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The largest natural numer $n$ for which $(33)!$ is divisible by $2^n$ is ?

$\begin{array}{1 1} 28 \\ 30 \\ 31 \\ 33 \end{array}$

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$(33)!=1.2.3.4.5................29.30.31.32.33$
$=(2.4.6.............32).(1.3.5.................31.33)$
$=(2.2.................16\:times).(1.2.3...........15.16).(1.3.5...............31.33)$
$=2^{16}.(2.4.6.........12.14.16).(1.3.5.......13.15).(1.3.5............31.33)$
$=2^{16}.(2.2.2.....8\:times).(1.2.3.....7.8).(1.3.5.....13.15).(1.3............31.33)$
$=2^{24}.(2.4.6.8).(1.3.5.7).(1.3.5.....13.15)(1.3.5.......31.33)$
$=2^{24}.2^4.(1.2.3.4).(1.3.5.7).(1.3.5.........13.15).(1.3.5............31.33)$
$=2^{28}.2^3.(1.3.)(1.3.5.7).(1.3......13.15).(1.3..........31.33)$
$=2^{31}.(1.3.)(1.3.5.7).(1.3......13.15).(1.3..........31.33)$
$\therefore$ Th least $n=31$
answered Aug 20, 2013 by rvidyagovindarajan_1
 

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