+218 Rick and Morty and playing a game involving 60 french fries. On a turn, Rick or Morty may eat 1 to 7 french fries. The player who eats the last french fry wins the game. Rick can't garuntee he'll win the game unless he goes first and removes x fries. The value of x is:
a) 1
b) 4
c) 2
d) 7
e) 6
+483 Let us call a position where you can win if you play correctly a winning position, and any other position a losing one. We can see that having 1-7 left is a winning position, and 8 is a losing position, etc. Notice that whatever amount \(x\) that Morty takes on a given turn, Rick can always take \(8-x\) french fries, and thus 8 will be taken in total. This means that if Rick can turn the number of french fries into a multiple of 8, Morty will be in a losing position, and so Rick will win by just taking \(8-x\) french fries each turn, lowering the number by 8 each time until it reaches 0. In order to turn 60 into a multiple of 8, you need to take away 4, so the answer should be \(\boxed{b) \space 4}\)
For winning of a game,
Must be obey the rules
of the game. i.e. last
french fries finished on
his turn.
Total french fries = 60
Rick or Morty may eat 1 to 7
french fries on a turn
Now,
Rules for win is the last
french fries eat on a turn.
If rick start to eat fries
than, for wining Rick must be
eat last french fries
So, the only possible way is
if he pick up 4 fries on his turn
than he win i.e 60/4 = 15 times
total turn out of the ricks turn
8. So he win
So, the option (b) is correct
+313 60 ÷ (1+7) = 60 ÷ 8 = 7...... 4
{Rick take four first then Marty takes N, and Rick
takes (8-N), which can ensure that the last one is
taken by Rick.}
