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Factorials - anyone know a really quick way

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by sudhir3127 » Thu Jul 31, 2008 8:50 am
My answer is 14.

here it goes..

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

10 + 3 + 1 = 14

let me know the OA,
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by pepeprepa » Thu Jul 31, 2008 9:02 am
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.
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by VP_RedSoxFan » Thu Jul 31, 2008 9:06 am
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.
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by [email protected] » Thu Jul 31, 2008 9:33 am
Thats a million thats great!
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by Ian Stewart » Thu Jul 31, 2008 2:08 pm
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,
Nice, but I think some didn't quite understand it :)

What sudhir did:

-count how many multiples of 3 there are between 1 and 30- there are 30/3, or ten multiples of 3.

-we need to count an extra 3 for the multiples of 3^2. How many of these are there? 30/9 = 3+remainder; there will be three multiples of 3^2.

-finally, we need to count yet another 3 for the multiples of 3^3. How many of these are there? 30/27 = 1+remainder; there will be only one multiple of 3^3.

10+3+1 = 14.

The approach does depend on the fact that the range of numbers begins from 1; with a different set of numbers, the calculations would be more awkward (in a range of 30 numbers, you can have one multiple of 3^3, or you can have two multiples of 3^3).

It's an elegant approach- nicely done, sudhir.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by parallel_chase » Thu Jul 31, 2008 3:05 pm
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.
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