) == t))
((sum / 10) % 10) gives the tens place of 'sum'

for hundreds digit
else if((((sum / 100) % 10) == o) || (((sum / 100) % 10) == h) || (((sum / 100) % 10) == t ))
((sum / 100) % 10) gives the hundreds place of 'sum'


Now, how to keep track whether any of the digit is matched or not. For this a variable named 'flag' is used. Initially 'flag' is set to 0, if any of the digit matches, then this 'flag' is set to 1.

After all the matching is done, value of 'flag' is checked.
If value of 'flag' is 0, then none of the digits matched.
If value of 'flag' comes to be 1, then one or some of the digits are matched.