Compound interest

Adam Zettel · September 22, 2026 · written with an agent

P r n times a year, t years A = ?

$1000 at 5% a year becomes $1628.89 in ten years if the interest is added once a year, and $1647.01 if it is added monthly. Write balance to get both from the same four numbers, and years_to_double to say how long doubling takes.

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Interest added $n$ times a year is $r / n$ each time, applied $n t$ times over $t$ years. Getting that substitution right is most of the exercise, and the two numbers above are far enough apart that a test fails if you use $r$ and $t$ where $r / n$ and $n t$ belong.

Monthly beats annual by $1647.01 - 1628.89 = 18.12$ dollars here. Weekly beats monthly by less, daily by less again, and no matter how often the interest is added the balance never passes $P e^{r t} = $1648.72$ . Once the tests pass, try a few values of $n$ to see it approach that limit.

years_to_double is the question the doubling-time post answers: $t$ is in the exponent, and taking the logarithm of both sides brings it down.

Your task

Write these 2 functions:

balance(p, r, n, t): The balance after t years: A = P\left(1 + \frac{r}{n}\right)^{nt}.
years_to_double(r): Years for money at annual rate r, compounded once a year, to double.

Make these tests pass 3 of 4 fail

$1000 at 5% for 10 years, compounded once a year, is $1628.89 assert_near: |1000 - 1628.89| = 628.89 > 0.01 (fails)
compounded monthly it is $1647.01 assert_near: |1000 - 1647.01| = 647.01 > 0.01 (fails)
no time, no interest: the balance is the principal (passes)
at 5% a year the money doubles in 14.21 years, and the balance then is 2P assert_near: |0 - 14.2067| = 14.2067 > 0.001 (and 1 more) (fails)

Printed

1000

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