Have You Lost Your Marbles?

The crux is to reach 6 marbles. After that you whatever the school bully may choose to pick up you can force him to pick the last marble. The only way to ensure that is by picking up 2 marbles in your first chance.

The trick is that you can use your opponent's move by making your move in such a way that you can form a concrete equation. The equation I created is, no matter how many marbles your opponent takes, you take the 4's compliment of it... i.e., if your opponent takes 3, you will take 4-3 = 1 marble. As a result, the number of marbles remaining in the heap after n rounds is A-4n (A is the initial number of marbles) So, your aim now is to make the number of marbles available to your opponent in the form of 4m+1. That way, one marble will be left with your opponent. There fore, if you take 2 marbles in your first move, your opponent will face a heap of 9 marbles which is in the form of 4m+1. All you need to do from now on is to keep taking the 4s compliment of your opponent.

I noticed that it was possible, no matter what number of marbles the bully picked up, it was possible for me to pick up a number of marbles that would allow us to pick up four marbles in our two turns combined. That is, if the bully picks any number of marbles, it is possible for me to pick a number that will make the total number of marbles picked up in our two turns combined, equal to four.

Ex: If the bully picks up 3 marbles, I can pick up 1 marble.

If the bully picks up 2 marbles, I can pick up 2 marbles.

If the bully picks up 1 marbles, I can pick up 3 marbles.

This kind of sequence can occur twice since 4 divides into 11 twice, with a remainder of 3. Since I want the bully to end up with one marble on his last turn, and 3-1=2, I will have to pick up two marbles in my first turn.

The following shows the sequence of turns that will occur.

First turn (Me): 2 marbles

9 marbles left

First turn (Bully): X marbles

9-X marbles left

Second turn (Me): 4-X marbles

5 marbles left

Second turn (Bully): Y marbles

5-Y marbles left

Last turn (Me): 4-Y marbles

1 marble left

Last turn (Bully): 1 marble

0 marbles left

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Note: X and Y are two random numbers. Their values can range from 1 to 3, as these are the possible number of marbles that the bully may pick up.

you need to bring it to the form 1 + 4x form after your pick... i.e. 1,5,9,13,.... because, once you reach such a number, you can pick [4 - (whatever he picked)] to bring ot to the 1+4x form whih will finally lead to 1 marble being left for him to pick!! putting him into detention

I can avoid the detection only if Insist him to take the last marble. Now, I can take one, two or three marbles, so in the last move of mine, I have to make sure there should be 2,3 or 4 marbles remain for me. Initially there are eleven marble If I take $2$ marbles then there remains $9$ marbles.

Now if he takes 1 marble there remains 8 marbles.

if he takes 2 marbles there remains 7 marbles.

if he takes 3 marble there remains 6 marble.

In your next move, depending on his move you have to pick up one, two or three marbles so that there remains 5 marble, so after that whatever is his move you can have him in your trap.

We can see that if our opponent plays first, we can always remove 4 balls in one turn : - If he removes 1 marble, I'll remove 3 - If he removes 2 marbles, I'll remove 2 - If he removes 3 marbles, I'll remove 1 Therefore, the solution is always to make him play when there are $4n+1$ marbles on the table, when n is an integer. Then I'll always use the strategie given above to go back to a " $4n+1$ " solution at his turn, until $n=0$ , and he has to remove the last marble. So, when we start with $n=11$ , I have to remove 2 marbles so that he plays when there are $4\times 2+1$ marbles. This solution works with every number of marbles except you have to let your opponent start if there are exactly " $4n+1$ " marbles.

First, we notice that if you receive 1 marble, you loose. Thus, if you receive 2, 3 or 4 marbles, you can win as you can reduce it to one and thus your opponent looses. Thus you loose if you receive 5, as you can only reduce it to 2, 3 or 4, and thus your opponent wins. Thus, if you start by taking two, your opponent can reduce it to 8, 7 or 6, and in each case you can reduce it to 5, and thus your opponent looses. Thus you are guaranteed to win if you take 2 at the beginning. If you take away 3 at the beginning, then your opponent can reduce it to 5 and thus you loose. The only option that fits these two facts is that you take 2 as your first move.

Quite simple. First, you must make sure that in one of the bully's turns (turn x) there must be exactly 5 marbles left on the table. He can take 3 at most, and you too, 1 at least, and you too. You can not take the last one. So if he takes 3 then you take 1; he takes 2 then you takes 2; he takes 1 then you takes 3. Then he loses.

Second, because the two of you can always take a total of 4 mables (he takes 1 you take 3, 2 and you 2, 3 and you 1). So in turn x-1 the must be 9 marbles left. It is the bully turn and you go first. So take 2 from 11 marbles.