Question

If 1000! = 3ⁿ × m, where m is an integer not divisible by 3, then n =?

• A. 498
• B. 298
• C. 398
• D. 98

Hint:

For factorial problems, use the formula for the highest power of a prime factor to break the problem into smaller steps.

Answer 1

The Correct Option is A

1. Formula for Highest Power of a Prime in Factorials: - The highest power of a prime \( p \) dividing \( n! \) is given by:

\[ n_p = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \dots \]

2. Substitute \( n = 1000 \) and \( p = 3 \):

\[ n_3 = \left\lfloor \frac{1000}{3} \right\rfloor + \left\lfloor \frac{1000}{3^2} \right\rfloor + \left\lfloor \frac{1000}{3^3} \right\rfloor + \dots \]

3. Compute each term:

\[ n_3 = \left\lfloor \frac{1000}{3} \right\rfloor + \left\lfloor \frac{1000}{9} \right\rfloor + \left\lfloor \frac{1000}{27} \right\rfloor + \left\lfloor \frac{1000}{81} \right\rfloor + \left\lfloor \frac{1000}{243} \right\rfloor + \left\lfloor \frac{1000}{729} \right\rfloor. \]

• \( \left\lfloor \frac{1000}{3} \right\rfloor = 333 \),
• \( \left\lfloor \frac{1000}{9} \right\rfloor = 111 \),
• \( \left\lfloor \frac{1000}{27} \right\rfloor = 37 \),
• \( \left\lfloor \frac{1000}{81} \right\rfloor = 12 \),
• \( \left\lfloor \frac{1000}{243} \right\rfloor = 4 \),
• \( \left\lfloor \frac{1000}{729} \right\rfloor = 1 \).

4. Sum up the terms:

\[ n_3 = 333 + 111 + 37 + 12 + 4 + 1 = 498. \]

5. Thus, \( n = 498 \).