If 33! Is divisible by 2ⁿ , then the maximum value of n is equal to:

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answer is C.

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Detailed Solution

$[\frac{33}{2}] + [\frac{33}{2^{2}}] + [\frac{33}{2^{3}}] + [\frac{33}{2^{4}}] + [\frac{33}{2^{5}}] + [\frac{33}{2^{6}}]$ $= 16 + 8 + 4 + 2 + 1 + 0$ =31
$∴ E_{2} (33!) = 31, ∴ n = 31$

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