All numbers that are neither divisible by 5 nor 11 were removed from a sequence of consecutive natural numbers from 1 to 10,000. A new sequence was formed. what is the value of the 2008th term of this sequence ?

Question

miracrown · Answer

All numbers on the list are either divisible by 5 or divisible by 11.
All terms in the sequence are either divisible by 5 or divisible by 11 based on the given information. 

there are 10000/5=2000 numbers divisible by 5 from 1 to 10000, and there are 10000/11=909 numbers divisible by 11 from 1 to 10000 
and there are 10000/55=181 numbers divisible by 55 from 1 to 10000.

So there are 2000+909-181=2728 numbers divisible by either 5 or 11.

hartnn · Answer

just to confirm my logic, is the answer 7360 ?

miracrown · Answer

then we count the numbers...

miracrown · Answer

yes, lol.

hartnn · Answer

S1 = set of number divisible by 5 
S2 = set of number divisible by 11
S3 = set of number divisible by 55 

so the required set will be S1 +S2 - S3 

in that set, every 15th term will be divisible by 55. 

so say we want to find 30th term, we divide 30/15 (we get '2') and the 30th term will be 55*2 = 110

similarly, we would easily get 1995th term
1995/15 = 133
and 55*133 = 7315 

when we know 1995th term, 2008th term is not difficult to get :)

miracrown · Answer

You are pretty quick on typing that. :)