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.