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In a set of numbers from 100 to 1000 inclusive, how many integers are [#permalink]
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13 May 2008, 18:56
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In a set of numbers from 100 to 1000 inclusive, how many integers are odd and do not contain the digit "5"? A. 180 B. 196 C. 286 D. 288 E. 324 M0312
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Last edited by Bunuel on 19 May 2015, 05:28, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.



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Re: In a set of numbers from 100 to 1000 inclusive, how many integers are [#permalink]
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13 May 2008, 19:54
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) 201=19 such numbers cause i already counted 155 twice.. 19*8=152 such numbers..plus 100 from 5XX..so total to be excluded 152+100450=198.. I would guess the number is 196..



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Re: In a set of numbers from 100 to 1000 inclusive, how many integers are [#permalink]
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13 May 2008, 20:11
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Should be 288. We have to find the total number of 3digit 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: In a set of numbers from 100 to 1000 inclusive, how many integers are [#permalink]
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24 Jan 2009, 14:36
Can someone tell my why we multiply 4*8*9, whats the logic behind multiplying versus another approach?
Thanks, Ali



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Re: In a set of numbers from 100 to 1000 inclusive, how many integers are [#permalink]
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24 Jan 2009, 16:07
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xALIx 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: In a set of numbers from 100 to 1000 inclusive, how many integers are [#permalink]
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18 May 2015, 16:43
x2suresh wrote: xALIx 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 Dear x2suresh, Clearly your approach sets the benchmark! Please share more of such examples. Thanks. My approach was cumbersome and time consuming:
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Re: In a set of numbers from 100 to 1000 inclusive, how many integers are [#permalink]
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19 May 2015, 05:29
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alimad wrote: In a set of numbers from 100 to 1000 inclusive, how many integers are odd and do not contain the digit "5"?
A. 180 B. 196 C. 286 D. 288 E. 324
M0312 Examine what digits these set members can contain: First digit (hundreds): 8 choices (1, 2, 3, 4, 6, 7, 8, 9  cannot be 0 or 5) Second digit (tens): 9 choices (0, 1, 2, 3, 4, 6, 7, 8, 9  cannot be 5) Last digit (units): 4 choices (1, 3, 7, 9  cannot be 0, 2, 4, 5, 6, 8) The answer is 8∗9∗4=32∗9=288. Answer: D.
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