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a
47
b
48
c
49
d
50
answer is C.
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Detailed Solution
Now,100!=1⋅2⋅3⋅…98⋅99⋅100=(1⋅2⋅4⋅5…98⋅100)(3⋅6⋅9⋅..96⋅99)=K⋅333(1⋅2⋅3…32⋅33)[∵ let K=1⋅2⋅4⋅5…98⋅100]=[K(1⋅2⋅4…31⋅32)]333⋅(3⋅9⋅12…30⋅33)=K1⋅333⋅311(1⋅2⋅3…10⋅11)∵ let K(1⋅2⋅4…31⋅32)=K1=K1(1⋅2⋅4…10⋅11)333⋅311(3⋅6⋅9⋅12)=K2333⋅311⋅34(1⋅2⋅3⋅4)=K3⋅333⋅311⋅34⋅3 ∵ let K2(1⋅2⋅3⋅4)=K3=K3⋅349Hence, exponent of 3 is 49