Converting decimal numbers to binary numbers can be done recursively. Write a function that takes a positive int argument and returns the corresponding binary number as an int composed of only 1s and 0s. For example, convert(5) returns the int: 101.
The insight for this problem comes from the fact that the rightmost digit of a decimal n is easy to calculate. It is the remainder when dividing by the base 2: n%2. To get the next rightmost digit, you take the resulting quotient, i.e., n2 = n/2, and find its remainder, n2%2, which is the next digit. Unfortunately, that generates the digits from right to left, and we want them from left to right. We could easily do that nonrecursively using string concatenation or string reversal, but in this exercise you are to use recursion to recursively calculate the digits. Effectively, you are letting recursion reverse the ordering: think in terms of calculating the rightmost digit as described previously, but then let recursion reverse the digits.
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