Problem Solving

What is the largest integer value of x that makes 30!/3^x an integer?

A)10
B)12
C)14
D)16
E)18

Thanks in advance!

My answer is 14.

here it goes..

30/3 + 30/3^2 + 30/3^3

10 + 3 + 1 = 14

let me know the OA,

You need to know the number of 3 you can have in 30!

You take the multiples of 3 and you see the number of 3 you have in, it takes 20sec with a paper.
3-->1
6-->1
9-->2
...
30-->1

It gives 14


Sudhir could you develop your point cause I couldn't use it, I don't catch.

I'm not entirely sure how sudhir3127 got the 14, but that is what I get as well by doing the following procedure:

If I'm going to take 30! and divide by a bunch of 3's, I would like to divide by all of the 3 factors in 30! to find the largest value for x.

I can figure this out by looking at all of the factors of 30! (30,29,28, etc) that will, themselves, have factors of 3 in them. At least that will be 30, 27, 24, ... , 6, 3. I can exclude examining any of the other factors of 30! (28, 19, ect.) as dividing them by 3 will not yield an integer.

These are the factors of 30! that have at least one 3 factor in them, some will have more than one 3 as a factor as follows.

3 = 3
6 = 2*3
9 = 3*3
12 = 2*2*3
15 = 5*3
18 = 2*3*3
21 = 7*3
24 = 2*2*2*3
27 = 3*3*3
30 = 2*5*3

Now, if I count up the 3's present in my chart, I get 14. I can divide 30! by 3 fourteen times and that will cancel out all of the threes in 30! but still leave me with an integer.

Hope this helps, let me know if I need to explain some more.

Thats a million thats great!

sudhir3127 wrote:My answer is 14.

here it goes..

30/3 + 30/3^2 + 30/3^3

10 + 3 + 1 = 14

let me know the OA,

This is sheer brilliance!!!!!!

Thanks a lot dude.