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# How many positive integers between 200 and 300 (both inclusi

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Senior Manager
Joined: 22 Dec 2009
Posts: 293
Re: Divisibility by 2,3 and 5  [#permalink]

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01 Feb 2010, 11:30
walker wrote:
jeeteshsingh wrote:
Hi Walker...... didnt get this approach..... request u to explain the same

The main idea is to consider 30 consecutive numbers rather than 100. 30=2*3*5. So, any pattern will repeat in 30 numbers. Let's say we have only 2 and 3 - the period will be 2*3:
[200, 201, 202, 203, 204, 205] [206, 207, 208, 209, 210, 211][212, 213, 214, 215, 216, 217]
(red) - a number that is not divisible by neither 2 or 3. We have 2 numbers out of 6 consecutive numbers.

The same here, but period is 30=2*3*5.

Definitely it is not the best approach but one of many.

Awesome Walker!!! Its incredible!!! Kudos X 10.... :cheer

Just one question... you considered.... the first 10 numbers stand alone.. but I guess the idea is that we can also pick the last 10 numbers... rite?
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Re: Doing set/number properties question quicker?  [#permalink]

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01 Feb 2010, 11:42
Yeah, you are right.
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Re: Divisibility by 2,3 and 5  [#permalink]

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01 Feb 2010, 14:55
GMAT TIGER wrote:
sdrandom1 wrote:
How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?

A. 3
B. 16
C. 75
D. 24
E. 26

Quick approach:
Number of integers between 200 and 300 = 300-200+1 = 101
Number of integers between 200 and 300 divisible by 2 = (300/2 - 200/2) +1 = 51
Number of odd integers between 200 and 300 divisible by 3 = (300/3 - 200/3) + 1 -17 = 18
Number of integers between 200 and 300 not divisible by 2 or 3 but by 5 = (300/5 - 200/5) +1 - 15 = 6

The number of integers between 200 and 300 not divisible by 2 or 3 or 5 = 101 - 51 - 18 - 6 = 26.

I don't understand how the Number of odd itegers between 200 and 300 divisible by 3 is 18. (300/3 - 200/3) + 1 =approx 35
Am I missing something
Senior Manager
Joined: 22 Dec 2009
Posts: 293
Re: Divisibility by 2,3 and 5  [#permalink]

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02 Feb 2010, 12:00
GMAT TIGER wrote:
sdrandom1 wrote:
How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?

A. 3
B. 16
C. 75
D. 24
E. 26

Quick approach:
Number of integers between 200 and 300 = 300-200+1 = 101
Number of integers between 200 and 300 divisible by 2 = (300/2 - 200/2) +1 = 51
Number of odd integers between 200 and 300 divisible by 3 = (300/3 - 200/3) + 1 -17 = 18
Number of integers between 200 and 300 not divisible by 2 or 3 but by 5 = (300/5 - 200/5) +1 - 15 = 6

The number of integers between 200 and 300 not divisible by 2 or 3 or 5 = 101 - 51 - 18 - 6 = 26.

I don't understand how the Number of odd itegers between 200 and 300 divisible by 3 is 18. (300/3 - 200/3) + 1 =approx 35
Am I missing something

Number of odd integers between 200 and 300 divisible by 3 = (300/3 - 200/3) + 1 -17 = 18

You forgetting that..... 17 = number of numbers which are divisible by both 3 and 2 (basically 6). Since numbers have been counted in 2's bucket above, the same shouldn't be counted again.. when accounting for 3. Hence 17 is being subtracted!
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Re: Doing set/number properties question quicker?  [#permalink]

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16 Apr 2011, 06:31
This is how I did and pls verify the reasoning.

There are total of 101 integers between 200 and 300. Out of this 51 are div by 2. The rest 50 are odd. These contain the integers divisible by 3 and 5.

There are 21 integers div by 5. This includes 11 integers div by 2 and 5. Hence (21-11) = 10 integers are odd. We should subtract this from above. 50 - 10 = 40 integers left.

There are 34 integers div by 3. This includes - 17 integers div by 3 and 2. And 7 integers div by 3 and 5. And 4 integers divisible by all 2,3,5 - Hence 34 - (17 + 7 - 4) integers are div by 3 alone. i.e. 14 div by 3 alone

40 - 14 = 26 integers left. They are div by neither 2,3 or 5.
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Re: Doing set/number properties question quicker?  [#permalink]

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16 Apr 2011, 22:09
300 = 200 + (n-1)2

n = 51

300 = 201 + (n-1)3

n = 99/3 + 1

= 34

17 (multiples of 6)

300 = 200 + (n-1)5

n-1 = 100/5

n = 21

multiples of 10 = 11

multiples of 15 = 300 = 210 + (n-1)15

n = 6 + 1 = 7

multiples of 30 = 4

Total = 27 + 13 + 4 + 7 + 14 + 3 + 7

= 75

101 - 75 = 26

I I found it useful to draw a Venn Diagram, and the working from inside out.
Attachments

No of Divisors.png [ 9.24 KiB | Viewed 9983 times ]

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Re: Doing set/number properties question quicker?  [#permalink]

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17 Apr 2011, 13:50
I like the set approach being used...
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Re: Doing set/number properties question quicker?  [#permalink]

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25 Aug 2013, 05:17
Here see another approach:

First select extreme numbers divisible by all three 2 and 3 and 5 --> 210 and 300.

Now 0 - 300 numbers not divisible by 2 or 3 or 5:

= 300 (1 - 1/2) (1 - 1/3) ( 1 - 1/5)
= 80

Now 0 - 210 numbers not divisible by 2 or 3 or 5:

= 210 (1 - 1/2) (1 - 1/3) ( 1 - 1/5)
= 56

Numbers not divisible by 2 or 3 or 5 between 210 - 300 = 80 - 56 = 24

Further there are two more numbers between 200 - 210 => 203 and 209 which are not divisible by 2 or 3 or 5.

Thus 24 + 2 = 26 Ans E.
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Re: How many positive integers between 200 and 300 (both inclusi  [#permalink]

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03 Feb 2019, 14:03
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Re: How many positive integers between 200 and 300 (both inclusi   [#permalink] 03 Feb 2019, 14:03

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