David has 80,000 dollars in his college fund, but he can only make one withdrawal for half of the remaining fund each month.

Let a₁, a₂ , a₃, ... a_n be the amount of withdrawal and
d₁, d₂ , d₃, ... d_n be the remaining balance of the fund for the 1^st, 2^nd, 3^rd, ... n^th month.

Each month, David can withdraw:

     a₁ = (60,000)(1/8)
     a₂ = (60,000)(1 - 1/8)(1/8) = (60,000)(7/64)
     a₃ = (60,000)(1 - 1/8 - 7/64)(1/8) = (60,000)(49/512)
      .
      .
      .
     a_n = (60,000)(1/8)(7/8)^n-1

The remaining balance becomes:

     d₁ = (60,000)(1 - 1/8) = (60,000)(7/8)
     d₂ = (60,000)(1 - 1/8 - 7/64) = (60,000)(49/64)
     d₃ = (60,000)(1 - 1/8 - 7/64 - 49/512)
         = (60,000)(343/512)
      .
      .
      .
     d_n = (60,000)(7/8)ⁿ

Thus, a sequence can be formed to represent the remaining balance of the fund:

     d_n = {60,000(7/8, 49/64, 343/512,..., (7/8)ⁿ)}

As the number of month increases, the balance remaining becomes closer and closer to 0. However, it is assumed that the college fund runs out when the remaining balance is one hundred dollars. Hence, the problem at hand is what value of n will give the sequence a limit of 100. That is,

     (60,000)(7/8)ⁿ = 100

Solving the above equation to obtain:

     n = 47.9

Hence, David has around 48 months to prepare and become financially independent.