Afaceinthematrix

Here is a fun math puzzle. It's moderately difficult; it took me a couple of hours to figure out. If you're interested in math, or like these types of puzzles, then here it is...

The Story:

King Bozo is hosting a party and he would like to play a game. He sets everyone up at a round table and he walks around eliminating every third person and the last person wins. So if one person plays, person 1 obviously wins because he's the only one there. If two people play, then one will win because it goes 2, 1, 2 (eliminated), one survives. If three people play, person 3 will win because it goes 2, 3, 1 (eliminated) 2, 3, 2 (eliminated). So basically it starts on the next person, and every third person is eliminated. To make it easier, I'll simply post the winning number for each amount of people that play up to 20.

1 1

2 1

3 3

4 1

5 3

6 5

7 7

8 1

9 3

10 5

11 7

12 9

13 11

14 13

15 15

16 1

17 3

18 5

19 7

20 9

.

.

.

It can go on forever.

The Problem:

Find an equation for which you can insert the amount of players playing (for x), and it will produce the winning number. I found this problem a while back and decided to post it. So basically you just need to find an equation for those set of numbers. If you put 20 in for x, the equation will produce 9, if you insert 14 in for x, you will get 13, if you insert 16 in for x you will get 1, etc.

Good luck! I'll post the answer in a few days if people are interested, and no one gets it. It took me a couple of hours to get it.

The Story:

King Bozo is hosting a party and he would like to play a game. He sets everyone up at a round table and he walks around eliminating every third person and the last person wins. So if one person plays, person 1 obviously wins because he's the only one there. If two people play, then one will win because it goes 2, 1, 2 (eliminated), one survives. If three people play, person 3 will win because it goes 2, 3, 1 (eliminated) 2, 3, 2 (eliminated). So basically it starts on the next person, and every third person is eliminated. To make it easier, I'll simply post the winning number for each amount of people that play up to 20.

1 1

2 1

3 3

4 1

5 3

6 5

7 7

8 1

9 3

10 5

11 7

12 9

13 11

14 13

15 15

16 1

17 3

18 5

19 7

20 9

.

.

.

It can go on forever.

The Problem:

Find an equation for which you can insert the amount of players playing (for x), and it will produce the winning number. I found this problem a while back and decided to post it. So basically you just need to find an equation for those set of numbers. If you put 20 in for x, the equation will produce 9, if you insert 14 in for x, you will get 13, if you insert 16 in for x you will get 1, etc.

Good luck! I'll post the answer in a few days if people are interested, and no one gets it. It took me a couple of hours to get it.