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In this question, we will be using Legendre's Formula ((E=[n/p]+[n/p^2]+...)) to count prime factors in any factorial.

As 135 = (3^3)*(5):

We will have to find out the highest power of 5 in 200! = [200/5]+[200/5^2]+[200/5^3] = 40+8+1 = 49

Similarly, we will have to find out the highest power of 27 in 200!. For the same, we will first have to find out the highest power of 3 in 200! (as the above formula is only valid for prime numbers)

Highest power of 3 in 200! = [200/3]+[200/3^2]+[200/ 3^3 ]+[200/3^4] = 66+21+7+2 = 96

Thus, the highest power of 27 in 200! = (3^3)^32

Since the highest power of 27 < highest power of 5, the highest integer k for which 135^k perfectly divides 200! = 32.

Hence, the correct answer is Option (D) 32. (Answer)
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What is the highest integer k for which 135^k perfectly divides 200! ?

A. 97
B. 49
C. 40
D. 32
E. 1
Video explanation:


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