m25#02 : Retired Discussions [Locked] - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 20 Feb 2017, 04:45

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m25#02

Author Message
Manager
Joined: 21 Nov 2010
Posts: 133
Followers: 0

Kudos [?]: 5 [0], given: 12

### Show Tags

01 Dec 2011, 23:55
There are 6 between 0 and 20 plus 1 for 10!. Answer is B.
Manager
Joined: 27 Oct 2011
Posts: 191
Location: United States
Concentration: Finance, Strategy
GMAT 1: Q V
GPA: 3.7
WE: Account Management (Consumer Products)
Followers: 5

Kudos [?]: 153 [0], given: 4

### Show Tags

03 Feb 2012, 21:36
7... its important to remember that 10! is a multiple of 3
_________________

DETERMINED TO BREAK 700!!!

Math Expert
Joined: 02 Sep 2009
Posts: 37036
Followers: 7230

Kudos [?]: 96112 [0], given: 10707

### Show Tags

04 Feb 2012, 08:39
duuuma wrote:
How many integers are divisible by 3 between $$10!$$ and $$10! + 20$$ inclusive?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

As soon as you realize that 10! itself is a multiple of 3, then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 form 0 to 20, inclusive?
$$# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1$$.

So, (18-0)/3+1=7

Hope it's clear.
_________________
Intern
Joined: 21 Feb 2012
Posts: 10
Followers: 0

Kudos [?]: 5 [0], given: 3

### Show Tags

06 Mar 2012, 18:45
Why do we consider 10! here as the Q asks between 10! and 10! +20 which doesn't include end points
Math Expert
Joined: 02 Sep 2009
Posts: 37036
Followers: 7230

Kudos [?]: 96112 [0], given: 10707

### Show Tags

06 Mar 2012, 20:54
mithun2vrs wrote:
Why do we consider 10! here as the Q asks between 10! and 10! +20 which doesn't include end points

Welcome to GMAT Club.

In fact we ARE told that 10! and 10!+20 should also be considered: "How many integers are divisible by 3 between 10! and 10!+20, INCLUSIVE?"

Hope it's clear.
_________________
Manager
Joined: 02 May 2012
Posts: 109
Location: United Kingdom
WE: Account Management (Other)
Followers: 0

Kudos [?]: 54 [0], given: 34

### Show Tags

18 Sep 2012, 07:00
1
This post was
BOOKMARKED
This question looks a lot more complicated than it actually is. Lets break it down:
1) 10! is a multiple of 3 b/c 10x9x8x7x6x5x4x3x2x1 (ignoring the 6 and 9 for simplicity)

2) Two multiples of n added together are also a multiple of n e.g. 3+6 = 9

3) Therefore the question is really how many multiples of 3 are between 0 and 20 (of which there are 6)

4) Then the final bit of housekeeping: add one because the '0' above is actually 10!, which is a multiple of 3.

So the total is 7
_________________

In the study cave!

Moderator
Joined: 01 Sep 2010
Posts: 3132
Followers: 806

Kudos [?]: 6764 [0], given: 1049

### Show Tags

03 Dec 2012, 07:17
Bunuel wrote:
duuuma wrote:
How many integers are divisible by 3 between $$10!$$ and $$10! + 20$$ inclusive?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

As soon as you realize that 10! itself is a multiple of 3, then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 form 0 to 20, inclusive?
$$# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1$$.

So, (18-0)/3+1=7

Hope it's clear.

Quote:
This rule applies to any number of consecutive integers: The product of k consecutive integers is always divisible by k factorial (K!).

between 0 and 20 we have 6 number + 1 from 10! = 7

As always, it seems difficult but is not at all
_________________
Intern
Joined: 31 Aug 2012
Posts: 47
GPA: 3.65
Followers: 0

Kudos [?]: 8 [0], given: 52

### Show Tags

03 Dec 2012, 09:31
duuuma wrote:
How many integers are divisible by 3 between $$10!$$ and $$10! + 20$$ inclusive?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

I'm glad to see This is the first arthematic question which I got it right on the first step in this forum,,,:)
Intern
Joined: 30 Aug 2012
Posts: 8
Followers: 0

Kudos [?]: 3 [0], given: 12

### Show Tags

04 Dec 2012, 13:39
10! is for sure divisible by 3
so 10!+ 20 will have 10!+3, 10!+6,10!+9,10!+12, 10!+15 & 10!+18 total 6 + 10! which is alreday divisible by 3
so 6+1 =7
Manager
Joined: 04 Oct 2013
Posts: 162
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Followers: 3

Kudos [?]: 105 [0], given: 54

### Show Tags

28 Nov 2013, 21:20
nerd wrote:
10! is for sure divisible by 3
so 10!+ 20 will have 10!+3, 10!+6,10!+9,10!+12, 10!+15 & 10!+18 total 6 + 10! which is alreday divisible by 3
so 6+1 =7

10! is divisible by 3 as 10! = 2 * 3 *4 * 5* 6* 7* 8* 9* 10
The nearest and farthest multiple in the range, 10! and 10! +18
The difference between the extremes 10! + 18 – 10! = 18
So number of multiples in the range (18/3) + 1 = 7
Hence, option (B) is the answer.
Re: m25#02   [#permalink] 28 Nov 2013, 21:20

Go to page   Previous    1   2   [ 30 posts ]

Display posts from previous: Sort by

# m25#02

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.