This has exactly the same requirements as assignment three
except that you must use a linked list to store each super
int's digits instead of the vector/array.

Plus one very small thing. Add the division operation.

That doesn't sound like a small thing, but because of this
requirement it is: Use binary chop search.

Ignoring negatives, you know that the result of an integer
division R = A / B, R must be at least 0 and at most A, so
you have the bounds for binary chop. The value at the middle
position, M, is what you hope to be the result. In what way
does the value of M * B tell you what to do?

Binary chop requires division by two (to find the mid-point)
and that is very easy even for super integers, but having
a linked list storing the digits in "reverse" order adds a
small complication. The solution is to reverse the order of
digits linked list then do the division by 2 the easy way,
then reverse the order of the result's digits.

Let's do 471928371 / 2. The correct result is 235964185 with 
a remainder of 1.

WHat's shown at each step
  remaining digits still to be divided by 2
  what's done at this step: 
      10 * carry + next digit is divided by 2
  digits of answer so far

471928371 remainder is 0
  10 * 0 + 4 is 4, divided by 2 gives 2 remainder 0
2

71928371 remainder is 0
  10 * 0 + 7 is 7, divided by 2 gives 3 remainder 1
2 3

1928371 remainder is 1
  10 * 1 + 1 is 11, divided by 2 gives 5 remainder 1
2 3 5

928371 remainder is 1
  10 * 1 + 9 is 19, divided by 2 gives 9 remainder 1
2 3 5 9

28371 remainder is 0
  10 * 0 + 2 is 12, divided by 2 gives 6 remainder 0
2 3 5 9 6

8371 remainder is 0
  10 * 0 + 8 is 8, divided by 2 gives 4 remainder 0
2 3 5 9 6 4

371 remainder is 0
  10 * 0 + 3 is 3, divided by 2 gives 1 remainder 1
2 3 5 9 6 4 1

71 remainder is 1
  10 * 1 + 7 is 17, divided by 2 gives 8 remainder 1
2 3 5 9 6 4 1 8

1 remainder is 1
  10 * 1 + 1 is 11, divided by 2 gives 5 remainder 1
2 3 5 9 6 4 1 8 5

And there we have it, the correct answer of 235964185
and even the final remainder 1 is the correct remainder
for the whole division.