A university student receives his statement for his tuition and notices that he doesn't have enough money to pay it all off at once. The student inquires about interest rates at his university and is told the following information:
You will be subject to interest charges of one per cent monthly on the amount owing from your last statement. The annual interest rate is $12.7$ per cent.
After paying all he has, he still owes his university $\$2420.$ Since he has a part time job in which he earns $\$600$ monthly, he decides to pay off $\$600$ every month on his balance.
How much money will he owe the university each month? Assume school starts in September. What month will the balance be completely paid off?
After the first month, the student will owe
$$2420 (1.01) = 2444.20$$
Then the student pays 600 dollars and owes 1844.20. With another compounding period, the student owes about 1862.64. The student pays 600 dollars and owes 1262.64. And so on.
Algebraically, the amount the student owes looks like, where $P$ is the principal and $m$ is the monthly payment,
$$[[[P (1+i) - m](1+i)-m](1+i)-m]....$$
which, when rearranged, becomes, for n compounding periods
$$P(1+i)^n - m \sum_{k=1}^{n-1} (1+i)^k = P (1+i)^n - m \frac{(1+i)^n-1}{i}$$
Let's say $n=4$ in your example; then the student owes
$$2420 (1.01)^4 - 600\frac{(1.01)^4-1}{0.01} \approx 82.02$$
The university would then compound interest once more, and the payoff amount would be $82.84$.
Here's a complete table of what the student will owe each month on this schedule:
$$\left( \begin{array}{cc} 1 & 1844.20 \\ 2 & 1262.64 \\ 3 & 675.27 \\ 4 & 82.02 \\ \end{array} \right)$$
ADDENDUM
You should know that the university charges $12.7\%$ interest annually, as $(1.01)^{12}-1 \approx 0.127$
Also, the student may pay off the loan completely in $n=4$ months by setting the above expression to zero to get a mothly payment that allows this. The result is
$$m = P \frac{i}{1-(1+i)^{-n}} = 620.20 $$