Consider the Arithmetic progression with first term 1, common difference 3 and last term 100. If A is any set of 19 distinct numbers from the arithmetic progression, show that there must always be two numbers whose sum is 104.

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- Sep 9th 2010, 09:04 PMacc100jtArithmetic Series
Consider the Arithmetic progression with first term 1, common difference 3 and last term 100. If A is any set of 19 distinct numbers from the arithmetic progression, show that there must always be two numbers whose sum is 104.

Thank you! - Sep 9th 2010, 11:35 PMCaptainBlack
The series is $\displaystyle a_n=2n-2,\ n=1, .. , 34$. That the $\displaystyle n$ and $\displaystyle $$ m$ th terms sum to $\displaystyle 104$ means:

$\displaystyle (3n-2)+(3m-2)=104$

or:

$\displaystyle n+m=36$

Now both of $\displaystyle n=17$ and $\displaystyle m=19$ are in any set of $\displaystyle 19$ consecutive integers between $\displaystyle 1$ and $\displaystyle 34$ inclusive, etc.. .

CB - Sep 10th 2010, 12:32 AMacc100jt
What happen if i pick 19 integers between 1 and 34 inclusive, but may not be consectutive, then n=17 and m=19 may not be in the set...hmm

- Sep 10th 2010, 01:34 AMCaptainBlack