number properties

This topic has expert replies
Source: — Problem Solving |

Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 am
Thanked: 230 times
Followed by:21 members

by shankar.ashwin » Fri Jan 13, 2012 6:28 am
There are 49 numbers divisible by 2 between 1 and 100

there are 33 numbers divisible by 3 between 1 and 100

there are 16 numbers divisible by 6 (common to multiples of 2 and 3)

Hence numbers not divisible by 2 and 3 = 100 - (49+33-16) = 34 IMO

User avatar
Community Manager
Posts: 1060
Joined: Fri May 13, 2011 6:46 am
Location: Utrecht, The Netherlands
Thanked: 318 times
Followed by:52 members

by neelgandham » Fri Jan 13, 2012 7:16 am
Shouldn't it be 98-(49+33-16)=32 ?
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Fri Jan 13, 2012 8:47 am
sud21 wrote:How many numbers between 1 and 100 are not divisible by 2 and 3?
30 31 32 33 34
The wording of this question needs to be clarified.
Not divisible by 2 AND 3 means NOT DIVISIBLE BY 6.
The intended meaning here seems to be not divisible by 2 OR 3.
The GMAT likely would say INTEGERS instead of numbers.
BETWEEN x and y generally means NOT INCLUDING x and y, but I suspect that the intended meaning here is between 1 and 100, INCLUSIVE.
If this question were to appear on the GMAT, the wording would make the intended meaning crystal clear.

The solution below presumes that the question intends to ask the following:
How many integers between 1 and 100, inclusive, are neither even nor a multiple of 3?

This is an overlapping groups problem.
The big idea is to SUBTRACT THE OVERLAP.

Total integers = Even integers + Multiples of 3 - Multiples of 6 + Integers neither even nor a multiple of 3.

When we count the even integers and the multiples of 3, the OVERLAP between the two groups -- the multiples of 6 -- is counted twice.
Hence, the overlap -- the multiples of 6 -- must be SUBTRACTED from the total, as shown in the equation above.

Total integers = 100.
Even integers = 50. (Half are even, half are odd.)
Multiples of 3 = 100/3 = 33. (Ignore the decimal. Since 3 can be divided into 100 at most 33 times, there are 33 multiples of 3 between 1 and 100, inclusive.)
Multiples of 6 = 100/6 = 16. (Ignore the decimal. Since 6 can be divided into 100 at most 16 times, there are 16 multiples of 6 between 1 and 100, inclusive.)

Plugging these values into the equation above:

100 = 50 + 33 - 16 + N
98 = 67 + N
N = 33.

The correct answer is D.

For a similar but trickier problem, check my post here:

https://www.beatthegmat.com/how-many-int ... 88553.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Master | Next Rank: 500 Posts
Posts: 382
Joined: Thu Mar 31, 2011 5:47 pm
Thanked: 15 times

by ArunangsuSahu » Fri Jan 13, 2012 12:00 pm
1. Step 1:

Total no of numbers divisible by 2

100=2+(n-1)*2
n=50

2. Step 2:

NUmbers divisible by 3 =33

3. Numbers divisible by 6=16

Not Divisible by 2 and 3= 100-(50+33-16)..By Set theory
=33