# Defying conventional calculus!

**Number Theory**Level 3

Which is the only positive integer whose \(k^\text{th}\) arithmetic derivative (where \(k \in \mathbb N\) ) is always equal to the number itself?

**Details and Assumptions:**

The arithmetic derivative function, denoted by \(n'\) is a function \(n: \mathbb{N} \cup \left\{0 \right\} \to \mathbb{N} \cup \left\{0 \right\}\) defined by

- \(n'=1\) for all prime numbers.
- \((ab)'=a' b+ a b'\) for \(a, b \in \mathbb{N} \cup \left\{0 \right\}\).
- \( n\overbrace { '''....'' }^{ { k \ times } } = (((n\overbrace { ')')'....')' }^{ { k \ times } } \)
- \(0'=1'=0\)

**Clarification:**

- The set of positive integers (or counting numbers or natural numbers) do not contain the member \(0\) in them.
- To be clear, it is asked to find a positive integer \(n\) which satisfies \[n = n' = n'' = n''' = \cdots = n\overbrace { '''....'' }^{ { k \ times } }\]