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# C Program to Find Sum of Squares of Digits of a Number

In this tutorial, we will learn about how to create a program in C that will find and print sum of squares of any two number and also will learn to find and print sum of squares of digits of any given number by user at run-time.

## C Find Sum of Squares of any Two Number

Let's first create a program in C that will ask from user to enter any two number to find and print sum of square of the given two numbers.

#include<stdio.h> #include<conio.h> int main() { int num1, num2, sqr1, sqr2, sum; printf("Enter any two numbers: "); scanf("%d%d", &num1, &num2); sqr1 = num1*num1; sqr2 = num2*num2; sum = sqr1 + sqr2; printf("Sum of their square = %d", sum); getch(); return 0; }

As the program was written under **Code::Blocks** IDE, therefore after successful build and run, here is the sample run:

Now supply any two number say **4** and **5** and press `ENTER`

key to see the result that will be the sum of square
of given two number. If you will square the two number say **4** and **5**, then you will get **16** and **25**. After summing
it up, you will get **16+25** equals **41** as result. Here is the second snapshot of the sample run:

Here are some of the main steps used in above program:

- Receive any two number as input from user (at run-time) in any two variables say
**num1**and**num2** - Calculate the square of both the number one by one and initialized its square to any two variables say
**sqr1**and**sqr2** - That is
**num1*num1**gets initialized to**sqr1**and**num2*num2**gets initialized to**sqr2** - Now calculate sum of two variable say
**sqr1**(square of first number) and**sqr2**(square of second number) - That is
**sqr1+sqr2**gets initialized to a variable say**sum**(holds summation value of square of the two given number) - Print the value of
**sum**at last of the program - For example, if user enters 4 and 5 as input
- Therefore,
**num1**holds 4 and**num2**holds 5 - Now
**num1*num1**or**4*4**or**16**gets initialized to**sqr1** - And
**num2*num2**or**5*5**or**25**gets initialized to**sqr2** - And
**sqr1+sqr2**or**16+25**or**41**gets initialized to**sum** - Print the value of
**sum**which is 41 as output

## C Find Sum of Squares of Digits of a Number

Now let's create a program that will ask from user to enter any number to find and print the sum of squares of its
digit. That is if user enters 342 as input, then the program will find the sum of square of its digit, that is
**(3*3) + (4*4) + (2*2)** or **9+16+4** or **29** will be the final result. Let's take a look at the program:

#include<stdio.h> #include<conio.h> int main() { int num, rem, sum=0; printf("Enter any number: "); scanf("%d", &num); while(num>0) { rem = num%10; sum = sum + rem*rem; num = num/10; } printf("\nSum of Square of Digits = %d", sum); getch(); return 0; }

Here is the final snapshot of the sample run:

Here are some of the main steps used in above program:

- Receive any number as input
- Never forgot to take a variable say
**sum**(holds sum of square of digits) with**0**as its initial value - That is declare and initialize 0 to
**sum**at start of the program - After getting input from user in a variable say
**num**, follow the steps given below - Create a while loop, runs until the value of
**num**is greater than 0 - Inside the loop, calculate the digit of that number using modulous operator (%)
- Using modulous operator or remainder operator, we will get the last digit of number
- Find the square of that digit and sum it up with
**sum**variable and then initialize it to**sum**variable - Then divide the value (num) with 10 to skip or delete the last digit
- For example, if user enters 2346 as input
- Therefore, at first run of the
**while**loop, as the value of**num**(2346) is greater than 0, therefore program flow goes inside the loop, and**num%10**or**2346%10**or**6**gets initialized to**rem**. Then**sum + rem*rem**or**0 + 6*6**or**36**gets initialized to**sum**. And at last,**num/10**or**2346/10**or**234**gets initialized to**num** - And then program flow goes inside the
**while**loop's condition statement, and as the value**234**is again greater than 0, therefore again the condition of**while**loop evaluates to be true, therefore program flow goes inside the loop, and do the same steps as told above - That is,
**num%10**or**234%10**or**4**gets initialized to**rem**. And**sum + rem*rem**or**36 + 4*4**or**36 + 16**or**52**gets initialized to**sum**. And at last again**num/10**or**234/10**or**23**gets initialized to**num** - Again at third run of the
**while**loop, as the condition statement of the loop, that is**num>0**or**23>0**evaluates to be true, therefore program flow again goes inside the**while**loop and follow the same procedure until the value of**num**becomes 0 - At last run of the
**while**loop, the variable**sum**holds its value as**6*6 + 4*4 + 3*3 + 2*2**or**36 + 16 + 9 + 4**or**65**as its final value, therefore 65 is the final output which is the result of sum of square of digits of given number that is 2346

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