SATHEE: Permutations And Combinations Question 37

Permutations And Combinations Question 37

Question: The exponent of 3 in $ 100\ ! $ is

Options:

A) 33

B) 44

C) 48

D) 52

Show Answer

Answer:

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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Permutations And Combinations Question 37

Question: The exponent of 3 in $ 100\ ! $ is

Options:

Answer:

Solution:

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