The largest natural number n such that 3^n divides 66! is: — Solution | JEE Mathematics

NVAEasyJEE 2023Applications of P&C

The largest natural number $n$ such that $3^n$ divides $66!$ is:

Answer

Correct answer:31

Standard Method

Given: We need the largest natural number $n$ such that $3^n$ divides $66!$ .

Find: The highest power of $3$ in $66!$ .

Step 1: Use Legendre’s formula.

The largest power of a prime $p$ dividing $n!$ is given by:

\left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \dots

Step 2: Apply for $p = 3$ and $n = 66$ .

\left\lfloor \frac{66}{3} \right\rfloor + \left\lfloor \frac{66}{9} \right\rfloor + \left\lfloor \frac{66}{27} \right\rfloor = 22 + 7 + 2 = 31

Therefore, the largest $n$ is $31$ .

Quick Tip

Given: We want the highest power of $3$ dividing $66!$ .

Find: The required exponent.

Quick Tip: To find the highest power of a prime dividing $n!$ , use successive divisions by powers of the prime.

Here, divide $66$ by $3, 9, 27$ and add the integer parts:

22 + 7 + 2 = 31

So, the required value is $31$ .

Stopping after $\left\lfloor \frac{66}{3} \right\rfloor = 22$ is incorrect because higher powers of $3$ also contribute additional factors. Continue with $\left\lfloor \frac{66}{9} \right\rfloor$ and $\left\lfloor \frac{66}{27} \right\rfloor$ as well.
Using ordinary division instead of floor values is wrong because Legendre’s formula counts only complete multiples. Always take the integer part in each term.
Continuing beyond the necessary terms without checking the power can cause confusion. Once the next power exceeds $66$ , the contribution becomes zero, so the sum stops there.

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