C program to check neon number; Through this tutorial, we will learn how to check neon number in c program using for loop, while loop and function.
Let’s use the following algorithm to write a program to check whether a number is neon or not; as follows:
- Take the number as input from the user
- Find the square of the number
- Find the sum of the digits in the square
- Compare the sum with the number. If both are equal then it is a Neon number
Programs to Check Neon Number
- C Program to Check Neon Number using For Loop
- C Program to Check Neon Number using While Loop
- C Program to Check Neon Number using Function
C Program to Check Neon Number using For Loop
#include <stdio.h> int main() { int i, num, square, sumDigits = 0; printf("Enter a number: "); scanf("%d", &num); square = num * num; for(i = 0; i<=square; i++) { sumDigits += square % 10; square /= 10; } if (num == sumDigits) { printf("It is a Neon number\n"); } else { printf("It is not a Neon number\n"); } return 0; }
The output of the above c program; as follows:
Enter a number: 10 It is not a Neon number
C Program to Check Neon Number using While Loop
#include <stdio.h> int main() { int num, square, sumDigits = 0; printf("Enter a number: "); scanf("%d", &num); square = num * num; while (square != 0) { sumDigits += square % 10; square /= 10; } if (num == sumDigits) { printf("It is a Neon number\n"); } else { printf("It is not a Neon number\n"); } return 0; }
The output of the above c program; as follows:
Enter a number: 9 It is a Neon number
C Program to Check Neon Number using Function
#include <stdio.h> int isNeon(int num) { int square, sumDigits = 0; square = num * num; while (square != 0) { sumDigits += square % 10; square /= 10; } if (num == sumDigits) { printf("It is a Neon number\n"); } else { printf("It is not a Neon number\n"); } } int main() { int n; printf("Enter a number: "); scanf("%d", &n); isNeon(n); return 0; }
The output of the above c program; as follows:
Enter a number: 1 It is a Neon number
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