A) 33
B) 44
C) 48
D) 52
Correct Answer: C
Solution :
Let \[E(n)\] denote the exponent of 3 in \[n\]. The greatest integer less than 100 divisible by 3 is 99. We have \[E(100\ !)=E(1\ .\ 2\ .\ 3\ .\ 4....99\ .\ 100)\] \[=E(3\ .\ 6\ .\ 9\ ....99)\]\[=E[(3\ .\ 1)(3\ .\ 2)(3\ .\ 3)........(3\ .\ 33)]\] \[=33+E(1\ .\ 2\ .\ 3......33)\] Now \[E(1\ .\ 2\ .\ 3......33)=E(3\,\ .\ 6\ .\ 9....33)\] \[=E[(3\ .\ 1)(3\ .\ 2)(3\ .\ 3)........(3\ .\ 11)]\] \[=11+E(1\ .\ 2\ .\ 3\ .....11)\] and \[E(1\ .\ 2\ .\ 3\ ....11)=E(3\ .\ 6\ .\ 9)=E[(3\ .\ 1)(3\ .\ 2)(3\ .\ 3)]\] \[3+E(1\ .\ 2\ .\ 3)=3+1=4\] Thus\[E(100\ !)=33+11+4=48\].
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