The order of an internal node in a B+ tree index is the

The order of an internal node in a B⁺ tree index is the maximum number of children it can have. Suppose that a child pointer takes 6 byte, the search field value takes 14 byte, and the block size is 512 byte. What is the order of the internal node?

Let the order of the internal node be X,
Then from the given,
(X – 1) 14 + 6X ≤ 512
⇒ 15X – 14 + 6X ≤ 512
20X ≤ 256
X = 26