\(18^8 = 3^{16}*2^8\)
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
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I don't understand where the last two 3's come from when I try with 36!.

I get 15 threes, not 17.

36/3 = 12

Here we have already divided 27 in 3 once, and there should be two 3s left from 27. The total is now 14.

Now I have divided 9 once, and there should be one 3 left. The total is now 15.

All the threes are gone. Where do you get the extra two of them?


Here is what I understand:

36/3 = integer 12 - one left!!!!!
33/3 = integer 11
30/3 = integer 10
27/3 = integer 9 - two left!
24/3 = integer 8
21/3 = integer 7
18/3 = integer 6 - one left!!!!!!
15/3 = integer 5
12/3 = integer 4
9/3 = integer 3 - one left!
6/3 = integer 2
3/3 = integer 1

AHA! Now I get it. I forgot to count the 3's left in 12 and 6 that came from 36 and 18, respectively. However, these 3's were not obvious from the solutions presented in this thread.