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

Question

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

shuvodip04 · Accepted Answer

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)

Archit3110 · Answer

multiples of 3 = 668
multiples of 4 = 501
multiplesof 12 = 167
so 668+501-167= 1002 
IMO D

Eulerelements · Answer

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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shuvodip04 · Answer

Eulerelements

thanks. i missed this point in my solution.Edited it.

ideree · Answer

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)

samarpan.g28 · Answer

multiples of 3: 3,6,9,....,2004
multiples of 4: 4,8,12,....,2004
multiples of 12: 12,24,....,2004

I used AP to find the no of terms in each case. last term=first term + (no of terms -1) * common difference.
mul of 3 has 668 terms, mul of 4 has 501 terms, mul of 12 has 167 terms.
But mul of 12 has mul of 3 and mul of 4 in it also. Therefore we need to subtract 167 from each of 668 and 501.
(668-167)+(501-167)=835. Option (C) is correct.

bumpbot · Answer

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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

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