Had trouble uploading the problem, so I just decided to type it:
Working on Problem: 31
WORK ---------------------MENTAL THINKING
Try it:
a) 1 5 7 10 11 15 16 20 24 25 29 31
b) 5 1015 20 25 30 31
c) 5 6 10 15 29 24 25 29 31
Try working it backwards. To be the person to reach 31, must land on 25 to ensure a win because if not- will end up landing on a number between 26-30. This leads to the other player adding to 31.
Try out hypothesis:
d) 1 3 7 8 13 15 1923 2529 31
(Eureka, it works!!) (PLUS, this works every time AS LONG AS you start off with the number 1.. and make sure you don't lose count.)
Try working from it backwards with consistent train of thought... If person who lands on 25 is the winner, then he/she would want to land on 19 to ensure a win. And keep doing the same pattern, and realize we're just subtracting six from the number each time. Ending up with the pattern of wanting to hit the numbers 7, 13, 19, 25, 31 in order to secure a win. In other words, we want to touch x = 1 (mod 6)
Question number two: What if 31 was some other number?
e) Reach 33: 3 8 9 10 14 20 21 22 27 28 33
f) Reach 24: 4 5 10 15 16 17 22
We can try to work backwards again or can we see if we can incorporate modular arithmetic? For example, if we want to reach 33, should we try x=3 (mod 6) where x is 33? Or if we want to reach 24, should we have x = 22= 4 (mod 6)
Question: What if it goes from 1-6?
g) Reach 31: 3 9 10 12 17 23 24 25 31
Well, since it was in mod 6 when it was from 1-5, is it now in mod 7? So let's set up our general equation and check: x = 31 = 3 (mod 7)
Question: What if the numbers were 1,3,5 or 2,3,7??
Ex) 1 2 7 10 13 18 19 22 25 26 31
Try out with 1, 3, 5... Can we still use generality? YES! Why? Because, as long as we start at 1, can always reach x = 1 (mod 6) because of how the numbers work--> at your turn, will be able to chose number that makes it become 7. To make this more clear, if we start with 1, then need to add 6 to become 7. If player 2 says any of the numbers (1, 3, 5), we can find the complement of it that adds up to 6- making it 7. 1&5, 3&3.
** Unfortunately, I got really confused if we extended it to 3 players and came to the conclusion that we can have 2 people gang up on the other player to make sure the 1 person does not win. But I'm not sure if we can have a guaranteed winner.
*** Tried to make a tree (like a probability tree) with testing out 2, 3, 7 to see all the possibilities. But it got really messy and I was unable to figure out the winning strategy if we only use number 2, 3, 7.
