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Prove that 33! is divisible by $2^{15}$ . What is the largest integer n such that 33! is divisible by $2^{n}$ ?

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None of the above

answer is C.

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Detailed Solution

Given, 33!=

33 \times 32 \times 31 \times 30 . . . . . . \times 2 \times 1

33!=

33 \times (2)^{5} \times 31 \times 30 \times 2^{4} . . . . \times 2^{3} . \times 2^{2} \times 3 \times 2^{1} \times 2^{0}

33!=

2^{15} \times 33 \times 31 \times 30 \times 29 \times . . . 17 \times 3 \times 1

Hence, 33! Is divisible by

2^{15}

Now, to find the max value of n, we divide the number by 2, 4 8, 16 and 32
We get

\frac{33}{2} + \frac{33}{4} + \frac{33}{8} + \frac{33}{16} + \frac{33}{32} = 16 + 8 + 4 + 2 + 1 = 31

Hence, n=31 and option 3 is the correct option

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