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200 and 300 (both inclusive)

by sanju09 » Tue Feb 24, 2009 4:41 am
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. 25
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by mjjking » Tue Feb 24, 2009 7:41 am
200-->300

numbers divisible by 2 => all the evn ones = 51
numbers divisible by 5 = all the 5 ones = 10
numbers divisible by 3 = all the ones divisible by three minus the even ones and the 5 ones = 14

total of numbers divisible by 2,3,5 = 51+10+14=75

what am I missing?
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by beater » Tue Feb 24, 2009 7:59 am
Can we solve this via combinatorics?
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Re: 200 and 300 (both inclusive)

by x2suresh » Tue Feb 24, 2009 8:12 am
sanju09 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. 25
divisble by 2 = 1+100/2 =51
divisble by 3 =1+99/3 = 34
divisble by 5 = 1+100/5 = 21

divisble by 6 (both 2 and 3)= 1+94/6 =16
divisble by 15 (both 3 and 5)= 1+90/15 =7
divisble by 10 (both 2 and 5)= 1+100/10 =11
divisble by 30 (both 2,3 and 5)= 1+90/30 =4


Nos. divisble by 2/3/5 = 51+34+21 - (16+7+11)+4 = 76

Anser = 101-76= 25
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Re: 200 and 300 (both inclusive)

by marcusking » Tue Feb 24, 2009 9:14 am
divisble by 2 = 1+100/2 =51
divisble by 3 =1+99/3 = 34
divisble by 5 = 1+100/5 = 21

divisble by 6 (both 2 and 3)= 1+94/6 =16
divisble by 15 (both 3 and 5)= 1+90/15 =7
divisble by 10 (both 2 and 5)= 1+100/10 =11
divisble by 30 (both 2,3 and 5)= 1+90/30 =4


Nos. divisble by 2/3/5 = 51+34+21 - (16+7+11)+4 = 76

Anser = 101-76= 25[/quote]

Bingo I got the same. Well done.
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by mjjking » Tue Feb 24, 2009 12:05 pm
could you explain why you added 1?
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by Mr2Bits » Tue Feb 24, 2009 12:12 pm
mjjking wrote:could you explain why you added 1?
200 and 300 are inclusive which gives you a range of 101 numbers.

You need to include "0" you have 0,2,4,6,8...100 which makes it 50+1 or 51
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by Alara533 » Tue Feb 24, 2009 3:13 pm
Could you please provide the source of the question, because I find all the answers wrong.

I found 26 digits which are not divisible by 5, 3 or 2. I even listed them down to make sure the answer is correct.

My approach -

There are total of 101 numbers
There are 51 multiples of 2 between 200 and 300 - both inclusive
There are 34 multiple of 3 between 200 and 300 - 17 odd and 17 even, out of this even numbers are already counted as the multiples of 2, then we need to count only the odd 17 numbers.
There are 21 multiples of 5 between 200 and 300 - 11 even and 10 odd. 11 even numbers are already counted in 2's multiple. Out of the 10 odd numbers 3 ( 225, 255 and 285) are 3's multiple, hence we have to count only the remaining 7 numbers.

ie. 51 + 17 + 7 = 75 => 75 numbers out of 101 are divisible by 2, 3 or 5.
So 101 - 75 = 26 numbers are not divisible by 2, 3, or 5.

Please let me know what am missing here!
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by DarkKnight » Tue Feb 24, 2009 6:32 pm
numbers divisible by 2 => all the evn ones = 51
numbers divisible by 5 = all the 5 ones = 10
numbers divisible by 3 = all the ones divisible by three minus the even ones and the 5 ones = 14

total of numbers divisible by 2,3,5 = 51+10+14=75

what am I missing?


mjjking.. you are not missing anything. So there are 75 numbers between 200 and 300 inclusive that are divisible by 2, 3 or 5. Hence, the 25 numbers are not divisible by 2 , 3 or 5.

Correct answer is 25.
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by krisraam » Tue Feb 24, 2009 7:32 pm
I agree with Alara33. The answer has to be 26.

My approach is similar to X2suresh.

I think suresh made a mistake in calculating the numbers divisible by 6

divisible by 6 = 1+(300-204)/6 = 17

Thanks
Raama
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Re: 200 and 300 (both inclusive)

by chintudave » Tue Feb 24, 2009 7:36 pm
[quote="marcusking"]divisble by 2 = 1+100/2 =51
divisble by 3 =1+99/3 = 34
divisble by 5 = 1+100/5 = 21

divisble by 6 (both 2 and 3)= 1+94/6 =16
divisble by 15 (both 3 and 5)= 1+90/15 =7
divisble by 10 (both 2 and 5)= 1+100/10 =11
divisble by 30 (both 2,3 and 5)= 1+90/30 =4


Nos. divisble by 2/3/5 = 51+34+21 - (16+7+11)+4 = 76

Anser = 101-76= 25[/quote]

Bingo I got the same. Well done.[/quote]

I did not understand the logic behind adding 1! Does this logic hold true all the time ? What will happen if the range is from 201 to 301 inclusive ?

Thanks in advance!
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by x2suresh » Tue Feb 24, 2009 8:56 pm
krisraam wrote:I agree with Alara33. The answer has to be 26.

My approach is similar to X2suresh.

I think suresh made a mistake in calculating the numbers divisible by 6

divisible by 6 = 1+(300-204)/6 = 17

Thanks
Raama
You are right.. I made a mistke.. it should 17

Agree with 26
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Re: 200 and 300 (both inclusive)

by x2suresh » Tue Feb 24, 2009 9:01 pm
chintudave wrote:
I did not understand the logic behind adding 1! Does this logic hold true all the time ? What will happen if the range is from 201 to 301 inclusive ?

Thanks in advance!
take for e.g.
find Numbers divisble by 3 between 200 to 300

First find least number that is divisble 3 and max number that is divisble 3

least number divisble by 3= 201
max number divisble by 3= 300

No. = 300-201/3 +1 (we need to add to include 201)
= 33+1 = 34

If are still not convinced.
try this..
how many integers are divisble by 3 between 201 and 202

lower =201
upper = 201
no = 0+1 (to include 201)

I hope it is clear now.
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by chintudave » Tue Feb 24, 2009 9:09 pm
Got it!

Thanks
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by mjjking » Wed Feb 25, 2009 12:56 am
so what's the OA? is 25 or 26? this is confusing... :idea:
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