HMMT-Feb 2017 comb, Problem 3

There are 2017 jars in a row on a table, initially empty. Each day, a nice man picks ten consecutive jars and deposits one coin in each of the ten jars. Later, Kelvin the Frog comes back to see that $N$ of the jars all contain the same positive integer number of coins (i.e. there is an integer $d>0$ such that $N$ of the jars have exactly $d$ coins). What is the maximum possible value of $N$ ?

Solution: Label the jars $1,2, \ldots, 2017$. I claim that the answer is 2014 . To show this, we need both a construction and an upper bound. For the construction, for $1 \leq i \leq 201$, put a coin in the jars $10 i+1,10 i+$ $2, \ldots, 10 i+10$. After this, each of the jars $1,2, \ldots, 2010$ has exactly one coin. Now, put a coin in each of the jars 2008, 2009, ..,2017. Now, the jars $1,2, \ldots, 2007,2011,2012, \ldots, 2017$ all have exactly one coin. This gives a construction for $N=2014$ (where $d=1$ ). Now, we show that this is optimal. Let $c_{1}, c_{2}, \ldots, c_{2017}$ denote the number of coins in each of the jars. For $1 \leq j \leq 10$, define \[ s_{j}=c_{j}+c_{j+10}+c_{j+20}+\ldots \] Note that throughout the process, $s_{1}=s_{2}=\cdots=s_{j}$. It is also easy to check that the sums $s_{1}, s_{2}, \ldots, s_{7}$ each involve 202 jars, while the sums $s_{8}, s_{9}, s_{10}$ each involve 201 jars. Call a jar good if it has exactly $d$ coins. If there are at least 2015 good jars, then one can check that it is forced that at least one of $s_{1}, s_{2}, \ldots, s_{7}$ only involves good jars, and similarly, at least one of $s_{8}, s_{9}, s_{10}$ only involves good jars. But this would mean that $202 d=201 d$ as all $s_{i}$ are equal, contradiction. $\fbox{2014}$.

Answer: 2014