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

 Nov 30, 2021
edited by MathyGoo13  Nov 30, 2021

C) 2

 Dec 1, 2021

Thank you. Can you please explain that as well?

MathyGoo13  Dec 1, 2021

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


        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

 Dec 2, 2021

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.}

 Dec 2, 2021

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