Sum Game

**souksfxc** · May 29th 2011, 05:45 AM

James is playing a card game called sum game. He holds 8 cards, numbered 1 to 8. Each time he places a card on the table he finds the sum of all the cards he has placed on the the table so far. To win the game , alternate sums must be divisible by two different numbers. For example, he plays the cards in order 5, 1, 4, 2, 8, 7, 3, 6, the sums are 5, 6,10, 12, 20, 27, 30, 36 which are alternately divisible by 5 and 3, starting with 5.

1. James now plays with 17 cards, numbered 1 to 17. SHow that he cannont place the cards so that the alternate sums are divisible by 4 and 9, starting with 4

2. James has cards numbered 1 to n. He places them so that the alternate sums are divisible by positive integers a and b, starting with a. Show that, if the card numbered 1 is placed in an even position, the maximum value for a + b is 2n + 1

**awkward** · May 29th 2011, 07:19 AM

Originally Posted by souksfxc

James is playing a card game called sum game. He holds 8 cards, numbered 1 to 8. Each time he places a card on the table he finds the sum of all the cards he has placed on the the table so far. To win the game , alternate sums must be divisible by two different numbers. For example, he plays the cards in order 5, 1, 4, 2, 8, 7, 3, 6, the sums are 5, 6,10, 12, 20, 27, 30, 36 which are alternately divisible by 5 and 3, starting with 5.

1. James now plays with 17 cards, numbered 1 to 17. SHow that he cannont place the cards so that the alternate sums are divisible by 4 and 9, starting with 4

2. James has cards numbered 1 to n. He places them so that the alternate sums are divisible by positive integers a and b, starting with a. Show that, if the card numbered 1 is placed in an even position, the maximum value for a + b is 2n + 1

1. Even if we drop the restriction that the first, third, fifth, etc. sums are divisible by 4, no such arrangement is possible. Here is why.

Let's say the cards played are numbered $x_1, x_2, x_3, ... , x_{17}$ and satisfy the condition that the sum of the first two, four, six, .., sixteen cards are all multiples of 9.

Working modulo 9, we must have
$x_1 + x_2 \equiv 0$
$x_1 + x_2 + x_3 + x_4 \equiv 0$
$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 \equiv 0$
...
$x_1 + x_2 + x_3 + \dots x_{16} \equiv 0$

So
$x_1 + x_2 \equiv 0$
$x_3 + x_4 \equiv 0$
$x_5 + x_6 \equiv 0$
...
$x_{15} + x_{16} \equiv 0$

Where is 9 in this list? There is no solution to $x_i + 9 \equiv 0$ with $1 \leq x_i \leq 17$ and $x_i \neq 9$ , so we must have $x_{17} = 9$ .

So all the integers from 1 to 17, except 9, must be among the first 16 numbers in the list: $x_1, x_2, x_3, \dots , x_{16}$ . Let's say $x_1$ and $x_2$ are "friends", as are $x_3$ and $x_4$ ,
$x_5$ and $x_6$ , ... $x_{15}$ and $x_{16}$ . Every number from 1 to 17, except for poor lonely 9, must have a friend. A pair of friends must sum to a multiple of 9.

Who are the possible friends of 1? Only 8 and 17. Let's assume 1's friend is 8. Then 17's friend must be 10 because 1 is already taken. Who can be 10's friend? Nobody is left; 8 and 17 are already taken. So 8 can't be 1's friend. It's left to the reader to show a similar contradiction arises if 1's friend is 17.

So no such arrangement is possible.

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