 Ceiling count for fixed-size batches

Suppose $N$ is a nonnegative integer count of items and each batch holds at most $k$ items, where $k$ is a positive integer. The smallest sufficient number of batches isIndeed, $bk\ge N$ exactly when the integer $b$ is at least $N/k$. If $N=0$, no batches are needed. If $N$ is a multiple of $k$, no partly filled batch is needed.

For $102$ items with capacity $20$ per batch, $b=\lceil 5.1\rceil =6$. Five batches can be full and the sixth holds two items, leaving $6\cdot 20-102=18$ unused places. This is a different question from the five complete groups counted by $\lfloor 102/20\rfloor$.

Writing the Euclidean division $N=kq+r$, $0\le r<k$, gives the equivalent rule $b=q$ when $r=0$, and $b=q+1$ when $r>0$. This explains the ceiling function's jump: one extra item beyond a complete multiple demands a whole additional batch. For exact integer arithmetic, the identity $b=\lfloor (N+k-1)/k\rfloor$ avoids a fractional intermediate value; an implementation still needs an integer representation large enough to hold $N+k-1$.