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

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

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New post 24 Mar 2019, 23:19
1
3
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

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

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Re: How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu  [#permalink]

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New post 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
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How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu  [#permalink]

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New post 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.
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How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu  [#permalink]

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New post 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
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How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu  [#permalink]

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New post 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
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Eulerelements

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

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New post 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)
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Re: How many numbers between 1 and 2005 are integer multiples of 3 or 4 bu   [#permalink] 25 Mar 2019, 15:49
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