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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"?

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..

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

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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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

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