$22^k$ Dividing 1000!!

By definition of double factorial , $1000!! = 1000 \times 998 \times \dots \times 4 \times 2$ Since each of the terms is divisible by $2$ , express $1000!!!$ as $\begin{array}{rl} 1000!! &= \left({\color{#D61F06}2} \times 500\right) \times \left({\color{#D61F06}2} \times 499\right) \times \dots \times \left({\color{#D61F06}2} \times 2\right) \times {\color{#D61F06}2}\\ &= {\color{#D61F06}2}^{500} \times 500! \end{array}$ We can apply the following formula ,

$d_p(n!) = \sum\limits_{i=1}^{\lfloor \log_p(n)\rfloor} \left\lfloor \dfrac{n}{p^i}\right\rfloor$

for prime number $p$

Since there are less terms of large prime numbers (i.e. $11$ ) than terms of small prime numbers (i.e. $2$ ) in the prime factorization of factorial numbers, it suffices to determine the value of $k$ by setting $p = 11$ and $n = 500$ . Then, $\lfloor log_p(n) \rfloor = 2$ , which gives $d_p(n!) = k = \boxed{49}$ .

In 1000!! there are :

1000 / 22 = 45 multiples of 22

1000 / 22^ 2 = 2 multiples of 22^ 2

1000 / 22^ 3 < 1 hence no multiple of 22 ^ n with n > 2

Moreover, since 22 = 2 × 11, we must also consider multiples of 11 different from those of 22 already counted :

1000 / 11^ 2 = 8 which we have to divide by 2 to get the even ones and by 2 again to leave out the multiples of 22 ^ 2 which gives us 2

Same as with 22, there are no multiples of 11^ n smaller than 1000 with n > 2

Hence we end up with : k = 45 + 2 + 2 = 49