Complete the method RECURSIVE productOfDigits() that computes the product of the digits of the given positive integer argument, n, except any digits that are 0. Note: productOfDigits() should return always return a non-zero answer since it will only be called with non-zero arguments and 0 digits are ignored. e.g., if n = 5403, the method will return: 5*4*3 = 60 (the 0 was ignored) note: the right-most (1's) digit of a positive integer, n, can be found using n%10 (5403 % 10 is 3, for example) the remaining digits (all but the 1's digit) can be found using n/10 (5403 / 10 is 540, for example) So use a recursive call to get the product of all the other digits and multiply it to the last digit if it's not a 0. The base case is when the number is only one digit, so a recursive call would not be needed.

//testing productOfDigits()

System.out.println(productOfDigits(412));

System.out.println(productOfDigits(2000056));

System.out.println(productOfDigits(90));