How to compute (2^32)/n using 32-bit integers

I have a 32-bit timer which I want to overflow after n steps. This means each step should be (2^32)/n. However, if I try to compute this number using 32-bit integers, the compiler complains that (1<<32) is larger than the type I'm using.

I could come quite close to the answer I'm looking for by instead doing something like (~0)/n. However, it bugs me that in this case I would not get the correct answer when n is a power of 2, which means in those cases it would take one extra step for the timer to overflow.

Is there a simple expression to calculate (2^32)/n using only 32-bit integers?

Solution

If you want the counter to overflow on precisely the nth step (when this is possible), then you need to compute ceil(2³² / n). Consider the two possible cases:

n is not a power of 2. In this case, n is not a factor of 2³², and the ceiling of the division is exactly one more than the floor. Moreover, (using truncating integer division, as in C or Java) floor(2³² / n) == floor((2³² - 1) / n). So the desired step is (2³² - 1)/n + 1.
n is a power of 2. In this case, n precisely divides 2³², and so (2³² - 1) / n will be one less than 2³² / n. So the desired step is (2³² - 1)/n + 1. Conveniently, that's the same value as in the first case.

Note: As @greybeard notes in a comment, there no guarantee that an appropriate step size exists if n > 2¹⁶. For larger n, the above procedure will compute the largest step size which guarantees that overflow will not occur before step n.