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M16#11

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Re: M16#11 [#permalink] New post 26 Jun 2012, 09:31
Nos. divided by 3 between 100 and 999 (including); starting 102,105....till 999 are {(999-102=897/3)+1 }=300
Therefore, nos. not divided by 300 between 100 & 999 = {(total nos between 100&999 ) minus nos. divided by 3}= [ {(999-100)+1}-300]=600

Ans-B
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Re: M16#11 [#permalink] New post 26 Jun 2013, 05:14
Simple.

Number of 3 digit numbers = 9(Hundreds's digit should be between 1-9)*10(ten's digit can be anything from 0-9)*10(Unit's digit can be anything from 0-9)= 900

All 900 numbers are in an order.i.e 100 to 999. Every 3rd number is divisible by 3.

Hence, numbers divisible by 3 = 900/3 = 300, other 600 are not divisible.
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Re: M16#11 [#permalink] New post 26 Jun 2013, 05:16
sshah0 wrote:
May be its a stupid question - but why are we not considering negative integers here? The questions says three digit integers!



When we check divisibility, we don't check negative numbers I guess that's the reason.
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Re: m16 #11 [#permalink] New post 26 Jun 2013, 06:10
TheSituation wrote:
I did 999-102 divided by 3 = 299

999-299 = 600.


I did it the same way except 999-299=700....if you count 0-99. 999-299=600 if you're only going down to 100 or 600+100(where you start) +299=999
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Re: M16#11 [#permalink] New post 26 Jun 2013, 06:24
I choose option B!

1) Every 1-in-3 integers is divisible by 3. (Every third number....)
2) In Total there are 900 3 digit numbers (100 to 999)
3) 1/3 rd of 900 are divisible by 3 --> 2/3 rd of 900 not divisible = 600 integers of 3 digit size not divisible by 3!

Hope it helps!
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Re: M16#11 [#permalink] New post 26 Jun 2013, 18:27
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100-999 = # of 3 digit integers.
999-100+1 = 900 (# of different 3 digit integers)

of these, we know that 2 of every 3 integers are not going to be divisible by 3.
so, 900 x 2 / 3 = 600
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Re: M16#11 [#permalink] New post 27 Jun 2013, 21:02
Answer B.

The way I approached this, is not basically using counting more along the principles of divisibility.

y = nx === if 'y' is divisible by 'x' then, that equation holds, where n is the number times it can be divided by 'x'.

So with that in mind largest 3 digit number = 999, divisibility by '3' => y = nx => 999 = n * 3 => n = 333
Similarly largest 2 digit number = 99, divisibility by 3 => 99= n*3 => n = 33

Hence the difference will give the number of the factors between 999 and 99 => 333-33 = 300

But it asks for number which isn't divisibile by 3 instead, hence total 3 digit numbers - number of factors of 3
= 900-300 = 600

Similarly if we do this for number 5 instead, it becomes:
Largest three digit number divisible by 5 = 995; hence quotient = 995/5 => 199
Largest two digit number divisible by 5 = 95 ; hence quotient = 95/5 => 19

Hence number of factors between 995 and 95 = 199-19 => 180
Numbers that aren't divisible by 5 = 900-180 => 720
Re: M16#11   [#permalink] 27 Jun 2013, 21:02
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