Question

The divBySum method is intended to return the sum of all the elements in the int array parameter arr that are divisible by the int parameter num. Consider the following examples, in which the array arrcontains {4, 1, 3, 6, 2, 9}.
The call divBySum(arr, 3) will return 18, which is the sum of 3, 6, and 9, since those are the only integers in arr that are divisible by 3.
The call divBySum(arr, 5) will return 0, since none of the integers in arr are divisible by 5.
Complete the divBySum method using an enhanced for loop. Assume that arr is properly declared and initialized. The method must use an enhanced for loop to earn full credit.
/** Returns the sum of all integers in arr that are divisible by num
* Precondition: num > 0
*/
public static int divBySum(int[] arr, int num)

Answer 1

Answer:

The complete method is as follows:

public static int divBySum(int[] arr, int num){        

     int sum = 0;

     for(int i:arr){

         if(i%num == 0)

         sum+=i;

     }

     return sum;

   }

Explanation:

As instructed, the program assumes that arr has been declared and initialized. So, this solution only completes the divBySum method (the main method is not included)

This line defines the method

public static int divBySum(int[] arr, int num){        

This line declares and initializes sum to 0

     int sum = 0;

This uses for each to iterate through the array elements

     for(int i:arr){

This checks if an array element is divisible by num (the second parameter)

         if(i%num == 0)

If yes, sum is updated

         sum+=i;

     }

This returns the calculated sum

     return sum;

   }