Community Event Puzzle 2 Stacks Giveaway - Ended!

Hello there

I planned on doing a giveaway for a long time, but I couldn’t think of anything special. But then I remembered this puzzle my brother told me.

In a flat with 6 floors and 6 apartments on each floor, are thieves making plans to rob a bank.
These thieves are wanted by the police because they are involved in another heist.
But due to privacy reasons is the police allowed to only look in 1 apartment a day. The thieves know that and they made up a pattern to outsmart the police.They start in a random apartment on a random floor. The second day they are in the apartment right or left from the apartment they were in the first day. Then the third day they are right or left from the apartment they were in the second day. And the fourth day left or right of the apartment they were the third day. The fifth day they are in the apartment above or under the apartment they were the fourth day.
That pattern repeats itself.
So the pattern short is: 3 times left or right, 1 time above or under.
The thieves will not leave the flat, and they can be in the same room more than once.
There aren't any securtity camera's around the flat and the police cannot camp outside of the flat to see where the thieves are going. they also cannot ask pedestrians.

The question is: what is the least possible days where the police has a 100% success chance of finding the thieves.
You must answer with the amount of days which sequence of searching they follow.
The person who is correct wins 2 stacks of le!
If your answer isn't correct you can join again

First hint:

Spoiler: Solution

So when the thief moves on one floor he moves 3 moves which means he moves either from even -> odd or from odd -> even during that entire floor.
So then when searching for even starting number (and floor) the police has to search
Floor 2: 2,3,4,5.
Floor 3: 5,4,3,2.
Floor 4: 2,3,4,5.
Floor 5: 5,4,3,2.
Then the thief will be caught using an even starting number and even floor number.

Now similar patterns for odd apartment and even floor.
So let's start over with Floor 2.

Floor 2: 5,4,3,2
Floor 3: 2,3,4,5
Floor 4: 5,4,3,2
Floor 5: 2,3,4,5

Now the thief will be caught when he starts on an even floor.

So now let's cover when the thiefs starts on an odd floor.
Using the same pattern as before this means starting at floor 5.

For finding odd apartment odd floor start:

Floor 5: 5,4,3,2
Floor 4: 2,3,4,5
Floor 3: 5,4,3,2
Floor 2: 2,3,4,5

Then to find for even apartment and odd floor start:

Floor 5: 2,3,4,5
Floor 4: 5,4,3,2
Floor 3: 2,3,4,5
Floor 2: 5,4,3,2

Now the thieves will be caught when combining these 4 patterns. And combinding them is just adding them onto each other since the start number and and number will be odd when the starting is odd and even when the starting is even.

And for the floors works as well.

After 16*4 days.

 

7 days idk xd

 

The police has a chance of finding the robbers in the right room every 1 in 54,022,500 days. First day, police have 36 rooms to chose from, 2nd day they have 35, 3rd day they have 35, 4th day they have 35 as well.

Idk? I don't got a scientific calculator to be sure to I'm going off rough math :(

 

10, 20, 30, 40, 50 etc. Because the Thieves can go right, left, right, left, up. Right left right left down. There for repeating the pattern of 10, 20, 30, 40, 50. They can also go left, right, left right.

 

sequence needed to enter

well as you are probably thinking you are wrong

they only go 3 times right or left before they go up or down

 

1 Day: The police would use the cameras and video footage outside the apartments on each floor to catch them leaving the apartment and what room they enter next. They can also have men stationed on each floor to catch them. This riddle only works if the thieves can get into the apartments in the first place. After they find the room they are in, they would select the room with 100% success.

 

oh ye, i forgot about that part
the police cannot wait for them to come out and there are no securtity camera's
i'll give you credit for finding this, i'll send you a pm.

 

6 days, because the robbers move in a left and right sequence, up or down. The police can check which flats have been occupied, and therefore can find out the pattern in 3 days.

 

how can the police check which ones were occupied?

 

Desk receptionist; people have to check in xd

 

got a working solution which gave 120 days as answer, with a sort of itterative formula for how they choose the next one - basically confirming they arent on a certain floor 1 floor at a time (takes 5 moves) assuming starting in either even or odd room on even or odd floor, then *4 for all those combinations

 

lol, they are thieves, why would they check in
btw it isn't a hotel

but they are 4 days on the same floor...
for example if they are on floor 1, on day 5 they move up to floor 2, starting at day 1 again

 

thats why its 5 days to check - the 5th is to see if they went up or down

 

ok 10 days, and the caretakers of the apartments still have to clean out rooms and collect trash every day, therefore they can still check which rooms have been occupied recently that people did not pay for.

 

well the answer isn't correct

it is a apartment, there are no caretakers

 

If you think about it from a certain perspective there would probably never be an 100% chance the police will know where the thieves are, even if they do figure out the pattern. It would always be 50% if the thieves moved around correctly.

I was thinking that the police could find evidence to suggest that the thieves have been in the room, but I'm assuming that's out of the picture.

 

ofcourse there's a 100% since they can only be in the same room twice. And that means theres a limited amount of patterns.

 

It never said they can only be in the same room twice. It actually just says more than once, which I'm assuming means infinite.

 

there is an answer where the police has a 100% chance of finding the thieves

ye that isn't the solution

they can be in the same room more then once, but not 2 days straight. i didn't say anything of only being able to be in the same room twice

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that is true

 

oh thought u said that earlier