What is the maximal value that we can have after 99 operations?

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we begin with the numbers $1,frac12 ,frac13,ldots frac1100$ written in a board.
We do the following operation : we delete $2$ numbers $a$ and $b$ from the board , and we remplace them with the number $a+b+ab$. what is the maximal value that we can have in the board after $99$ operations



here is what i think
if we have for example numbers $a$ and $b$ and $c$ in the board if we choose $a$ and $b$ first we will have two numbers , $(a+ab+b,c)$ then we will have the number $a+b+c+ab+bc+ac+abc$
and if we choose $a$ and $c$ first we will have $a+c+ac$ and $b$ , then $a+b+c+ab+bc+ac+abc$ ...... i think that whatever numbers we choose we will have the same number at the end , but we need to prove it , and we need to know the number that will remain.






combinatorics maxima-minima combinatorial-game-theory combinatorial-proofs







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  • 1




    Show that the operation $astar b=a+b+ab$ is commutative and associative. This proves that the order you do the operations does not matter.
    – Mike Earnest
    Sep 10 at 19:12










  • Is $n$ supposed to be $99$?
    – Michael Hoppe
    Sep 10 at 19:17










  • If $n<99$, then you will not have enough values and the process would not make sense. If $n>99$, then you can wind up with $dfrac1n$ left, which would be the minimal number. If $n=99$, then as @MikeEarnest suggests, you can show that the operation is commutative and associative to show that the order of operations does not matter.
    – InterstellarProbe
    Sep 10 at 19:20










  • Are you looking for a maximal or minimal value?
    – Michael Hoppe
    Sep 10 at 19:23






  • 1




    The hints above are good. An alternative way of solving it is to first show, with $astar b equiv a+b+ab$, that $a_1star a_2 star ldots star a_n = (a_1+1)(a_2+1)cdots (a_n+1) - 1$
    – Winther
    Sep 10 at 20:31














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we begin with the numbers $1,frac12 ,frac13,ldots frac1100$ written in a board.
We do the following operation : we delete $2$ numbers $a$ and $b$ from the board , and we remplace them with the number $a+b+ab$. what is the maximal value that we can have in the board after $99$ operations



here is what i think
if we have for example numbers $a$ and $b$ and $c$ in the board if we choose $a$ and $b$ first we will have two numbers , $(a+ab+b,c)$ then we will have the number $a+b+c+ab+bc+ac+abc$
and if we choose $a$ and $c$ first we will have $a+c+ac$ and $b$ , then $a+b+c+ab+bc+ac+abc$ ...... i think that whatever numbers we choose we will have the same number at the end , but we need to prove it , and we need to know the number that will remain.






combinatorics maxima-minima combinatorial-game-theory combinatorial-proofs







share|cite|improve this question



















  • 1




    Show that the operation $astar b=a+b+ab$ is commutative and associative. This proves that the order you do the operations does not matter.
    – Mike Earnest
    Sep 10 at 19:12










  • Is $n$ supposed to be $99$?
    – Michael Hoppe
    Sep 10 at 19:17










  • If $n<99$, then you will not have enough values and the process would not make sense. If $n>99$, then you can wind up with $dfrac1n$ left, which would be the minimal number. If $n=99$, then as @MikeEarnest suggests, you can show that the operation is commutative and associative to show that the order of operations does not matter.
    – InterstellarProbe
    Sep 10 at 19:20










  • Are you looking for a maximal or minimal value?
    – Michael Hoppe
    Sep 10 at 19:23






  • 1




    The hints above are good. An alternative way of solving it is to first show, with $astar b equiv a+b+ab$, that $a_1star a_2 star ldots star a_n = (a_1+1)(a_2+1)cdots (a_n+1) - 1$
    – Winther
    Sep 10 at 20:31












up vote
0
down vote

favorite









up vote
0
down vote

favorite











we begin with the numbers $1,frac12 ,frac13,ldots frac1100$ written in a board.
We do the following operation : we delete $2$ numbers $a$ and $b$ from the board , and we remplace them with the number $a+b+ab$. what is the maximal value that we can have in the board after $99$ operations



here is what i think
if we have for example numbers $a$ and $b$ and $c$ in the board if we choose $a$ and $b$ first we will have two numbers , $(a+ab+b,c)$ then we will have the number $a+b+c+ab+bc+ac+abc$
and if we choose $a$ and $c$ first we will have $a+c+ac$ and $b$ , then $a+b+c+ab+bc+ac+abc$ ...... i think that whatever numbers we choose we will have the same number at the end , but we need to prove it , and we need to know the number that will remain.






combinatorics maxima-minima combinatorial-game-theory combinatorial-proofs







share|cite|improve this question















we begin with the numbers $1,frac12 ,frac13,ldots frac1100$ written in a board.
We do the following operation : we delete $2$ numbers $a$ and $b$ from the board , and we remplace them with the number $a+b+ab$. what is the maximal value that we can have in the board after $99$ operations



here is what i think
if we have for example numbers $a$ and $b$ and $c$ in the board if we choose $a$ and $b$ first we will have two numbers , $(a+ab+b,c)$ then we will have the number $a+b+c+ab+bc+ac+abc$
and if we choose $a$ and $c$ first we will have $a+c+ac$ and $b$ , then $a+b+c+ab+bc+ac+abc$ ...... i think that whatever numbers we choose we will have the same number at the end , but we need to prove it , and we need to know the number that will remain.






combinatorics maxima-minima combinatorial-game-theory combinatorial-proofs




combinatorics maxima-minima combinatorial-game-theory combinatorial-proofs






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Sep 10 at 19:26

























asked Sep 10 at 19:11









contestant IMO 2020

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  • 1




    Show that the operation $astar b=a+b+ab$ is commutative and associative. This proves that the order you do the operations does not matter.
    – Mike Earnest
    Sep 10 at 19:12










  • Is $n$ supposed to be $99$?
    – Michael Hoppe
    Sep 10 at 19:17










  • If $n<99$, then you will not have enough values and the process would not make sense. If $n>99$, then you can wind up with $dfrac1n$ left, which would be the minimal number. If $n=99$, then as @MikeEarnest suggests, you can show that the operation is commutative and associative to show that the order of operations does not matter.
    – InterstellarProbe
    Sep 10 at 19:20










  • Are you looking for a maximal or minimal value?
    – Michael Hoppe
    Sep 10 at 19:23






  • 1




    The hints above are good. An alternative way of solving it is to first show, with $astar b equiv a+b+ab$, that $a_1star a_2 star ldots star a_n = (a_1+1)(a_2+1)cdots (a_n+1) - 1$
    – Winther
    Sep 10 at 20:31












  • 1




    Show that the operation $astar b=a+b+ab$ is commutative and associative. This proves that the order you do the operations does not matter.
    – Mike Earnest
    Sep 10 at 19:12










  • Is $n$ supposed to be $99$?
    – Michael Hoppe
    Sep 10 at 19:17










  • If $n<99$, then you will not have enough values and the process would not make sense. If $n>99$, then you can wind up with $dfrac1n$ left, which would be the minimal number. If $n=99$, then as @MikeEarnest suggests, you can show that the operation is commutative and associative to show that the order of operations does not matter.
    – InterstellarProbe
    Sep 10 at 19:20










  • Are you looking for a maximal or minimal value?
    – Michael Hoppe
    Sep 10 at 19:23






  • 1




    The hints above are good. An alternative way of solving it is to first show, with $astar b equiv a+b+ab$, that $a_1star a_2 star ldots star a_n = (a_1+1)(a_2+1)cdots (a_n+1) - 1$
    – Winther
    Sep 10 at 20:31







1




1




Show that the operation $astar b=a+b+ab$ is commutative and associative. This proves that the order you do the operations does not matter.
– Mike Earnest
Sep 10 at 19:12




Show that the operation $astar b=a+b+ab$ is commutative and associative. This proves that the order you do the operations does not matter.
– Mike Earnest
Sep 10 at 19:12












Is $n$ supposed to be $99$?
– Michael Hoppe
Sep 10 at 19:17




Is $n$ supposed to be $99$?
– Michael Hoppe
Sep 10 at 19:17












If $n<99$, then you will not have enough values and the process would not make sense. If $n>99$, then you can wind up with $dfrac1n$ left, which would be the minimal number. If $n=99$, then as @MikeEarnest suggests, you can show that the operation is commutative and associative to show that the order of operations does not matter.
– InterstellarProbe
Sep 10 at 19:20




If $n<99$, then you will not have enough values and the process would not make sense. If $n>99$, then you can wind up with $dfrac1n$ left, which would be the minimal number. If $n=99$, then as @MikeEarnest suggests, you can show that the operation is commutative and associative to show that the order of operations does not matter.
– InterstellarProbe
Sep 10 at 19:20












Are you looking for a maximal or minimal value?
– Michael Hoppe
Sep 10 at 19:23




Are you looking for a maximal or minimal value?
– Michael Hoppe
Sep 10 at 19:23




1




1




The hints above are good. An alternative way of solving it is to first show, with $astar b equiv a+b+ab$, that $a_1star a_2 star ldots star a_n = (a_1+1)(a_2+1)cdots (a_n+1) - 1$
– Winther
Sep 10 at 20:31




The hints above are good. An alternative way of solving it is to first show, with $astar b equiv a+b+ab$, that $a_1star a_2 star ldots star a_n = (a_1+1)(a_2+1)cdots (a_n+1) - 1$
– Winther
Sep 10 at 20:31















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