Prove that $(⌊ 33)$ is divisible by $2^{16}$ . What is the largest integer $n$ such that $(⌊ 33)$ is divisible by $2^{n}$ ?

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Solution

$\begin{matrix} o n e n t o f 2 i n 33! g r e a t e s t i n t e r g e r f u n c t i o n \leftarrow [\frac{33}{2}] + [\frac{33}{4}] + [\frac{33}{8}] + [\frac{33}{16}] + [\frac{33}{32}] = 16 + 8 + 4 + 2 + 1 = 33 i n 33!, 2^{33} w i l l c o m e t h e n w e c a n s a y 33! 18 d i v i s i b l e b y 2^{16} . h e n c e, l e s t v a l u e o f n i s 33. \end{matrix}$

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