1. ## So confused

Rell who should go first, and how many you should take when it is your turn if you yeally, really want to win. (ex. I will go first and take 3 and leave multipules of 4.)

There are 77 objects on the table. Each player, in turn, may remove 1, 2, 3,..., 9 objects from the starting pile. The player who takes the last object loses. Find the winning strategy.

Can someone figure is out?! I need help please!

2. What may I ask is "Rell"

I assume it's the person's choice which number he picks (during any of his turns he could take off 1-9)

My strategy, keep taking 9 off each time until you have only a litte amount of tiles left. Then what you do is take off enough tiles so that your opponent is left with 11 tiles, then you win.

3. Hello, t-lee!

This is a classic (old) problem.

There are 77 objects on the table.
Each player, in turn, may remove 1, 2, 3,..., 9 objects from the pile.
The player who takes the last object loses.
Find the winning strategy.

You play first and take 6 objects.

After than, when your opponent takes n objects, you take 10 - n objects.
. . (If he takes 3, you take 7. .If he takes 8, you take 2.)

He will be left with the last object.

4. Neat trick. Now that you know how it's done, you can go back and see how well the person who gave this problem to you knows it. For example, instead of 77 objects, start with any number (that does not end with a 0 or a 1), and set the same scenario.

For 88, for example, you remove 7 the first time to leave 81 remaining. For 106, you remove 5 to leave 101 remaining. For subsequent pairs of moves (with your opponent going then you going to constitute a pair of moves), you make sure that 10 are removed for each set of moves, as Soroban said, and you will always leave your opponent with 1 object. For 88, there would be 8 pairs of moves before your opponent had the 1 left over. For 106, there would be 10 pairs of moves before you opponent had the 1 left over.