Find the largest integer $k$ such that $22^k \mid 1000!!$ .

**
Notation:
**
$!!$
denotes the
double factorial
notation. For example,
$10!!=10\times8\times6\times4\times2$
.

The answer is 49.

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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 ,

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 $d_p(n!) = k = \boxed{49}$ .