Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
The maximum value of $$n$$ such that $$2^n$$ divides $$21!$$ is
The power of 2 in $$21!$$ is $$\left[\frac{21}{2}\right] + \left[\frac{21}{4}\right] + \left[\frac{21}{8}\right] + \left[\frac{21}{16}\right] = 10 + 5 + 2 + 1 = 18$$. So the maximum $$n$$ is 18.
Was this solution helpful?
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.