B+ Tree(order of Internal node)

Question

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 $3$ bytes, the search field value takes $17$ bytes, and the block size is $1024$ bytes. The order of the internal node is_______.

Comments

suppose order N of internal nodes

so N(child pointer)+(N-1)(Search key) <= block size

    3N+17N-17<=1024

 So N = 52

Answer 1

Block pointer size  = 3B
Search key size = 17B
Block size = 1024B
Let the order of the internal node of the B+ tree be p.
Therefore, p*Block pointer size + (p-1)*Search key size <= Block size
=> p*3 + (p-1)*17 <=1024
=> p<=52.05
=> p=52

Order of the internal node of the given B+ tree is 52.

Answer 2

suppose order N of internal nodes 

so N(child pointer)+(N-1)(Search key) <= block size 

    3N+17N-17<=1024

 So, N = 52