discrete mathematics Two people play a game as follows There

(discrete mathematics)

Two people play a game as follows. There are N coins in a pile, and a positive integer k is fixed for the game. The two players alternate, in turn taking a number of coins from the pile, the number of coins taken being at least 1 and at most k. The objective is to be the person taking the last coins. Show that if N is not divisible by k+1 then the first player has a strategy that always wins.

I received this as an answer:

LET THE TWO PLAYERS ARE A AND B

GIVEN THAT THERE ARE N COINS IN PILE

Can someone please explain this to me?

SO IT IS NOT DIVISIABLE BY k+1

Solution

GIVEN THAT N IS NOT DIVISIBLE BY K+1.

SO IF THE FIRST PLAYER PLANS TO KEEP K+1 COINS ON THE BOARD WHEN HIS TURN COMES ...
THEN HE IS BOUND TO WIN , SINCE THE OTHER PLAYER HAS TO TAKE AT LEAST ONE AND AT MOST K.THEN THE FIRST PLAYER IN HIS TURN CAN TAKE OUT ALL K COINS OR ONE COIN ON THE BOARD AND WIN ..

SO THE STRATEGY SHALL BE TO REACH THAT STAGE IN THE LAST BUT ONE ROUND .SO HE SHOULD SEE THAT THERE IS MULTIPLE OF [K+1] ON THE BOARD WHEN HE DRAWS FIRST . THAT IS HE SHOULD TAKE OUT...[ N - P*(K+1)] > 0 .WHERE P IS AN INTEGER IN HIS TURN FIRST TO GET THE STRATEGIC POSITION AND THERE AFTER CONTINUE TO KEEP THE PILE IN MULTIPLES OF K+1.

OBVIOUSLY THE GIVEN CONDITION THAT ....

\"if N is not divisible by K+1 then the first player has a strategy that always wins\"..