GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Oct 2019, 04:56

### 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

# How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58428
How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu  [#permalink]

### Show Tags

24 Mar 2019, 23:19
1
3
00:00

Difficulty:

85% (hard)

Question Stats:

40% (02:22) correct 60% (02:35) wrong based on 50 sessions

### HideShow timer Statistics

How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not 12 ?

(A) 501
(B) 668
(C) 835
(D) 1002
(E) 1169

_________________
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5031
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu  [#permalink]

### Show Tags

24 Mar 2019, 23:24
1
Bunuel wrote:
How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not 12 ?

(A) 501
(B) 668
(C) 835
(D) 1002
(E) 1169

multiples of 3 = 668
multiples of 4 = 501
multiplesof 12 = 167
so 668+501-167= 1002
IMO D
Manager
Joined: 05 Oct 2017
Posts: 101
Location: India
Schools: ISB '21, IIMA , IIMB
GPA: 4
WE: Analyst (Energy and Utilities)
How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu  [#permalink]

### Show Tags

Updated on: 25 Mar 2019, 10:13
2
Bunuel wrote:
How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not 12 ?

(A) 501
(B) 668
(C) 835
(D) 1002
(E) 1169

Between 1 and 2005 first and last number divisible by 3 are 3 and 2004.

Numbers divisible by 3 = $$(2004-3)/3 + 1$$ = 668

Between 1 and 2005 first and last number divisible by 4 are 4 and 2004.

Numbers divisible by 4 = $$(2004-4)/4 +1$$ = 501

Between 1 and 2005 first and last number divisible by 12 are 12 and 2004.

Numbers divisible by 12 = $$(2004-12)/12 + 1$$ = 166 + 1 = 167

Since numbers divisible by 3 and 4 both contains numbers divisible by 12 we have to subtract it twice to get numbers divisible only by 3 or 4

Numbers divisible by 3 or 4 but not by 12 = 668+501 -167 -167= 835 . Ans (C)

Originally posted by shuvodip04 on 25 Mar 2019, 02:46.
Last edited by shuvodip04 on 25 Mar 2019, 10:13, edited 1 time in total.
Intern
Joined: 08 Dec 2017
Posts: 15
How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu  [#permalink]

### Show Tags

25 Mar 2019, 09:51
1
2004/3= 668

2004/4=501

2004/12=167

The 668 includes multiples of 12. So does the 501. They are double counted. You want to remove them altogether. If you only subtract 167 you counting multiples of 12 once.

Correct calculation: 668+501-167-167=835

Method 2:

1,2,3,4,5,6,7,8,9,10,11,12

13,14,15,16,17,18,19,20,21,22,23,24

For the first set numbers 3,4,6,8,9 all satisfy the condition. So 5/12.

For the next set of 12 numbers 15,16,18,20,21. Again 5/12

For every 12 numbers, 5 satisfy the condition.

2004(5/12)=835
Posted from my mobile device
Manager
Joined: 05 Oct 2017
Posts: 101
Location: India
Schools: ISB '21, IIMA , IIMB
GPA: 4
WE: Analyst (Energy and Utilities)
How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu  [#permalink]

### Show Tags

25 Mar 2019, 10:09
Eulerelements wrote:
2004/3= 668

2004/4=501

2004/12=167

The 668 includes multiples of 12. So does the 501. They are double counted. You want to remove them altogether. If you only subtract 167 you counting multiples of 12 once.

Correct calculation: 668+501-167-167=835

Method 2:

1,2,3,4,5,6,7,8,9,10,11,12

13,14,15,16,17,18,19,20,21,22,23,24

For the first set numbers 3,4,6,8,9 all satisfy the condition. So 5/12.

For the next set of 12 numbers 15,16,18,20,21. Again 5/12

For every 12 numbers, 5 satisfy the condition.

2004(5/12)=835
Posted from my mobile device

Eulerelements

thanks. i missed this point in my solution.Edited it.
Intern
Joined: 23 Dec 2017
Posts: 3
Re: How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu  [#permalink]

### Show Tags

25 Mar 2019, 15:49
Numbers divisible by 12 => 2005/12=167
Numbers divisible by 3 => 2005/3=668
Numbers divisible by 4 => 2005/4=501

BE CAREFUL NOW!

Numbers divisible by 3 but not by 12 => 668-167=501
Numbers divisible by 4 but not by 12 = > 501-167=334

Numbers divisible by 3 or 4 but not by 12 => 501+334=835 (Correct answer is C)
Re: How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu   [#permalink] 25 Mar 2019, 15:49
Display posts from previous: Sort by