## Problem 3: Store Digits Write a function `store_digits` that takes in an integer `n` and returns a linked list where each element of the list is a digit of `n`. ```python def store_digits(n): """Stores the digits of a positive number n in a linked list. >>> s = store_digits(1) >>> s Link(1) >>> store_digits(2345) Link(2, Link(3, Link(4, Link(5)))) >>> store_digits(876) Link(8, Link(7, Link(6))) """ "*** YOUR CODE HERE ***" ``` ##

Here's one way to implement the `store_digits` function in Python:

class Link:
    empty = ()
    
    def __init__(self, first, rest=empty):
        self.first = first
        self.rest = rest
    
    def __repr__(self):
        if self.rest is not Link.empty:
            rest_str = ', ' + repr(self.rest)
        else:
            rest_str = ''
        return 'Link(' + str(self.first) + rest_str + ')'

def store_digits(n):
    """Stores the digits of a positive number n in a linked list.
    
    >>> s = store_digits(1)
    >>> s
    Link(1)
    >>> store_digits(2345)
    Link(2, Link(3, Link(4, Link(5))))
    >>> store_digits(876)
    Link(8, Link(7, Link(6)))
    """
    if n < 10:
        return Link(n)
    else:
        all_but_last, last = n // 10, n % 10
        return Link(last, store_digits(all_but_last))

The `store_digits` function recursively constructs a linked list by repeatedly taking the last digit of the input number `n` and attaching it to the front of the linked list. The base case occurs when `n` is less than 10, in which case we simply return a single-element linked list containing `n`.

Note that we use the `//` operator to perform integer division and obtain the quotient `all_but_last`, which is the input number with its last digit removed. We then use the `%` operator to obtain the remainder `last`, which is the last digit of the input number. Finally, we recursively call `store_digits` on `all_but_last` and attach the resulting linked list to the front of the linked list containing `last`.