Credit Card Mathematics Tidbit (10/21/2001)

Prepared by:

Joseph Malkevitch
Mathematics and Computing Department
York College (CUNY)
Jamaica, NY 11451

Email: (for additions, suggestions, and corrections)

At the start of the fall semester, it common to see near or on college campuses representatives of various credit card companies trying to encourage students to get a credit card by giving them free merchandise, tee shirts, and the like. Why does it pay credit card companies to do this?

The credit card industry makes money by lending people money and collecting interest on the money borrowed. They also make money from the merchants who use the services of the credit card company to help merchants assist in the selling of their products.

Interest is the money paid to a lender that is paid by the borrower of money, which is to be repaid in the future, for being able to have the money now. The longer one borrows money for the more money one has to pay back. Simple interest is computed by multiplying the interest rate times the time period. It is common to quote interest as a percent per year, the annual rate of interest. For example, one might borrow money at 8% annual interest or 18% annual interest. It is common when one borrows money on a credit card for the money that is borrowed to be repaid in a series of monthly payments. In the steady state situation, the amount that a borrower will owe the following month is the amount he/she owed the previous month, plus the amount of interest due for the month for having borrowed the money diminished by the amount that the borrower repaid that month. A borrower can avoid all interest charges by repaying the total amount due from the purchases made during the month. If the money is not repaid however, the interest rates are often quite high. Furthermore, and that is what we propose to investigate here, the amount of time that a borrower needs to repay the whole loan can go on for a very long time if the borrower chooses the option of repaying the minimum amount allowed for each payment. One common practice is to set the minimum amount that must be paid in a given month as 2% of the previous balance. (On some credit cards, this minimum amount is 2% of the unpaid balance rounded down to the nearest dollar.) As we shall see, if one chooses to only pay this minimum amount each time, the amount of time to pay off one's credit card debt can go on for a long time.

Let us construct a mathematical model for credit card debt. Suppose we use B(n) to denote the size of the unpaid balance on a credit card after n time periods, which we will take to be measured in months. Thus, the size of the initial loan will be B(0) which we will also denote by L. We will use i to denote the monthly rate of interest, and thus, the monthly interest rate is annual rate divided by 12. (For example, if the annual interest rate is 12% then i is .01.) If the minimum payment due is 2% of the unpaid balance, we can obtain the following difference equation for the situation:

B(n+1) = B(n) + iB(n) -.02B(n)


B(n+1) = (.98 + i)B(n).

This equation can be solved easily:

For simplicity, use m to denote .98 + i.

Since B(1) = mB(0), and B(2) = mB(1), etc., we can conclude that:

B(n) = mnB(0). (*)


If one borrows $800 at 18% annual interest and one pays the minimal amount possible each time, what amount would be left to pay off after one year?


Since the annual interest is 18 percent, the monthly interest is .18/12=.015.

Furthermore, we want to know how much money is left to repay in one year, 12 months, thus, we want the value of B(n) with n = 12.

Hence, B(12) = (.99512)x(800) = $753.30. This is little progress on retiring your debt from one year of payments.

If you think about the situation here you will see that if one always pays the minimum amount, but for the fact that there is a minimum currency amount of 1 cent the process could go on forever!

If one picks a particular fraction of the original loan, one can compute the amount of time that it takes to pay down your credit card balance to this amount. Say that this fraction is k. Thus, given a loan of size L we want to know when the loan balance will be reduced to kL. Using equation (*) we have:

kL = mnL

Hence k = mn

Now if one computes log10 of both sides one gets:

log10 k = n log10 m

Solving for n and replacing m by (.98 + i ) one gets:

n = (log10k)/(log10(.98 + i))

Here, we have used logarithms base 10 but one can use any base. Most calculators use base 10 logarithms.

Note that what is going on here is exactly like the situation in studying the half-life or other fraction for the decay of a radioactive compound.

A much better way for someone who has a credit card debt to repay the debt is to pick an amount that he/she can reasonably repay each month and always pay that amount or more. This amount can serve as a guide for how much one can borrow so as not to be in debt for a very long time. Many people borrow more money than they can reasonably repay and this means that they sometimes never get out from under the debt they incur.

One can study the repayment of a fixed amount M using mathematics as well. Here is the equation for this situation:

B(n+1) = B(n) +iB(n) - M.

This equation can be solved by:

a. Find the general solution of B(n+1) = (1+i)B(n)

b. Find a particular solution of B(n+1) = B(n) +iB(n) - M

c. Solve the initial value problem: for B(0) = L.

The solution of this problem is similar to and related to problems that come up in the mathematics of finance and involve the payment of annuities and mortgages. An annuity is a regular sequence of payments of a fixed size which are paid (or collected) for a fixed period of time at a fixed interest rate. Using simple properties of the geometric series, one can do calculations involving annuities to find the value of the regular payment, the value today of a sequence of regular payment of a particular amount, find the amount of a regular payment to build up a certain amount in the future.

Here are typical problems one can solve with these methods.

1. If I obtain a mortgage on a house for $200,000 what would be the monthly payment to retire this amount at 7% annual interest (monthly payments) over a period of 30 years?

2. If a winner of a lottery must be sent checks of $50,000 a year for 20 years (one million dollars) how much is the cost of an annuity to make these payments if the current interest that can be earned is 8% annual interest.

3. At the birth of my child, I want to make monthly payments so that I will have $70,000 to pay the child's tuition when she is 18 years old. If the annual interest rate is 7% what is the size of the monthly payment to achieve this goal?

4. John's pension fund is worth $400,000. What size monthly annuity would he receive if he wanted monthly payments for 20 years and the interest rate is 5% annual interest?

Many years ago it was common for undergraduate mathematics departments to offer courses in the mathematics of finance. These courses disappeared in part because some of the topics were treated in the Finite Mathematics for Business students taught by mathematics. This material is often limited to a discussion of annuities and compound interest, without any of the more detailed models that had been the courses of yesteryear. In recent years, there has been a renaissance of interest in mathematical applications to finance. Much of this interest was inspired by the development of what has come to be called the Black-Scholes Equation. This work, developed by Fisher Black, a mathematician, and Scholes and Merton revolutionized the options market by developing a reasonable method pricing certain types of options. Perhaps surprisingly, the work involved the solution of a partial differential equation. Much research is now being done in extending the theory of how to price options, and to the development of many new financial instruments that are designed to minimize investment risk.


Hart, W., Mathematics of Investment, (fifth edition), D.C. Heath, Lexington, 1975.

Simpson, T. and Z. Pirenian, B. Crenshaw, Mathematics of Finance, Part II, (3rd edition), Prentice-Hall, Englewood Cliffs, 1951

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