which r the first 10 amstrong numbers??
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Just generate the Fibonacci numbers one by one, and print each number's last digit ie number%10.
/*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; } }
class Armstrong{ public static void main(String args[]) { int num,rem,qub,sum=0,i; for(i=0; i<=999; i++) { num=i; sum=0; while(num>0) { rem=num%10; qub=rem*rem*rem; sum=sum+qub; num=num/10; } if(sum==i) { System.out.println("Print 1 to 1000 Armstrong Number",sum); } } } }
#include<iostream> int main() { int i=0; while(i++<10) std::cout<<i*i<<std::endl; }
/*Program to find whether given no. is Armstrong or not. Example : Input - 153 Output - 1^3 + 5^3 + 3^3 = 153, so it is Armstrong no. */ class Armstrong{ public static void main(String args[]){ int num = Integer.parseInt(args[0]); int n = num; //use to check at last time int check=0,remainder; while(num > 0){ remainder = num % 10; check = check + (int)Math.pow(remainder,3); num = num / 10; } if(check == n) System.out.println(n+" is an Armstrong Number"); else System.out.println(n+" is not a Armstrong Number"); } }