Rita and Sam play the following game with n sticks on a table. Each mu

The question defines a game- whoever removes the last stick wins. You can remove up to 5 sticks, so if you get to a situation where there are 5 or fewer sticks left and it's your turn, you are certain to win. If you are in a situation where there are 6 sticks and it's your turn, you have to remove at least one stick, and you can't remove all 6, so your opponent certainly is left with between 1 and 5 sticks, and certainly wins. So really what you'd want to do is force your opponent into a situation where he or she has 6 sticks. You can do that if you have between 7 and 11 sticks. And you will automatically have between 7 and 11 sticks on your turn if you force your opponent to have 12 sticks- which you can do if you have between 13 and 17 sticks, and so on. As long as you can force, after your turn, the number of sticks on the table to be a multiple of 6, you can be certain to win if you play correctly. The only way Rita loses, if she goes first, is if the number of sticks on the table is a multiple of six when the game starts. That is, for the answer choices, only D) defines a situation where Rita is certain to lose (as long as Sam plays well).

Or, the longer explanation- let's look at some of the possibilities here:

If n=7, and Rita goes first, Rita will remove just 1 stick. Sam is facing 6 sticks now, so whatever he does, Sam is going to lose, Rita is going to win.

If n=12, and Rita goes first, after Rita's turn, there will be between 7 and 11 sticks left. Sam can remove enough to leave Rita with 6 sticks, so Rita is going to lose, Sam is going to win.

If n=22, and Rita goes first, Rita will remove 4 sticks. There are 18 left. No matter what Sam does, he leaves between 13 and 17 on the table. Rita can now remove enough to leave 12 sticks. After Sam moves, there are between 7 and 11- Rita removes enough to leave 6 left. etc. Sam loses, Rita wins.

And so on.

Hi Bunnel,
Please explain this:

N = 12, here 1 and 2 shows steps in a game: rita picks 5 first, out of remaining 7 sam can pick a maximum of 5, which leaves 2 sticks after round one. On her next chance rita can pick 2 and win.

R S
1 5 5
2 2 > Rita wins

similarly:
R S
1 4 5
2 3 > Rita wins
R S
1 2 5
2 5 > Rita wins
R S
1 2 2
2 5 3 > Sam wins
R S
1 2 3
2 5 2 > Sam wins

So both can win when n=12.
I agree for n=6, but not for n=12.

That;s not correct.

Both players can win BUT if the number of sticks on a table is a multiple of 6, then the second player will win in any case IF the player is smart enough.

n=12: no matter how many sticks will be removed by the first player 1, 2, 3 , 4 or 5, the second one can remove 5, 4, 3, 2, or 1, RESPECTIVELY so that to leave 6 sticks on the table.

so what is the generalisation in such questions or we just have to analyze everytime?

To have a sure shot winner, you need complimentary moves. You have to analyze to figure out the complimentary move every time, of course.

The trick is to rephrase the question in more general terms. In this case it would be: What is the number that can always be divided into even number of times when each division can be up to 5. The answer is one greater than 5 which is 6 because whatever be the first value chosen, the second value can be chosen such that 6 can always be divided into two. The same idea can be extended to the multiples of 6 such that they can always be divided even number of times given that each division can be from 1 to 5.

Let’s go through the choices.

A. 7

Since Rita goes first, she can remove only 1 stick. (If she removes any other number of sticks, then Sam will be able to win on his next turn.) Then there are 6 sticks left, and no matter how many sticks Sam removes, Rita will always win (for example, if Sam removes 2 sticks, Rita can then remove 4 sticks for the win). Since we want Sam to win, choice A is not the correct answer.

B. 10

Since Rita goes first, she can remove 4 sticks. Then there are 6 sticks left, and no matter how many sticks (up to the 5-stick limit) Sam removes on his turn, Rita will win on her next turn. Thus, there is no way Sam can win.

C. 11

Since Rita goes first, she can remove 5 sticks. Then there are 6 sticks left, and just as in the explanation for answer choice B, there is no way Sam can win.

D. 12

Since Rita goes first, she can remove any number of sticks, from 1 to 5 (inclusive). Then Sam can remove a number of sticks so that there are 6 sticks left. For example, if Rita removes 2 sticks, Sam should remove 4 sticks so that there are 6 sticks lefts. Once there are 6 sticks left and since it’s now Rita’s turn, no matter how many sticks she removes, Sam will always win (for example, if Rita removes 5 sticks, Sam can then remove 1 stick for the win).

Answer: D

If there are 6 sticks on the table before a turn, then after that turn, there will be between 1 and 5 sticks on the table, and the next player can win. So if Sam can, no matter how Rita plays, force Rita to chose from 6 sticks at some point, Sam can always win (assuming he plays perfectly). No matter how many sticks Rita chooses on her first turn, Sam can always choose a specific number of sticks so that, combined, they remove exactly six sticks from the table (if Rita removes one, Sam can remove five, if Rita removes two, Sam can remove four, and so on). So if there are 12 sticks to begin with, no matter how many Rita removes on her first turn, Sam can remove enough sticks on his turn so there are 6 sticks left on the table. Then Sam can always win. So the answer is 12.

I still didn't understand. This is not a probability question to guess how many they will pick at a time. It's 1,2,3, 4 or 5. Confusing!

Posted from my mobile device

Hi - This is a reasoning question.

Question:
Rita and Sam play the following game with n sticks on a table. Each must remove 1, 2, 3, 4 or 5 sticks at a time on alternate turns, and no stick that is removed is put back on the table. The one who removes the last stick (or sticks) from the table wins. If Rita goes first, which of the following is a value of n such that Sam can always win no matter how Rita plays?


Explanation:

Let us first decide the winning strategy for Rita:
In order to win, Rita (R) must be able to pick the last stick.

In order to decide the winning strategy, Rita must be able to make something constant in the game based on which he would do her calculations.

For any number of sticks that Sam picks, say k, Rita can always pick (6 – k) sticks in her next pick; thus maintaining the sum of the number of sticks as {k + (6 – k)} = 6, i.e. a constant.
For example: If Sam picks 1, Rita can pick 6 – 1 = 5; if Sam picks 2, Rita can pick 6 – 2 = 4; and so on ... if Sam picks 5, Rita can pick 6 – 5 = 1.

Thus, if the number of sticks is 52 (let us assume any random number to see what happens), we would have: We remove as many multiples of 6 as possible from 52, we would be left with 4 sticks. Rita needs to pick all 4 of these in the beginning so that she can finish the game. The sequence in which the sticks are picked would have been:




Now, let us decide: how can Rita lose? This can only happen if the total number of sticks is a multiple of 6. In that case, Rita HAS to pick some number in the first draw; and then, Sam will use the above strategy and ensure Rita loses. The diagram below explains this situation:

Let the number of sticks be 24 (a multiple of 6) and Rita picks k in the first attempt:




The only multiple of 6 in the options is 12

Answer D

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the options 10 11 12 are quite close and confuse one . nice question! identifying who leaves at n=6 wins. that is key!