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In a set of numbers from 100 to 1000 inclusive, how many

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Director
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In a set of numbers from 100 to 1000 inclusive, how many [#permalink] New post 13 May 2008, 17:56
In a set of numbers from 100 to 1000 inclusive, how many integers are odd and do not contain the digit "5"?



180
196
286
288
324
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Re: Sets [#permalink] New post 13 May 2008, 18:54
alimad wrote:
In a set of numbers from 100 to 1000 inclusive, how many integers are odd and do not contain the digit "5"?



180
196
286
288
324


lets see..in all there are 450 odd numbers!

of which 105, 115, 125, 135..10 such number per 100 numbrs 10+10(from 15X) 20-1=19 such numbers cause i already counted 155 twice..
19*8=152 such numbers..plus 100 from 5XX..so total to be excluded 152+100-450=198..

I would guess the number is 196..
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Re: Sets [#permalink] New post 13 May 2008, 19:11
Should be 288.
We have to find the total number of 3-digit odd numbers not having 5 as a digit.
Units digits will be among 1,3,7,9
Tenth digits will be among 0,1,2,3,4,6,7,8,9
Hundredth digits will be among 1,2,3,4,6,7,8,9
So total numbers = 4*9*8 =288
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Re: Sets [#permalink] New post 24 Jan 2009, 13:36
Can someone tell my why we multiply 4*8*9, whats the logic behind multiplying versus another approach?

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Re: Sets [#permalink] New post 24 Jan 2009, 15:07
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x-ALI-x wrote:
Can someone tell my why we multiply 4*8*9, whats the logic behind multiplying versus another approach?

Thanks,
Ali


we need to find odd numbers without digit 5 from 100 to 1000 (i.e same as odd numbers without digit 5 from 100 to 999)

let say number= XYZ
Z= units digit
Y= tenths digit
X= Hundredth's digit

For any 3 digit number to be odd, Unit digit must be odd
So Z can be filled with 1,3,7,9 (we are exlcuding digit 5) = 4 ways.
Y can be filled with 0,1,2,3,4,6,7,8,9 (we are ecluding digit 5) = 9 ways
X can be filled with 1,2,3,4,6,7,8,9 (we are exlcuding digits 0 and 5 .. ) = 8 ways

No of ways= 4*9*8= 288
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Re: Sets [#permalink] New post 24 Jan 2009, 15:11
x2suresh wrote:

we need to find odd numbers without digit 5 from 100 to 1000 (i.e same as odd numbers without digit 5 from 100 to 999)

let say number= XYZ
Z= units digit
Y= tenths digit
X= Hundredth's digit

For any 3 digit number to be odd, Unit digit must be odd
So Z can be filled with 1,3,7,9 (we are exlcuding digit 5) = 4 ways.
Y can be filled with 0,1,2,3,4,6,7,8,9 (we are ecluding digit 5) = 9 ways
X can be filled with 1,2,3,4,6,7,8,9 (we are exlcuding digits 0 and 5 .. ) = 8 ways

No of ways= 4*9*8= 288


Enough said, very well put!
:thanks
Re: Sets   [#permalink] 24 Jan 2009, 15:11
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