How do you write a program to print Armstrong numbers between 1 and 100 using for loop?

Answer 1

/*Program to find Armstrong number between 1 to N*/

int main() {

int n = 0, remainder, sum = 0, i = 0, noDigits = 0, isArm = 0;

char ch[60] = {0};

printf("Find the Arm Strong Numbers between 1 to N");

scanf("%d", &n);

for(i = 1; i<n; i++)

{

isArm = i;

itoa(isArm, ch, 10);

noDigits = strlen(ch);

while(isArm)

{

remainder = isArm%10;

isArm=isArm/10;

sum= sum+pow(remainder, noDigits);

}

if(sum == i)

printf("\nArm Strong Nos are %d\n", i);

sum = noDigits = 0;

}

}

Answer 2

if the sum of the cubic value of each digit of a number is equal to the original number , then that particular number is an Armstrong number :-)

Actually, Armstrong numbers are the sum of their own digits to the power of the number of digits.

Answer 3

#include(stdio.h)

#include(conio.h)

int main( )

{

int no, temp, rem, sum;

clrscr( );

printf("Armstrong numbers between 1 and 1000 are:\n");

for(no=1; no<=1000; no++)

{

temp=no;

sum=0;

while(temp>0)

{

rem=temp%10;

sum=sum+(rem*rem*rem);

temp=temp/10;

}

if(no==sum)

printf("\n%d", no);

}

getch( );

return 0;

}

Answer 4

#include <iostream>

#include <math.h> // for std::pow()

unsigned int get_length(unsigned int num,const unsigned int base=10)

{

unsigned int len=1;

while(num && (num/=base))

++len;

return( len );

}

bool is_armstrong(const unsigned int num,const unsigned int base=10)

{

unsigned int len=get_length(num,base);

unsigned int sum=0;

unsigned int tmp=num;

while(tmp)

{

sum+=(unsigned int)std::pow((double)(tmp%base),(double)len);

tmp/=base;

}

return(num==sum);

}

int main()

{

std::cout << "Armstrong series (base 10):";

for(unsigned int num=0; num<=0xffffffff; ++num)

if(is_armstrong(num))

std::cout << " " << num;

std::cout << std::endl;

return(0);

}

Answer 5

You will need:

• A computer
• The computer language you will write in
• The understanding of Armstrong Numbers
• The number you want to find
• How the number is presented to the program as the input
• Where the output of the program should go
• How to put all together as a program

Answer 6

class armstrong

{

public void m()

{

int s,d,i,n;

for(i=1;i<=500;i++)

{

s=0;

n=i;

while(n>0)

{

d=n%10;

s=s+(d*d*d);

n=n/10;

}

if(s==i)

{

System.out.println(s+" ");

}

}

}

}

Answer 7

/*Program to check for armstrong number */

#include<stdio.h>

#include<math.h>

main()

{

int a, b, z,x, sum = 0;

printf("enter the number to check if it is armstrong number\n");

scanf("%d", &x);

z=x;

if ((z >= 100) && (z <= 900))

{

a = z % 10;

z = z / 10;

b = z % 10;

z = z / 10;

sum = pow(a, 3) + pow(b, 3) + pow(z, 3);

if (x== sum)

printf("a is an armstrong number.\n");

else

printf("not a armstrong number\n");

}

else

printf("armstrong number must be of three digit\n");

}

/*posted by prabhat5052@gmail.com*/

Answer 8

class ArmstrongNumbers{

public static void main(String[] args){

int num, temp, digit1, digit2, digit3;

System.out.println("Armstrong numbers between 1 and 500: ");

num = 1;

while (num <= 1000){

digit1 = num - ((number / 10) * 10);

digit2 = (num / 10) - ((number / 100) * 10);

digit3 = (num / 100) - ((number / 1000) * 10);

temp = (digit1*digit1*digit1) + (digit2*digit2*digit2) + (digit3*digit3*digit3);

if (temp == num){

System.out.println("Armstrong Number:"+ temp);

}

num++;

}

}

}

Answer 9

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