Summing to 15
Alice and Bob alternately choose numbers, one at a time, from the set:
1, 2, 3, 4, 5, 6, 7, 8, 9
The first to select 3 numbers whose sum is 15, wins the game.
If Alice goes first, it there a strategy which insures her win?
1, 2, 3, 4, 5, 6, 7, 8, 9
The first to select 3 numbers whose sum is 15, wins the game.
If Alice goes first, it there a strategy which insures her win?
Labels: SharedPuzzle
posted by Zaux at
10:48 AM
9 Comments:
I think this is right but I'm not sure.
These are the only ways to make 15 using those with the numbers only once:
195
285
375
645
186
276
348
249
and permutations of thereof the numbers appear the following times:
1= twice
2= three times
3= twice
4= three times
5= four times
6= three times
7= twice
8= three times
9= twice
As I see it if she picks 5 first she can pick any number not restricting any options further on. this wont completely mean a win but it will mean definitely no loss.
does that count?
I'd just about given up, but then realised that the rules don't say Alice (or Bob) can't choose four or five numbers.
It can't be done if only three numbers can be selected, because whatever two Alice starts with, Bob simply removes the one that Alice needs on his second go.
I don't know what the solution is though.
The best I can come up with is if she starts with 3 and he chooses 8, then she'll win with 3 7 2 6 and he has 8 5 4. The mirror image of this works too, if she picks 7 3 8 4 and he gets 2 5 6.
If she starts with 3 and he does 7, it'll come to a draw at best.
If I have time, i'll work on it some more..
you guys are bacically right ... there is no strategy to insure a victory for Alice.
Personally I think only EB deserves any credit.
Was this a trick question? Bit mean if is as it takes quite a lot of effort to think it through. Not that I've put much effort in.
This is the correct answer: NO
Start with 5. The opponent can pick any number, say 9, it does not matter. Now you can pick another number, say 7. Now to play defense, your opponent must use the 3 to prevent you from getting 15. Now you can pick another number, say 2. Once again, your opponent is forced to play defense by picking 8.
With this pattern, there are 4 "pairs" of numbers: 4 & 6, 3 & 7, 2 & 8, 1 & 9 where if you pick one, the opponent must pick the other. It is easy to see that 15 can be achieved without the number 5 by the following combinations:
8+1+6=15
6+7+2=15
4+9+2=15
8+3+4=15
All of these equations contain a number from the 2-8 pair and the 4-6 pair, thus if your opponent starts with any of these, the result is always a draw, analogous to a cat's game in tic-tac-toe where you start with the middle and the opponent picks a corner. If they start with a number from the 3-7 or 1-9 pairs, you are guaranteed a victory, as if in the tic-tac-toe the opponent started with a side square instead of a corner.
I will not talk about the case of starting with a number other than 5, but it easy easy to see that playing with a competent opponent will always result in a draw. This may be easier to see if you look at a 3x3 magic square.
Therefore, there is no ensured method of victory.
Is it assumed that each number can only be chosen once, because if not Alice can choose 5 three times.
If I'm not mistaken the rules do not say she can't, so I see this way as being right
And sorry if someone said this already. I'm quite tired
Hi jake. I nearly offered that solution.
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