The discussion revolves around a mathematical problem involving 20 pairwise distinct positive integers, each less than 70, and the claim that among their pairwise differences, there must be at least four equal numbers. The scope includes mathematical reasoning and proof exploration.
While some participants agree on the reasoning leading to the conclusion, there is no explicit consensus on the overall approach or the validity of the assumptions made, leaving the discussion somewhat unresolved.
Participants have not fully resolved the mathematical steps or assumptions required to establish the proof definitively. The discussion reflects various approaches and reasoning without a clear consensus on the methodology.
caffeinemachine said:Given $20$ pairwise distinct positive integers each less than $70$. Prove that among their pairwise differences there are at least four equal numbers.
Alexmahone said:Total number of pairwise differences $\displaystyle =\binom{20}{2}=190$. Each difference must belong to the set $\{1,\ 2,\ \ldots,\ 68\}$: there are only 68 distinct values for each pairwise difference.
I don't know how to proceed...
caffeinemachine said:How does "Challenging Puzzles.." forum work?
Am I supposed to post the solution(in case no one is able to solve it) within a day, a week or what?
Let the $20$ numbers be ordered as $a_1 < a_2 < \ldots < a_{20}$.Alexmahone said:If no one replied, you could do that. But since I've posted an attempt, you could just give me a hint. :)
caffeinemachine said:Let the $20$ numbers be ordered as $a_1 < a_2 < \ldots < a_{20}$.
Consider the differences:
$a_{20}-a_{19}, a_{19}-a_{18}, \ldots, a_2-a_1$
Alexmahone said:Assume, for the sake of argument, that among the pairwise differences there are at most 3 equal numbers.
$(a_{20}-a_{19})+(a_{19}-a_{18})+\ldots+(a_2-a_1)\ge 1+1+1+2+2+2+\ldots+6+6+6+7=3*6*7/2+7=70$
$a_{20}-a_1\ge 70$
$a_{20}\ge 70+a_1>70$ (contradiction)
So, among the pairwise differences there are at least 4 equal numbers.