The integers from 1 to 100 inclusive are each written on a : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 24 Jan 2017, 06:53

### 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

# The integers from 1 to 100 inclusive are each written on a

Author Message
TAGS:

### Hide Tags

Manager
Joined: 15 Jul 2004
Posts: 73
Followers: 3

Kudos [?]: 26 [1] , given: 0

The integers from 1 to 100 inclusive are each written on a [#permalink]

### Show Tags

09 Apr 2005, 00:13
1
KUDOS
9
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

57% (02:30) correct 43% (01:23) wrong based on 785 sessions

### HideShow timer Statistics

The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

A. 83%
B. 67%
C. 50%
D. 33%
E. 17%
[Reveal] Spoiler: OA

_________________

===========================
Let us make hay while the sun shines.
Don Quixote. Part i. Book. iii. Chap. xi.

VP
Joined: 30 Sep 2004
Posts: 1488
Location: Germany
Followers: 6

Kudos [?]: 327 [1] , given: 0

### Show Tags

09 Apr 2005, 00:44
1
KUDOS
67%...

50/100 (all multiples of 2) + 33/100 (all multiples of 3) - 1650/10000 (multiples of 3 and 2) =6650/10000=67%...
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy
Followers: 1

Kudos [?]: 23 [1] , given: 0

### Show Tags

09 Apr 2005, 01:39
1
KUDOS
odd multiples of 3 are 17/100
3 3+6 3+6+6 3+6+6+6 etc
50/100+17/100=67/100
Senior Manager
Joined: 15 Mar 2005
Posts: 419
Location: Phoenix
Followers: 2

Kudos [?]: 26 [2] , given: 0

### Show Tags

09 Apr 2005, 02:20
2
KUDOS
1
This post was
BOOKMARKED
Nither even nor a multiple of 3.

Even numbers between 1 and 100 are 2, 4, 6 ... 100 (50 numbers).

Numbers that are multiple of 3 are 3, 6, 9 ... 99. However, all even multiples of 3 have been counted already in even numbers. Therefore, this set includes 3, 9, 15, 21 ... 99. (17 numbers).

Thus total numbers that are to be excluded = 50 + 17 = 67.

Therefore probability that a randomly picked number is nither even nor a multiple of 3 = 33%
_________________

Who says elephants can't dance?

Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy
Followers: 1

Kudos [?]: 23 [0], given: 0

### Show Tags

09 Apr 2005, 02:23

Options >=50% should be ruled out soon because numbers range from 1 to 100, and thus subtracting 50 even you can't have more than 49 (you know at least that 3 is an odd multiple of 3, son you have at least 1 in 50 remaining numbers)

This question is asking nothing more than:
"how many odd numbers are in the first 100 positive ones that are not multiple of 3?"
Senior Manager
Joined: 15 Mar 2005
Posts: 419
Location: Phoenix
Followers: 2

Kudos [?]: 26 [0], given: 0

### Show Tags

10 Apr 2005, 02:37
thearch wrote:

Options >=50% should be ruled out soon because numbers range from 1 to 100, and thus subtracting 50 even you can't have more than 49 (you know at least that 3 is an odd multiple of 3, son you have at least 1 in 50 remaining numbers)

This question is asking nothing more than:
"how many odd numbers are in the first 100 positive ones that are not multiple of 3?"

Careful theArch, that wall is rather fragile !!!

_________________

Who says elephants can't dance?

Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy
Followers: 1

Kudos [?]: 23 [1] , given: 0

### Show Tags

10 Apr 2005, 02:46
1
KUDOS
However, I was deceived by christoph, because I calculated 67/100 and then I looked at answers already given, and forgot to subtract. This wouldn't happen in the actual test
Manager
Joined: 17 Dec 2004
Posts: 72
Followers: 2

Kudos [?]: 13 [1] , given: 0

### Show Tags

10 Apr 2005, 05:49
1
KUDOS
33%

100 slips of paper: 50 are even, and 17 are odd multiples of 3

100-(50+17)=33
Director
Joined: 18 Feb 2005
Posts: 673
Followers: 1

Kudos [?]: 6 [0], given: 0

### Show Tags

10 Apr 2005, 05:59
thearch wrote:

Options >=50% should be ruled out soon because numbers range from 1 to 100, and thus subtracting 50 even you can't have more than 49 (you know at least that 3 is an odd multiple of 3, son you have at least 1 in 50 remaining numbers)

This question is asking nothing more than:
"how many odd numbers are in the first 100 positive ones that are not multiple of 3?"

This happens with many of us....We get there to the answer and forget subtracting.......GMAT has both the answers as a trap and we should be careful enough not to fall in it........
Senior Manager
Joined: 15 Mar 2005
Posts: 419
Location: Phoenix
Followers: 2

Kudos [?]: 26 [0], given: 0

### Show Tags

11 Apr 2005, 11:24
Zem wrote:
but how you found the number: 17?

Zem, we are trying to find the numbers that are nither divisible by 2 nor 3.

If you take the numbers that are divisible by 2 away, you've taken all even numbers between 1 and 100 away. They are (100-1)/2 = 50 numbers.

We could repeat the same process with 3. However, there are numbers that are common to 2 and 3, and we don't have to take them out (they were accounted for in the even numbers list anyway).

So we see what are these numbers we're looking at (odd multiples of 3).
They are 3, 9, 15, 21 and so on.

There're a plathora of ways to figuring out how many such numbers would exist between 1 and 100. I'd list out two here.

One is the way of sequences.

the last number in this list would be 99.
99 = 3 + (n - 1) 6 => n = 17.

Else do a 100/16 = 16.xxx. 16*6 = 96. but since it starts at 3, the 16th multiple would be 99. 16 multiples, and add one (for 3) and you've 17.

Thus we have the number 17.

Hope that helps.
_________________

Who says elephants can't dance?

VP
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 45 [0], given: 0

### Show Tags

11 Apr 2005, 19:30
I think that in this problem it's easier to find directly the prob than to use the formula 1-(opposite prob)

from 1 to 100 : 100 numbers
even numbers : 50
multiples of 3 : 33 (3*1, 3*2, ...3*33) however in those 33 numbers there are already even numbers and you have to be careful to not repeat the even numbers you've already taken out so finally you will get 17 odd and 16 even (you can see that everytime 3 is multiply by an even number the final result is obviously even so 1/2 of the multiples of 3 will be even and 1/2 odd, howver the last one is odd so there is an additionnal odd one)

17 odd multiples of 3 + 50 even numbers = 67

100-67 = 33
Manager
Joined: 22 Jan 2012
Posts: 90
Location: India
Concentration: General Management, Technology
GMAT 1: Q39 V29
GPA: 3.3
WE: Engineering (Consulting)
Followers: 6

Kudos [?]: 103 [1] , given: 9

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

### Show Tags

18 Mar 2012, 19:56
1
KUDOS
Simple to solve using P(n) = 1-X concept

P = 1 - [ P(even 0r multiples of 2)+P(multiples of 3) - P(multiples of 2 and 3 or multiples of 6)]
P = (1 - [ 50/100+33/100 - 16/100]) = 33%
_________________

Press +1 Kudos rather than saying thanks

Ill be posting good questions as many as I can...

Towards Success

Math Expert
Joined: 02 Sep 2009
Posts: 36625
Followers: 7105

Kudos [?]: 93642 [2] , given: 10583

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

### Show Tags

19 Mar 2012, 02:48
2
KUDOS
Expert's post
4
This post was
BOOKMARKED
Zem wrote:
The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

A. 83%
B. 67%
C. 50%
D. 33%
E. 17%

# of multiples of 2 in the range (100-2)/2+1=50 (check this: totally-basic-94862.html#p730075);
# of multiples of 3 in the range (99-3)/3+1=33;
# of multiples of both 2 and 3, so multiples of 6, in the range (96-6)/6+1=16 (to get the overlap of above two sets);

Hence there are total of 50+33-16=67 numbers from 1 to 100, inclusive, which are multiples of 2 or 3, which means that the probability of selecting neither multiple of 2 nor multiple of 3 is 1-67/100=33/100=33%.

Hope it's clear.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13542
Followers: 578

Kudos [?]: 163 [0], given: 0

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

### Show Tags

30 Sep 2013, 10:26
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 62

Kudos [?]: 594 [0], given: 355

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

### Show Tags

10 Dec 2013, 05:19
Zem wrote:
The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

A. 83%
B. 67%
C. 50%
D. 33%
E. 17%

Bunuel is this approach correct?

Ballparking from 1 to 10 there are only 3 numbers that satisfy these conditions: 1,5 and 7. Then 3/10 so approximately 30% only answer choice that is close is D

If you do it for other ranges you are basically going to have 4,3,4,3,4,3 so in theory it would be something like 3.5/10 or between 30% and 35%

Cheers
J
Senior Manager
Affiliations: CrackVerbal
Joined: 03 Oct 2013
Posts: 356
Location: India
GMAT 1: 780 Q51 V46
Followers: 71

Kudos [?]: 310 [0], given: 5

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

### Show Tags

10 Dec 2013, 05:22
Zem wrote:
The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

A. 83%
B. 67%
C. 50%
D. 33%
E. 17%

Multiple of 3: 100/3 = 33 numbers
Multiple of 2: 100/2 = 50 numbers

Multiple of 6: 100/6= 16 numbers

Total numbers divisible by 2 or 3: 50 + 33 - 16 = 67

Neither = 100 - 67 = 33

Probability = 33/100

Pushpinder Gill
_________________

For 5 common traps on CR check here :
http://gmatonline.crackverbal.com/p/gmat-critical-reasoning

For sentence correction videos check here -
http://gmatonline.crackverbal.com/p/gmat-sentence-correction

Learn all PS and DS strategies here-
http://gmatonline.crackverbal.com/p/mastering-quant-on-gmat

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13542
Followers: 578

Kudos [?]: 163 [0], given: 0

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

### Show Tags

25 Dec 2014, 21:48
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 31 Oct 2015
Posts: 37
Followers: 0

Kudos [?]: 2 [0], given: 53

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

### Show Tags

03 Jan 2016, 08:46
Even numbers = 100/2 = 50
Numbers divisible by 3 = 99/3 = 33
Even or divisible by 3 =
50 + 33 - duplicates(divisible by 3 and even)
Duplicates = 32/2 = 16
Even or divisible by 3 = 50 + 33 - 16
= 67
Numbers that are not even or divisible by 3 = 100 - 67 = 33
BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 1905
Followers: 50

Kudos [?]: 375 [0], given: 457

Re: The integers from 1 to 100 inclusive are each written on a [#permalink]

### Show Tags

14 Mar 2016, 00:12
Even => 50
Multiples of 3 => 33
multiples of 6 => 16
Not cases => 67
cases =33
P(cases) = 33/100
percentage =33
hence D
_________________

Mock Test -1 (Integer Properties Basic Quiz) ---> http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1676182

Mock Test -2 (Integer Properties Advanced Quiz) --->http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1765951

Mock Test -2 (Evens and Odds Basic Quiz) --->http://gmatclub.com/forum/stonecold-s-mock-test-217160.html#p1768023

Give me a hell yeah ...!!!!!

Re: The integers from 1 to 100 inclusive are each written on a   [#permalink] 14 Mar 2016, 00:12
Similar topics Replies Last post
Similar
Topics:
2 Each of the integers from 0 to 9, inclusive, is written on a 2 15 Apr 2014, 15:45
8 If S is the product of the integers from 1 to 100, inclusive 6 27 Feb 2013, 12:39
17 Each of the integers from 0 to 9, inclusive, is written on a 9 11 Apr 2012, 08:26
Each of the integers from 0 to 9, inclusive, is written on a 2 24 Jul 2011, 02:48
5 Each of the integers from 0 to 9, inclusive, is written on a 7 17 Aug 2008, 01:51
Display posts from previous: Sort by