Financial mathematics I was given the following information:

A young couple decide that they want to buy a house for 220000. They
  have saved a deposit of 80000 and are confident that they can pay 2500
  per month, starting in 3 months' time to amortise their loan. The
  current interest rate is 20% p.a., compounded monthly.

The question:

What will the amount of the last payment (less than 2500) be?

Now I thought all payments would be 2500, if the loan is to be amortised?
 A: If $P$ is your payment and $L$ is the loan amount, then 
$$L=P\left(\frac{1-\frac{1}{(1+i)^n}}{i}\right)$$
You are solving for $n$, since you already know $L, P,$ and $i$.  $n$ will tell you the term of the loan, but it will most likely not be a natural number.
Then set
$$L_n=P\left(\frac{1-\frac{1}{(1+i)^{\lfloor{n}\rfloor}}}{i}\right)$$
(I substituted $\lfloor{n}\rfloor$ back in the expression).
Your drop payment should be $L-L_n$
Note, you have to convert your annual percentage to monthly.  Use the following conversion:
$$(1+i)^n=\left(1+\frac{i^{(12)}}{12}\right)^{12n}$$
Finally, it says that you are waiting 3 months to begin payments so assuming interest is accruing, your initial $L$ should be $140,000(1+.0153095)^3=146,528.932.$  
(since your monthly interest using the above formula is $1.53095\%$)
So you have this expression
$$146,528.095=2,500\left(\frac{1-(1.0153095)^{-n}}{.0153095}\right)$$
Solving for $n$ you get $n=149.806$.  This tells us that our $150$th payment is not a full payment.  Thus, $\lfloor{n}\rfloor=149$.  Plugging this into the right hand side and solving for $L_n$ give us
$$L_n=2,500\left(\frac{1-\frac{1}{(1.0153095)^{149}}}{.0153095}\right)=146,322.291$$
Finally, $L-L_n=146,528.932-146,322.291=206.641$
Hence, your final payment is $\$206.641$.
