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Armstrong number check

Discussion in 'C' started by adroit89, Jan 7, 2007.

  1. adroit89

    adroit89 New Member

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    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();
    }
     
  2. shabbir

    shabbir Administrator Staff Member

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    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
     
  3. Peter_APIIT

    Peter_APIIT New Member

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    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.
     
  4. Peter_APIIT

    Peter_APIIT New Member

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    Thanks for your help.
     

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