Test your logical thinking!

[Trun]

Problem 1 (solved):

You want to go in a castle with guards who will let you in only if you know the password. You have no fucking idea what the password is, so you decide to stay in distance and see what others who want to go in say when asked for the password.

After a short time, a visitor arrives.

Guard: 8.
Visitor: 5.
Guard: Welcome.

Guard: 11.
Visitor: 6.
Guard: Welcome.

Guard: 4.
Visitor: 4.
Guard: Welcome.

You already know that you can do it, so you go bravely forward.

Guard: 7.
You: ........

Problem 2 (solved):

You have 7 copper and 1 golden coins which are absoultely similar in appearance. How can you find the golden one with two weighings, having a scale with no weights?

Problem 3 (solved but open for discussion):

How many eggs you need to throw in order to see which is the highest floor of a 100-floor building, where they won't break if they fall? You are locked in the building and can't pick them up and throw them again even if they aren't broken.

Problem 4:

How many throws you need to make to see the same thing if you have only two eggs? You can pick them up and throw them again if they aren't broken.


I have one more which I haven't solved yet, if I solve it, I'll post it :lol:

[albosomething]

1st one 5?

[Trun]

1st one 5?

Yes.

The rest are a bit more complicated, especially the last one.

[albosomething]

the others can't quite understand them , some of them have obvious answers, dunno if its my englishs fault or my brains fault

[Trun]

the others can't quite understand them , some of them have obvious answers, dunno if its my englishs fault or my brains fault

What do you mean with obvious?

[Vorpal]

2: gold is heavier than silver, right?
so you basically have to put each coin on the scale one by one and find out which one is the heaviest
or I understood the question wrong way

first, you place one coin on each weighing, and repeat til you find the heaviest one

[Trun]

2: gold is heavier than silver, right?
so you basically have to put each coin on the scale one by one and find out which one is the heaviest
or I understood the question wrong way

first, you place one coin on each weighing, and repeat til you find the heaviest one

You can do only 2 weighings. To weigh each pair you will need 28 which is pretty much.

But it is true that if you have a golden and a copper coin with the same volume the golden one is heavier.

[Stefan_Dusan]

for gold one, are you allowed to weigh them by dropping the coins one at a time? And can you use that gold is twice as heavy as copper per volume?

[albosomething]

put 4 on one side, 4 on the other, the one that is heavier has the golden one in it..

[albosomething]

than put 2 of them in each .. and the one that is heaviest has the golden one.

fuck yall and your 200 iQ

[Trun]

for gold one, are you allowed to weigh them by dropping the coins one at a time? And can you use that gold is twice as heavy as copper per volume?

Actually, in the original problem the metal of the coins isn't specified; you only know one weights different from the others. Could be heavier, could be lighter. It is irrelevant for the solution.

What do you mean by dropping one at a time?

[Stefan_Dusan]

put 4 on one side, 4 on the other, the one that is heavier has the golden one in it..

Yes but you still don't have the gold coin, then you can divide the 4 into 2 by 2 and select heavier again, but again you still have one copper and one gold, and you need to weigh again, so 3 weighings total.

I can't figure out to get it to 2 unless you can drop the coins one by one instead of all at once.

[Vorpal]

You can do only 2 weighings. To weigh each pair you will need 28 which is pretty much.

But it is true that if you have a golden and a copper coin with the same volume the golden one is heavier.

2 times only?
well, first time you place them 4:4, you select the group of heavier ones
so now you have only 4 unknown coins, then you place them 2:2, and select the group of heavier ones again
damn, still 2

[Trun]

put 4 on one side, 4 on the other, the one that is heavier has the golden one in it..


than put 2 of them in each .. and the one that is heaviest has the golden one.

fuck yall and your 200 iQ

You're in the right track. But this isn't the solution because this way you will need 3 weighings.

[pelikarski]

First is Seven = 5
Second is killing me. Can I use digital weighing scale? :D

[Stefan_Dusan]

You weigh 3 vs 3 and leave 2 on the side, thats how you do it.

[Trun]

Fourth is the most interesting if you ask me. Maybe because I needed to look it up in the internet to see the answer. Ouch.

Third is my creation. I hope the answer I have in mind is right.

After you solve the remaining three I have some others. Lqlqlq.


You weigh 3 vs 3 and leave 2 on the side, thats how you do it.

Yes.

If one threesome is heavier than the other, the golden is in it; you weigh two of these three. If they weight the same, the third is the golden. Otherwise the golden is the heavier of the two.

If the threesomes are equal, you weigh the other two coins; the heavier is...yesh.

Is using the word threesome in the case right by the way? :D

[albosomething]

in the second one you just see which one has the biggest impact i guess

[albosomething]

the fourth one u just throw one from the 99th and 100th floor? lol

[HellLander87]

Solution for 2.I ll ask my jewish neighbour to take a pick.:D

[albosomething]

what type of a fucking dinasour egg would it have to be to not break after throwing that sucker from the 100th floor

[Vorpal]

an egg of svendc

[Trun]

what type of a fucking dinasour egg would it have to be to not break after throwing that sucker from the 100th floor

Imagine it to be a snooker ball, for example. The original problem is Taipe 101 and two snooker balls.


an egg of svendc

You make fun of him but imagine the frustration if he answers all correctly :D

[albosomething]

50/25/12/6/3/2/1


6 or 7 throws

[Trun]

50/25/12/6/3/2/1


6 or 7 throws

I was thinking about 7. It should be something like, if you have n floors, the answer is [lg2(n)]+1. This is what I expect, anyways. This problem is my creation :p

[albosomething]

it was a bit confusing the question.


with 2 throws.. now

[Trun]

it was a bit confusing the question.


with 2 eggs.. now

Fixed.

[albosomething]

the 4th one the easiest ..


you just throw and pick up until the 1st one breaks lol

starting from the bottom ofc

[Trun]

the 4th one the easiest ..


you just throw and pick up until the 1st one breaks lol

starting from the bottom ofc

Nope.

[Mazik]

Problem 3:

How many eggs you need to throw in order to see which is the highest floor of a 100-floor building, where they won't break if they fall? You are locked in the building and can't pick them up and throw them again even if they aren't broken.


Only one. Check how much the weight of the egg is and search at google with your phone when you are inside the building how much Newton an egg can withstand.
It is apparently 35 N.

Then calculate it with the classic potential energy and kinetic energy formulas.

:rolleyes:

[Vorpal]

Only one. Check how much the weight of the egg is and search at google with your phone when you are inside the building how much Newton an egg can withstand.
It is apparently 35 N.

Then calculate it with the classic potential energy and kinetic energy formulas.

:rolleyes:

winner :rolleyes:

[albosomething]

Only one. Check how much the weight of the egg is and search at google with your phone when you are inside the building how much Newton an egg can withstand.
It is apparently 35 N.

Then calculate it with the classic potential energy and kinetic energy formulas.

:rolleyes:

stfu, faggot

[Trun]

I have three more ready.

[albosomething]

post the new ones the last one is too complicated math, dont have the patience to do it

[Stefan_Dusan]

I don't get the last one. What do you mean see the same thing twice?

[Trun]

Here is the answer of the fourth.

http://datagenetics.com/blog/july22012/index.html