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Given integer $n$ and $k$ such that $0 < k \leq n$ and $n$ piles of stones,two players alternate turns. By one move it is allowed to choose any $k$ pilesand remove exactly one stone from each. The player who has to move but cannotis the loser. in the normal version of the game and (s)he is the winner in themis\`ere version. Cases $k=1$ and $k = n$ are trivial. For $k=2$ the game wassolved for $n \leq 6$. For $n \leq 4$ the Sprague-Grundy function wasefficiently computed (for both versions). For $n = 5,6$ a polynomial algorithmcomputing P-positions was obtained for the normal version. \newline Then, forthe case $k = n-1$, a very simple explicit rule that determines the Smithremoteness function was found for the normal version of the game: the playerwho has to move keeps a pile with the minimum even number of stones; if allpiles have odd number of stones then (s)he keeps a maximum one, while the $n-1$remaining piles are reduced by one stone each in accordance with the rules ofthe game. \newline Computations show that the same rule works efficiently forthe mis\`ere version too. The exceptions are sparse and are listed in Section2. Denote a position by $x = (x_1, \dots, x_n)$. Due to symmetry, we can assumewlog that $x_1 \leq \ldots \leq x_n$. Our computations partition all exceptionsinto the following three families: $x_1$ is even, $x_1 = 1$, and $x_1 \geq 3$is odd. In all three cases we suggest explicit formulas that cover all foundexceptions, but this is not proven.

On remoteness functions of k-NIM with k+1 piles in normal and in mis\`ere versions