The highest power of $9$ dividing $99!$ completely is

### Concept & Formula To find the highest power of a composite number (like $9$) in a factorial ($n!$), we must first break it down into its prime factors. Since $9 = 3^2$, we need to find the highest power of the prime number $3$ in $99!$, and then halve it. We use Legendre's Formula to find the highest power of a prime $p$ in $n!$: $$E_p(n!) = \lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor + \lfloor \frac{n}{p^3} \rfloor + ...$$ ### Step-by-Step Solution * **Given:** * Factorial: $99!$ * Divisor: $9$ (which is $3^2$) * **Calculation / Deduction:** 1. Find the highest power of $3$ in $99!$ using Legendre's Formula by successively dividing $99$ by $3$ and adding the quotients (ignoring remainders). 2. $\lfloor 99 / 3 \rfloor = 33$ 3. $\lfloor 33 / 3 \rfloor = 11$ 4. $\lfloor 11 / 3 \rfloor = 3$ 5. $\lfloor 3 / 3 \rfloor = 1$ 6. Total power of $3 = 33 + 11 + 3 + 1 = 48$. So, $99!$ contains $3^{48}$. 7. Since we need the highest power of $9$, and $9 = 3^2$, we group the powers of $3$ in pairs. 8. $3^{48} = (3^2)^{24} = 9^{24}$ ### Exam Strategy & Shortcut Use the successive division method on the prime factor $3$. Divide $99$ by $3$, write the integer quotient below it, and repeat until the quotient is $0$. Sum the quotients: $33 + 11 + 3 + 1 = 48$. Since $9$ is $3$ squared, immediately divide $48$ by $2$ to get $24$. This takes less than $20$ seconds. ### Common Pitfall The most common mistake is applying Legendre's formula directly with $9$ (i.e., $99/9 = 11$, $11/9 = 1 \implies 12$). Legendre's formula *only* works for prime numbers. You must always reduce the divisor to its prime factors first. ### Final Answer Therefore, the correct answer is 24.