Question

fun questions


1)There are 49 lamps set in a circle, and a tool that can toggle the state (on/off) of any set of 5 consecutive lamps. Tintin has to apply this tool several times to toggle the state of a single lamp. What are the minimum possible no. of times Tintin would have to use this tool to achieve this?

2)There are 2 boxes, one having 15 apples and the other having 12 apples. Tintin and Sharktooth can eat equal number of apples from both the boxes or any number of apples from any of the boxes. Whoever eats the last apple will win, Tintin will go first. How many apples will Tintin eat in his first move to win?

Answer 1

My answer: Tintin is smart.. there for can figure it out by himself for i watched Tintin ( the anime) as a little kid and he solved every mysterious clues.

Answer 2

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Answer 3

2) All of them
1) I don't want to deal with that many numbers

Answer 4

2) whoever reaches the condition (2,1) he wins the game.so tintin first picks up 9 apples from both boxes,so it will be (6,3) apples. in any move from sharktooh tintin can reach the situation (2,1) or (1,2) so he wins the game

Answer 5

If you go all the way around the circle turning on as many lights as possible you'll eventually get to 50 lights turned on. But since there are only 49 unique spots, that means one light was toggled twice so it was untouched. So I can think of "toggling all lights except for 1" as a single "move" which costs 10 uses of the tool.

If I focus that move on two separate points, it will end up toggling all the points twice except for the two points chosen which will each get toggled once. So the net effect that doing 20 uses of the original tool ends up getting us 2 separate points turned on with nothing else changed.

Since this strategy I've come up with can only turn on two lights at a time, the total number of lights created in this way will always be an even number so you can't uniquely turn on one light this way.

Maybe there's some other strategy to take, since this doesn't work on its own. Plus just cause I find a way doesn't guarantee that it's the minimum either.

Answer 6

One strategy I'm considering is apply the 5 tool twice but moved over by 1 so that you end up with a net of 2 lights on with 4 lights in between unaffected. Now you can apply this "move" to itself multiple times to get two lights turned on that are separated by \(5n+4 \mod 49\) with \(n\) being the number of times this move is used, which counts for 2 uses of the tool, so \(2n\) tool uses in all.

Kinda fun just playing around but it looks like that on its own won't be super helpful either since it only produces two lights. But it's like a telescoping series on a ring or something so idk just fooling around.

Now that I'm thinking about it, forget all this and going to the previous post, I can use that strategy and turn on 4 arbitrary points, then use the regular tool to turn them all off while leaving the light I do want on. That means I can do it in \(41\) moves. But I don't know if that's the minimum.