CODEWITHSUNDEEP
javaloops
Fahim Alam Shakie
Fahim Alam Shakie

Reputation: 5

I cannot get my for loop for Armstrong number in Java to work

So I'm trying to find every Armstrong number between 1-500. I have to only use a For Loop for this, I know people can also use While loops. I'm not sure what im doing wrong here. Its only giving me 375, but I know numbers like 153 is also an armstrong number. `

Here my code :

public class ArmstrongNumber {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        //Armstrong numbers 1-500

        int D1;
        int D2; 
        int D3;



        for (int number = 1; number <= 500 ; number++) {
            int temp=number;
            D1=number%10; //ones

            temp= temp/10;
            D2=number%10; //tens

            temp= temp/10;
            D3=number%10; //hundreds

            if( ((D1*D1*D1) + (D2*D2*D2) + (D3*D3*D3)) == number)
            {
        System.out.println(number + " is a Armstrong number");
            }

        }


        }

    }

Upvotes: 0

Views: 145

Answers (3)

WJS
WJS

Reputation: 40062

Armstrong numbers are numbers where the sum of the digits, each raised to the power of the number of digits in that number equals the number. Your method above will miss some numbers because you are not considering the number of digit of all numbers.

So the digits 1 thru 9 are each Armstrong numbers since each raised to the power of 1 is that number.

Here is one way to check for all numbers up to a certain point.

// the digits raised to the power of 1
int[] powers = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

int nextPower = 10;

for (int r = 1; r <= 500; r++) {
    int n = r;
    int sum = 0;
    while (n > 0) {
        // sum of the powers
        sum += powers[n % 10];
        // if at any time the sum exceeds the number
        // break out of the loop
        if (sum > r) {
            break;
        }
        n /= 10;
    }
    if (sum == r) {
        System.out.println(r + " is an Armstrong number.");
    }
    // check to see if time to transition to next power.
    if (r == nextPower - 1) {
        nextPower *= 10;
        // and increase each digits power
        for (int i = 2; i < powers.length; i++) {
            powers[i] *= i;
        }
    }
}

Prints

1 is an Armstrong number.
2 is an Armstrong number.
3 is an Armstrong number.
4 is an Armstrong number.
5 is an Armstrong number.
6 is an Armstrong number.
7 is an Armstrong number.
8 is an Armstrong number.
9 is an Armstrong number.
153 is an Armstrong number.
370 is an Armstrong number.
371 is an Armstrong number.
407 is an Armstrong number.

Upvotes: 0

kozmo
kozmo

Reputation: 4491

You have to use remainder operation (%) to tmp:

for (int number = 1; number <= 500; number++) {
     int temp = number;
     D1 = temp % 10; //ones

     temp = temp / 10;
     D2 = temp % 10; //tens

     temp = temp / 10;
     D3 = temp % 10; //hundreds

     if (((D1 * D1 * D1) + (D2 * D2 * D2) + (D3 * D3 * D3)) == number) {
          System.out.println(number + " is a Armstrong number");
     }
}

)))

Upvotes: 0

Mobina
Mobina

Reputation: 7109

You should write D2 = temp%10 and D3 = temp%10 (temp instead of number).

Upvotes: 1

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