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May 25th, 2017, 12:17 AM   #1
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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
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May 25th, 2017, 01:00 AM   #2
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I think you misunderstood the definition of "thick".
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May 25th, 2017, 01:44 AM   #3
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Quote:
Originally Posted by skipjack View Post
I think you misunderstood the definition of "thick".
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.
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May 25th, 2017, 01:10 PM   #4
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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 $S$?
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June 1st, 2017, 09:15 PM   #5
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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
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June 1st, 2017, 09:44 PM   #6
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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.
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June 1st, 2017, 09:48 PM   #7
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Quote:
Originally Posted by skipjack View Post
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.
Okay, I got it. Then Option A and B can satisfy this condition. How can we check Option D ?

Sorry I have mistaken.

Last edited by Lalitha183; June 1st, 2017 at 10:08 PM.
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June 1st, 2017, 10:00 PM   #8
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Quote:
Originally Posted by Lalitha183 View Post
How can we check Option A ?
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.
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June 1st, 2017, 10:12 PM   #9
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Quote:
Originally Posted by Maschke View Post
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 $2$.
So in the long run the difference between the powers of $2$ dissatisfy this condition.
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June 1st, 2017, 10:14 PM   #10
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Quote:
Originally Posted by Lalitha183 View Post
So in the long run the difference between the powers of $2$ dissatisfy this condition.
It's not that long, it happens as early as the 11th term of the sequence.
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