Complexity of inserting an integer to an empty hashtable What's the complexity of inserting an integer $n$ to an empty hash table? My guess is that it's $O(\log n)$ as we require to scan every bit of $n$ to compute its hash. However, I  often see it as $O(1)$. Am I mixing up things here ?
The answer depends on your model of computation.    In other words, it depends on how you count the number of basic operations, or on what you consider to be a "basic operation".

In the transdichotomous model or Word RAM model, you can do computation on an entire word in $O(1)$ time, and the word size is guaranteed to be large enough to hold the integer (and, in particular, at least $\log n)$.  An operation on a single word counts as a single operation, even though you're operating on a bunch of bits.  Consequently, there exist hash functions that can hash the number $n$ in $O(1)$ time, in these models of computation.

In the bit complexity model, it takes at least $\log n$ time to read an integer with $\log n$ bits, so you're right that hashing $n$ will take $\Omega(\log n)$ time.

You might find it useful to read https://en.wikipedia.org/wiki/Transdichotomous_model and https://cstheory.stackexchange.com/q/2656/5038.

Ultimately, the time to compute a hash will depend on the hash function (some hashes might be even slower than the above), but this might be enough to clear up your particular question.