The largest n in mathbbN$n \in \mathbb{N}$, for which 7^n$7^n$ divides 101!, is:

### Related Formula textLegendre's Formula: The exponent of a prime p text in N! text is given by: $$\text{Legendre's Formula: The exponent of a prime } p \text{ in } N! \text{ is given by: }$$ E_p(N!) = leftlfloor fracNp rightrfloor + leftlfloor fracNp^2 rightrfloor + leftlfloor fracNp^3 rightrfloor + ldots$$E_p(N!) = \left\lfloor \frac{N}{p} \right\rfloor + \left\lfloor \frac{N}{p^2} \right\rfloor + \left\lfloor \frac{N}{p^3} \right\rfloor + \ldots$$ ### Core Logic To find the maximum power n$n$ such that 7^n$7^n$ divides 101!$101!$, we need to find the exponent of the prime 7$7$ in the prime factorization of 101!$101!$. ### Step 1: Apply Legendre's Formula n = leftlfloor frac1017 rightrfloor + leftlfloor frac1017^2 rightrfloor + leftlfloor frac1017^3 rightrfloor + ldots$$n = \left\lfloor \frac{101}{7} \right\rfloor + \left\lfloor \frac{101}{7^2} \right\rfloor + \left\lfloor \frac{101}{7^3} \right\rfloor + \ldots$$ n = leftlfloor frac1017 rightrfloor + leftlfloor frac10149 rightrfloor + leftlfloor frac101343 rightrfloor$$n = \left\lfloor \frac{101}{7} \right\rfloor + \left\lfloor \frac{101}{49} \right\rfloor + \left\lfloor \frac{101}{343} \right\rfloor$$ n = 14 + 2 + 0$n = 14 + 2 + 0$ n = 16$n = 16$ ### Pattern Recognition For prime p$p$ in N!$N!$, iteratively divide N$N$ by p$p$ taking only integer parts and sum them. Fast mental math: 101 div 7 = 14$101 \div 7 = 14$, 14 div 7 = 2$14 \div 7 = 2$. 14+2=16$14+2=16$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations Class 11 Maths: Number Theory