Let $a_{1}, \cdots, a_{44}$ be natural numbers such that \[0<a_{1}<a_{2}< \cdots < a_{44}<125.\] Prove that at least one of the $43$ differences $d_{j}=a_{j+1}-a_{j}$ occurs at least $10$ times.
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Peter
16.10.2007 02:29
Assume each appeared less than $ 10$ times, then $ a_{44}\ge 9\cdot 1+9\cdot 2+9\cdot 3+9\cdot 4+8\cdot 5 = 130$, contradiction. (strange how the problem isn't sharp, it would have been sharp if the sequence went up to $ a_{43}$, right? )
Isogonics
23.12.2012 08:01
Peter wrote: (strange how the problem isn't sharp, it would have been sharp if the sequence went up to $ a_{43}$, right? ) Because there are 43 differences, not 44 differences. Then it must be $ a_{44}\ge 9\cdot 1+9\cdot 2+9\cdot 3+9\cdot 4+7\cdot 5 = 125$ .
Woett
02.04.2023 17:51
The problem is still not as sharp as it could be. If we assume that for every $m \in \{1,2,3,4\}$ there are at most $9$ indices $j$ such that $a_{j+1} - a_j = m$, then we get: \begin{align*} a_{44} &= a_1 + \sum_{j=1}^{43} (a_{j+1} - a_j) \\ &\ge 1 + 9 \cdot 1 + 9 \cdot 2 + 9 \cdot 3 + 9 \cdot 4 + 7 \cdot 5 \\ &= 126 \end{align*}