Here is a simple Java program that uses a do-while loop to compute the sum of the squares of the digits of a number:

```java
import java.util.Scanner;

public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter a number:");
int n = scanner.nextInt();
int sum = 0;
do {
int digit = n % 10;
sum += digit * digit;
n /= 10;
} while (n 0);
System.out.println("The sum of the squares of the digits is: " + sum);
}
}
```

Here's how it works:

1. The program first prompts the user to enter a number.
2. It then enters a do-while loop, which continues as long as the number `n` is greater than 0.
3. Inside the loop, it calculates the last digit of `n` by taking `n % 10` (the remainder when `n` is divided by 10).
4. It then adds the square of this digit to the `sum`.
5. It then removes the last digit from `n` by doing `n /= 10` (which is equivalent to `n = n / 10`).
6. Once `n` is no longer greater than 0, the loop ends and the program prints out the sum of the squares of the digits.

Question

Here is a simple Java program that uses a do-while loop to compute the sum of the squares of the digits of a number:

```java
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.println("Enter a number:");
        int n = scanner.nextInt();
        int sum = 0;
        do {
            int digit = n % 10;
            sum += digit * digit;
            n /= 10;
        } while (n > 0);
        System.out.println("The sum of the squares of the digits is: " + sum);
    }
}
```

Here's how it works:

1. The program first prompts the user to enter a number.
2. It then enters a do-while loop, which continues as long as the number `n` is greater than 0.
3. Inside the loop, it calculates the last digit of `n` by taking `n % 10` (the remainder when `n` is divided by 10).
4. It then adds the square of this digit to the `sum`.
5. It then removes the last digit from `n` by doing `n /= 10` (which is equivalent to `n = n / 10`).
6. Once `n` is no longer greater than 0, the loop ends and the program prints out the sum of the squares of the digits.

Knowee AI · Accepted Answer

Here is a simple Java program that uses a do-while loop to compute the sum of the squares of the digits of a number:

```java
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.println("Enter a number:");
        int n = scanner.nextInt();
        int sum = 0;
        do {
            int digit = n % 10;
            sum += digit * digit;
            n /= 10;
        } while (n > 0);
        System.out.println("The sum of the squares of the digits is: " + sum);
    }
}
```

Here's how it works:

1. The program first prompts the user to enter a number.
2. It then enters a do-while loop, which continues as long as the number `n` is greater than 0.
3. Inside the loop, it calculates the last digit of `n` by taking `n % 10` (the remainder when `n` is divided by 10).
4. It then adds the square of this digit to the `sum`.
5. It then removes the last digit from `n` by doing `n /= 10` (which is equivalent to `n = n / 10`).
6. Once `n` is no longer greater than 0, the loop ends and the program prints out the sum of the squares of the digits.

Question

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