May 25th, 2017, 12:17 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Subset of N
A subset $S$ of $N$ is said to be thick if among any 2016 consecutive positive integers, at least one should belong to $S$. Which of these subsets are thick ? A) The set of the geometric progression $ {2,2^2,2^3,...}.$ B) The set of the arithmetic progression ${1000,2000,3000...}.$ C) {$\displaystyle n \in N/n > 2016$} D) The set of all composite numbers. My answer is Option C. Please check and let me know Thanks 
May 25th, 2017, 01:00 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
I think you misunderstood the definition of "thick".

May 25th, 2017, 01:44 AM  #3 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  In any sequence of 2016 consecutive terms, $S$ should contain at least one term. How could we check it? Please help. Last edited by skipjack; May 25th, 2017 at 01:11 PM. 
May 25th, 2017, 01:10 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
That's not what the problem refers to. It refers to "any 2016 consecutive positive integers", such as, for example, 514, 515, 516, . . . , 2528, 2529. Is at least one of those integers in $S$ for a given definition of $S$?

June 1st, 2017, 09:15 PM  #5 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2 
Does it mean : At least one out of any 2016 consecutive positive number should exist in the sets given ? If so, Option C has consecutive Natural numbers above 2016. Is it wrong ? Please explain 
June 1st, 2017, 09:44 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
Consider, for example, the 2016 consecutive positive integers 1, 2, 3, . . . , 2016. Not a single one of them is greater than 2016, so Option C is wrong.

June 1st, 2017, 09:48 PM  #7  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Quote:
Sorry I have mistaken. Last edited by Lalitha183; June 1st, 2017 at 10:08 PM.  
June 1st, 2017, 10:00 PM  #8 
Senior Member Joined: Aug 2012 Posts: 1,528 Thanks: 364  Among $2, 2^2, 2^3, \dots$ there will soon be a point after which all the consecutive differences are greater than $2016$. In fact $2^{12}  2^{11} = 2048$ so from $2^{11} + 1$ through $2^{11} + 2016$ there is a run of $2016$ consecutive natural numbers that contains no power of $2$.
Last edited by Maschke; June 1st, 2017 at 10:03 PM. 
June 1st, 2017, 10:12 PM  #9  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Quote:
 
June 1st, 2017, 10:14 PM  #10 
Senior Member Joined: Aug 2012 Posts: 1,528 Thanks: 364  

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