How will Y lose the game  ?X and Y are playing a game. There are eleven coins on the table and each player must pick up at least one coin but not more than five in a turn. The person picking up the last coin loses. X starts. How many should X pick up to start to ensure a win no matter what strategy Y employs 

N

X picks 4.
Y picks N$N$ coins.
X removes 6−N$6-N$ coins.
There is now 1 coin on the table because we removed 4+N+6−N=10$4+N+6-N=10$.

Generalization.

There are K$K$ coins on the table and one player can pick as many as M coins at once.

Case 1:
M >= K. This is trivial. Player 1 picks K−1$K-1$ coins, leaving 1 coin and wins.
Case 2.1:
M<K$M<K$ and K≢1(mod (M+1))$K\not\equiv 1(\text{mod}(M+1))$
player 1 picks P$P$ coins in such a way that (K−P)=K1≡1(mod (M+1))$(K-P)=K_1\equiv 1(\text{mod}(M+1))$
player 2 picks N$N$ coins (N<M$N<M$).
Player 1 picks M+1−N$M+1-N$ coins leaving again the number of coins on the table K2≡1(mod (M+1))$K_2\equiv 1(\text{mod}(M+1))$
Repeat this process until Kq=1$K_q=1$ and player 1 wins.
Case 2.2
M<K$M<K$ and K≢1(mod (M+1))$K\not\equiv 1(\text{mod}(M+1))$
Player 2 can follow the same strategy as player 1 followed in case 2.1 and win.

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