A maths question

From the given, we can see that the monthly interest rate is 1%.
Suppose that Tom repays $C every month, then:
Amount he still owes after his first repay = 100000 x 1.01 - C
Amount he still owes after his second repay = (100000 x 1.01 - C) x 1.01 - C = 100000 x (1.01)2 - 1.01C - C
Amount he still owes after his third repay = (100000 x (1.01)2 - 1.01C - C) x 1.01 - C = 100000 x (1.01)3 - (1.01)2C - 1.01C - C
So, we can see that after his nth repay, he still owes:
100000 x (1.01)n - (1.01)n-1C - (1.01)n-2C - ... - 1.01C - C
= 100000 x (1.01)n - C[(1.01)n-1 - (1.01)n-2 - ... - 1.01 - 1]
= 100000 x (1.01)n - [(1.01)n - 1]C/(1.01 - 1)
= 100000 x (1.01)n - 100[(1.01)n - 1]C
= 100000 x (1.01)n - 100 x (1.01)nC + 100C
= 100 x (1.01)n x (1000 - C) + 100C
If he pays back all the debt after the nth repay, then:
100 x (1.01)n x (1000 - C) + 100C = 0
100 x (1.01)n x (C - 1000) = 100C
(1.01)n = C/(C - 1000)
n = {log [C/(C - 1000)]}/log 1.01
By trial, we can find out that C should be 1260 so that the total amount that he pays will not be more than $200 000.
Then
n = log (1260/260)/log 1.01 = 158.6
Thus a total of 159 months he will need for repayment.
For the first 158 months, he has paid 1260 x 158 = 199080
After that, amount he still owes for the last month is:
100 x (1.01)158 x (1000 - 1260) + 100 x 1260 = 757.9
Thus, total amount he would pay = 199080 + 757.9 = 199837.9

2009-07-12 11:50:39 補充：
Making C as the subject, we have:
C = 1000 x (1.01)^n / [(1.01)^n - 1]
So when n = 158, C = 1261.99
and the total amount paid = 1261.99 x 158 = 199393.72

2009-07-12 11:50:50 補充：
When n = 159, C = 1258.72
and the total amount paid = 1258.72 x 159 = 200136.6

Therefore he should choose to repay for 158 months.

Thanks for the hints from nelsonywm2000.

The FIXED amount is $1,261.99
158 x 1261.99 = $199,393.72