To check whether entered number is armstrong number or not. Code: #include<stdio.h> #include<conio.h> #include<math.h> main() { int n,sum=0,rem=0,cube=0,n1,i; clrscr(); printf("enter a number"); scanf("%d",&n); n1=n; while(n!=0) { rem=n%10; cube=pow(rem,3); sum=sum+cube; n=n/10; } if(sum==n1) printf("angstrom"); else printf("not angstrom"); getch(); }
You are printing the wrong output. Code: if(sum==n1) printf("angstrom"); else printf("not angstrom"); It should be armstrong number and not angstrom. For those who dont know what it is. Armstrong number: An n-digit number equal to the sum of the nth powers of its digits. There are no two-digit Armstrong numbers; and there are four three-digit Armstrong numbers: 153, 370, 371, and 407. 1x1x1 + 5x5x5 + 3x3x3 = 153
I have read the above program and try to understand it but i cannot understand all. I only can calculate the 53 without the 1 and 370 without the 3. I don't know why. Your explanations is greatly appreciated by me and others. n=n/10; -- Is it the 153 calculate 3 first, then 5 and finally 1. Sorry for my stupidness. Your help is greatly appreciated by me an others.
