# Add four or seven

We propose to play the following game for two players: Starting from scratch, each player in his turn adds 4 or adds 7 to the result that the other player has obtained. If the result that a player gets in his turn ends in two zeros the player wins.

Is there any strategy to win the game?

Extracted from the problemate.blogspot.com.es page

#### Solution

These types of games should always start at the end, to study how you can ensure victory by dividing situations into values ​​with which you win and values ​​with which you lose.

In our case, we must start the study from 100, which is the first time someone can win. Obviously, 93 and 96 are positions from which we can win, so we must prevent our opponent from having them. Looking for a number with which to force our rival to return that value, we soon find 89. If our rival adds 4, he returns 93, and if he adds 7, 96. In either of them, we win. So in reality, whoever leaves a 89 wins.

Reasoning back, we found 78, 67, 56, 45, 34, 23, 12 and 1. If we managed to pass 1 to our opponent, we would surely reach 100 and win.

Unfortunately, we must start by adding 4 or 7. Then we must not aim at 100. If we repeat the system (note that it consists of subtracting 11 each time), we reason from 200 that we would have to start from 2, to reach 300, from 3 and! finally !, to reach 400, from 4.

Now let's check if our system works. We start adding 4. Our opponent makes his play, and we make the opposite, so that we will reach 15. We will continue like this, adding the opposite play to that made by our rival. When we get close to 100, we pass the opponent 4 + 11 * 8 = 92, so that he cannot achieve 100, if not 96 or 99. We add the opposite number, reaching 103, and continue. The next dangerous step will be 103 + 8 * 11 = 191, in which our rival can reach 195 or 198. Again we join to reach 202. The last dangerous step will be 202 + 8 * 11 = 290, in which our opponent can go to 294 or 297. In any case we will arrive at 301. The end of the game will arrive at 301 + 8 * 11 = 389, in which we pass this number to the rival, so that we returns 393 or 396, winning in any case to get 400.