10 can be split in to 5 * 2
In 30! 5 is the bottle neck that will limit 10's powers and not 2 since there are more 2's (every 2nd number is divisible by 2 whereas only every 5 the number is divisible by 5)
There is kind of a shortcut to finding this.
Divide 30 by increasing powers of 5 till u get 0 as the integer quotient and add all the integer quotients in the process
i.e 30/5 + 30/5^2 +30/5^3
= 6 + 1 + 0 (we stop here)= 7
Refer to this post (it will be more clear) - https://www.beatthegmat.com/ps-confused- ... html#92483. Ian has provided an explanation to this shortcut
Now to the actual problem:
Stmt I
10^ d is factor of F = 30!
By doing what I mentioned in bold above we see 10^7 is the highest power of 10 that will divide 30! (d can be any number from 0 to 7)
INSUFF
Stmt II
d>6
We have no idea how D relates to F
INSUFF
Combining I and II
we know d = 7
SUFF
C)
Consecutive Integer Problem
This topic has expert replies
Source: Beat The GMAT — Data Sufficiency |
-
iamcste
- Legendary Member
- Posts: 940
- Joined: Tue Aug 26, 2008 3:22 am
- Thanked: 55 times
- Followed by:1 members
https://www.manhattangmat.com/forums/if- ... t2234.htmlhypik21 wrote:Whats the quickest way to solve this?
If D is a positive integer and F is the product of the first 30 integers, what is the calue of D.
1. 10^d is a factor of F
2. d> 6
OA: c
This is the expert way!
