\(18^8 = 3^{16}*2^8\)
thus the number should atleast be divisible by 3^16
look for option
highest power of 3 dividing the numbers will be
a) 33! will be divisible by =\( \frac{33}{3}+\frac{33}{9}+\frac{33}{27} = 15\) (nope)
b) 36! = \( \frac{36}{3}+\frac{36}{9}+\frac{36}{37 }\)= 12+4+1= 17 thus this will be divisible
hence B
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Solution

• Prime factorization of \(18 = 2*3^2\)

o \(18^8 = 2^8*3^{16}\)
• For, \( \frac{n! }{18^8} \) to be an integer, prime factorization of n! must contains power of \(3 ≥ 16\)

o If n! contains \(3^{16}\) then power of 2 in n! must be greater than 16. So, we need not to be worried about the powers of 2.
Now, let us check the answer options:
• Option A. \(n= 33\)

o Powers of 3 in 33! \(= [\frac{33}{3}] + [\frac{33}{9}] + [\frac{33}{27}] = 11+3+1 = 15 \) which is less than 16.

 Here, [] represents the greatest integer function.
o So, A cannot be the answer.
• Option B. \(n = 36\)

o Powers of 3 in \(36! = [\frac{36}{3}] + [\frac{36}{9}] + [\frac{36}{27}] = 12+4+1 = 17\) which is greater than 16.

 Here, [] represents the greatest integer function.
o Hence, B can be the answer.

Since all other options are greater than 36 and we have asked for smallest integer n, so we need not to check further options.
Thus, the correct answer is Option B.
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Hi
Can I have some other qick solutions to this tricky problem?

https://www.youtube.com/watch?v=fxJBwPCEMUA&t=602