Ron and Martin are playing a game with a bowl containing 39 marbles. Each player takes turns removing 1, 2, 3 or 4 marbles from the bowl. The person who removes the last marble loses. If Ron takes the first turn to start the game, how many marbles should he remove to guarantee he is the winner?

After thinking about this, I am not sure how to do it.

Because there are 39 marbles, Ron needs to pick a number of marbles in which he can stick to a pattern to guaruntee a win. But how....

Also confirming if I am a robot is so annoying... I know its too prevent advertisement attacks, but like once you confirm it once on a registered account, it should stop asking.

CalculatorUser Sep 22, 2019

#1**+1 **

No matter how many marbles he takes with his first choice Ron can still lose if he is not playing strategically.

I guess we are suppose to assume that Ron is playing the ultimate game and so it Martin.

I have got nothing more.

Melody Sep 22, 2019

#2**0 **

Yes agreed!! thats why Im confused... The answer was apparently 3 but I don't know why.

CalculatorUser
Sep 22, 2019

#3**+3 **

**Ron and Martin are playing a game with a bowl containing 39 marbles. Each player takes turns removing 1, 2, 3 or 4 marbles from the bowl. The person who removes the last marble loses. If Ron takes the first turn to start the game, how many marbles should he remove to guarantee he is the winner?**

Ron first takes **3** marbles.

Than, if Martin takes x marbles, Ron takes 5-x marbles, this repeats 7 times.

The last marble is then for Martin.

39 = 3 + 7*5 + 1

heureka Sep 22, 2019

#4**+1 **

Why 5-x marbles? Is it because that guaruntees martin will choose 4, 3, 2, or 1 marbles?

CalculatorUser
Sep 22, 2019

#5**+2 **

**Why 5-x marbles? Is it because that guaruntees martin will choose 4, 3, 2, or 1 marbles?**

**Hello CalculatorUser **

No, it is because that guaruntees that after Martin and Ron the sum of removed marbles is **5**.

or in other words Ron supplements each time to 5.

Here is the course

\(\begin{array}{|r|r|l|} \hline \text{Martin} & \text{Ron} & \text{sum} \\ \hline & 3 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ \{4,3,2,1\} & \{1,2,3,4\} & 4+1 = 5,\ 3+2=5,\ 2+3=5,\ 1+4 = 5 \\ 1 \\ \hline \end{array}\)

heureka
Sep 22, 2019

#6**0 **

Eventually....at some pont, the captcha robot thingie quits happening....I do not know what the tripping point is.... ~ EP

ElectricPavlov Sep 22, 2019