20! is divisible

[sanju09]

If 20! is divisible by 2^x, what is the greatest possible value of x?
A. 20
B. 18
C. 17
D. 15
E. 10



OA B

[Sher1]

18

20! has can be broken down as follows

20 = 2x2x5
18 = 2x9
16 = 2x2x2x2
so on

add the twos up and you get 18

so 20! should be divisible by 2^18

[sanju09]

18

20! has can be broken down as follows

20 = 2x2x5
18 = 2x9
16 = 2x2x2x2
so on

add the twos up and you get 18

so 20! should be divisible by 2^18

Well done Lion#1!

Any short-cut?

[Sher1]

Can't think of a short cut but would love to know if anyone can figure out

[lilu]

This shortcut was described by Ian awhile ago.
When you need to find the highest power of a factor of a certain number (in this case we need to find the highest power of 2 that is a factor of 20!), we need to follow these steps:

20!/2^x=20/2^1+20/2^2+20/2^3+20/2^4-->
20/2 is 10
20/4 is 5
20/8 is appr. 2 (need to take integers only)
20/16 is appr. 1

You can't take 2^5 because that is going to be greater than 20,so need to take numbers that will be factors of the number in question!!

So, now add the results-->10+5+2+1=18


That is the answer.
You can try this for different kinds of numbers and you'll see that this shortcut works and takes less than a minute to find the answer.

[KICKGMATASS123]

cool method!

Thanks for sharing Lilu

Shruti

[dwilliams05]

i obviously have a looong way to go to conquer this quant in order to reach that 700 range. you people are good!! thanks for the help

[Musiq]

You can call this a "short cut" if you want...but the below method really uses number properties:

There are 10 even numbers in the first 20 integers.....so you need atleast 10 2's to account for the division.

Then you have numbers like 4, 8 and 16 which are multiple powers of 2.You have already counted one of the 2's in these numbers when you counted the even numbers (since 4, 8 and 16 are also even).

Therefore the greatest value of X = 10 + 1 + 2 + 3 = 18.