# Can we find the numbers for which the minimum of the net result is maximum?

A and B play a game.A selects one number from the set {1,2,..,9} at first and supplies it to B.B puts a plus or minus sign before the number(this act is visible to A).The process is repeated twice such that B can place a minus sign before a number only once. and A may select the same number without any restriction twice or thrice.

Then,can we find the three numbers such that the minimum result obtained by placing the minus sign before one of them is greater than any other 3 numbers selected by A?

Thank you.

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In the title you say "net result" -- what does that refer to? In the body, you only speak of the "result obtained by placing the minus sign before one of them". Since I think the problem would be trivial if you just mean the number itself, I'm guessing you intended to introduce a sum somewhere? – joriki Aug 2 '11 at 8:24
@joriki: my guess is that the OP intends it to be A picks a number, B puts a plus or minus, A picks another number, B puts the other sign, then A puts a third number and the resulting expression is evaluated. Sabyasachi, is this correct? – El'endia Starman Aug 2 '11 at 8:27
@El'endia -- I see, that makes sense -- I was thinking of these as signs, not as operators. – joriki Aug 2 '11 at 8:29

A should pick $7$, $5$, $1$ and achieves $11$. We can deduce this by evaluating the value of each move, starting at the back.
When B has used the minus sign, all remaining moves are worth $9$ for A. If B still has the minus option in the last move, A must select $1$, and the value of the move is $-1$. If B still has the minus option in the penultimate move and A selects $n_2$, B can either use the minus, for a value of $-n_2+9$, or not use it, for a value of $n_2-1$. A must select such that these two are the same, so $-n_2+9=n_2-1$ and $n_2=5$, and the value of the two moves is $4$. In the first move, if A selects $n_1$, B can either use the minus, for a value of $-n_1+18$, or not use it, for a value of $n_1+4$. So A must select $n_1$ such that $-n_1+18=n_1+4$, and thus $n_1=7$, and the value of the game is $11$.