This is a C Program to print armstrong number from 1 to 1000.

This C Program print armstrong number from 1 to 1000.

An Armstrong number is an n-digit base b number such that the sum of its (base b) digits raised to the power n is the number itself. Hence 153 because 1^{3} + 5^{3} + 3^{3} = 1 + 125 + 27 = 153.

Here is source code of the C Program to print armstrong number from 1 to 1000.

The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/* * C Program to Print Armstrong Number from 1 to 1000 */ #include <stdio.h> main() { int number, temp, digit1, digit2, digit3; printf("Print all Armstrong numbers between 1 and 1000:\n"); number = 001; while (number <= 900) { digit1 = number - ((number / 10) * 10); digit2 = (number / 10) - ((number / 100) * 10); digit3 = (number / 100) - ((number / 1000) * 10); temp = (digit1 * digit1 * digit1) + (digit2 * digit2 * digit2) + (digit3 * digit3 * digit3); if (temp == number) { printf("\n Armstrong no is:%d", temp); } number++; } }

In this C program, we are printing the Armstrong number from 1 to 1000. An Armstrong number is an n-digit base b number such that the sum of its (base b) digits raised to the power n is the number itself. Hence, 153 because 1^{3} + 5^{3} + 3^{3} = 1 + 125 + 27 = 153. We are initializing the number variable value as 001.

Using while loop checks the condition that the value of ‘number’ variable is less than or equal to 900. If the condition is true then, execute the iteration of the loop. The ‘digit1’ variable is used to compute the division of the value of ‘number’ variable by 10 and multiply the resulting value by 10 and subtract the multiplied value with the value of ‘number’ variable.

The ‘digit2’ variable is used to compute the division of the value of ‘number’ variable by 100, multiply the resulting value with 10, divide the value of ‘number’ variable by 10 and subtract the multiplied value from the divided value.

The ‘digit3’ variable is used to compute the division of the value of ‘number’ variable by 100 and also by 1000, then multiply the value divided by 100 with 10 and subtract both the values. The ‘temp’ variable is used to multiply the value of digit1, digit2, digit3 variables to the power of 3 respectively. Compute the summation of all the three multiplied values. If condition statement is used to check the value of ‘temp’ and ‘number’ variables are equal, if the condition is true print the Armstrong number.

Output: $ cc pgm44.c $ a.out Print all Armstrong numbers between 1 and 1000: Amstrong no is:1 Amstrong no is:153 Amstrong no is:370 Amstrong no is:371 Amstrong no is:407

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