| | Question 2

Find the maximum value of n such that 671! is perfectly divisible by 45n.

(A)

163

(B)

164

(C)

165

(D)

166

Answer: (C)
Explanation:

Prime Factor of 45= 3²x5 
We will count the number of 3^2 and 5 in 671!, and which one is lesser in number would be the answer. 
No of 3’s= 671/3 + 671/9 + 671/27 + 671/81 + 671/243 
= 223 + 74 + 24 + 8 + 2 
= 331 
No of 3²= 331/2 = 165
No of 5= 671/5 + 671/25 + 671/125 + 671/625 
= 134 + 26 + 5 + 1 = 166
165 will be the answer because 3² is lower in number than 5.

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