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What is the number of integers from 1 to 1000 (inclusive)

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CEO
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What is the number of integers from 1 to 1000 (inclusive) [#permalink] New post 16 Nov 2007, 01:58
What is the number of integers from 1 to 1000 (inclusive) that are not divisible by 11 nor by 35?

A. 884
B. 999
C. 123
D. 124
E. 231
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Manager
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 [#permalink] New post 16 Nov 2007, 02:05
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Answer should be A) 884

Total multiples of 35 between 1 & 1000 = 28 (35*28=980)
Total multiples of 11 between 1 & 1000 = 90 (90*11=990)
Total multiples of 11 and 35 between 1 & 1000 = 2 (35*11 = 385 & 770)

Therefore 28+90-2=116

1000-116 = 884.
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 [#permalink] New post 16 Nov 2007, 02:31
Agree. Ans is A).
Took me 3 mins to solve though :(
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 [#permalink] New post 16 Nov 2007, 18:26
is there a shorter way to figure out the #s of multiples, especially for 35 ?
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Re: 7.14 Counting [#permalink] New post 16 Nov 2007, 20:27
bmwhype2 wrote:
What is the number of integers from 1 to 1000 (inclusive) that are not divisible by 11 nor by 35?

A. 884
B. 999
C. 123
D. 124
E. 231


Yes A. Could solve in 1 min..

I went this way.. There are 9 multiples of 11 from 1 - 100 so from 1 -1000 they would be ~90.
There are 2 multiples of 35 in 1-100 hence from 1-1000, they will be ~20.. So total will be ~110.. Only A comes close .. I would have been screwed choices were closer ..
:lol: :lol:
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 [#permalink] New post 16 Nov 2007, 20:42
pmenon wrote:
is there a shorter way to figure out the #s of multiples, especially for 35 ?


This is how i did it.

Find any easy-to-calculate multiple of 35 close to 1000.
e.g: 35 * 30 = 1050.

Work from there to get 980 = 35*28.

Hope this helps.
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Re: 7.14 Counting [#permalink] New post 16 Nov 2007, 22:44
bmwhype2 wrote:
What is the number of integers from 1 to 1000 (inclusive) that are not divisible by 11 nor by 35?

A. 884
B. 999
C. 123
D. 124
E. 231


Answer is A.

Total ints. = 1000

# of ints. divisible by 11 = 1000/11 = 90
# of ints. divisible by 35 = 1000/35 = 28
# of ints. divisible by both 11 & 35 = 1000/(LCM of 11, 35) = 1000/385 = 2

# of ints. not divisible by either = 1000 - (90+28) + 2 = 884
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 [#permalink] New post 18 Nov 2007, 18:44
Late, but A
The same method as jbs
  [#permalink] 18 Nov 2007, 18:44
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