If 1000! = 3^n times m where m is an integer | vedclass

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(A) To find the exponent of a prime $p$ in the prime factorization of $N!$, we use Legendre's Formula: $E_p(N!) = \sum_{k=1}^{\infty} \left[ \frac{N}{

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$498$

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(A) To find the exponent of a prime $p$ in the prime factorization of $N!$, we use Legendre's Formula: $E_p(N!) = \sum_{k=1}^{\infty} \left[ \frac{N}{p^k} \right]$.Here, $N = 1000$ and $p = 3$.$E_3(1000!) = \left[ \frac{1000}{3} \right] + \left[ \frac{1000}{9} \right] + \left[ \frac{1000}{27} \right] + \left[ \frac{1000}{81} \right] + \left[ \frac{1000}{243} \right] + \left[ \frac{1000}{729} \right]$.Calculating each term:$\left[ 333.33 \right] = 333$$\left[ 111.11 \right] = 111$$\left[ 37.03 \right] = 37$$\left[ 12.34 \right] = 12$$\left[ 4.11 \right] = 4$$\left[ 1.37 \right] = 1$.Summing these values: $333 + 111 + 37 + 12 + 4 + 1 = 498$.Therefore, $n = 498$.

If $1000! = 3^n \times m$ where $m$ is an integer not divisible by $3$, then $n = $

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