Odd integers not containing 5 in a set

jzebra10 wrote:How should you solve this problem, other than finding the actual numbers?

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

_ _ _

the unit's digit can be filled in 4 ways with any of 1,3,7,9 digits.
the hundred's digit can be filled in 8 ways (5,0)
the ten's digit can be filled in 9 ways(5)

so we have 8*9*4 ways = 288 ways.

user123321

I don't see why you want to ensure the digits are not repeated. The question says odd numbers with no 5:

Hundreds digit cannot be 0 or 5. Units digit has to be one of 1, 3, 7, 9. No 5 anywhere.

8*9*4=288.

Not sure whether I missed something about repeating digits.

oops you're right. I am changing the post now. thanks for pointing out.

user123321

Anurag@Gurome's method is the best one.
here is one more method, though not very elegant (I even dare to say, that the method below is a real nightmare )

so, total of odd numbers from 101 to 999 inclusive is (999-101)/2+1=450
(another method to find total-since we need only odd numbers, our last digit can be 1,3,5,7 or 9(total 5), tens digit can be anything from 0 to 9(total 10), hundreds digit is anything from 1 to 9 (total =9) . so total of odd numbers =5*10*9=450)

now we need to find out all numbers that contain digit 5 and exclude them-

first exclude all odd numbers from 501 to 599 (599-501)/2+1=50

now let's find numbers from 105 to 195 (we have 105; 115; 125; ...195. so, the difference is 10)
195=105+10(n-1) n=10
since we also have total of 8 ranges 105-195;205-295;305-395;405-495;605-695;705-795;805-895;905-995, 10*8=80

do not forget, that we also have numbers 151 153 157 159 (same with other ranges 2***-9***) total 4*8=32

so, the result is 450-50-80-32=288

p.s. it is an awful method but if u want to refresh ur knowledge of arithmetic progression, or u r simply a "pervert" , u can use this method ))).