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); } } } }
/*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"); } }
#include<iostream> int main() { int i=0; while(i++<10) std::cout<<i*i<<std::endl; }
Armstrong Circle Theatre - 1950 The Numbers Racket 10-13 was released on: USA: 13 April 1960
By adding the differences of the previous two numbers.
The sum of the first 10 even numbers is 110.
The sum of the first 10 counting numbers (1-10) is 51.
The sum of the first 10 odd whole numbers is 100.
The sum of the first 10 natural numbers is 51, with an average of 5.1
The numbers 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 are the first ten composite numbers.
The first ten positive numbers total 55.
Free credit card numbers are generated automatically with computer software that generate numbers that will comply with the MOD 10 algorithm or also knows as the Luhn algorithm.
An Armstrong is equal to 1*10^-10 metres
Benjamin Armstrong is 5' 10".
Emma Armstrong is 5' 10".