Endless Paradigm

First one: What is the half of 2 + 2?


>.>....



Second one: A man lives in floor 16 of a building. He goes down on the elevator every day to floor 1 to get to his work. Then when he gets back he can only go to floor 12, there he must get off and walk all the way to his floor. Why? What is the cause of this? He does not stop in Floor 12 because he wants to, he has to. Also, he doesn't have any compromises there.


Third one: You have 50 coins, one of these coins weighs more then all the others. They all look exactly the same. You have a balance (The things that are two plates and you play 10 things on one side and 9 on the other and it tells you that the side with 10 weighs more), using it only 4 times, find the coin that weighs in more then the others. You can only use the balance, nothing else. Also, post your full formula. By the way, it's possible.

the second one is because hes only tall enough to reach the button for floor 12

roberth Wrote:the second one is because hes only tall enough to reach the button for floor 12

Good job. What about the first one? It's just basic math dude =/... Also, the third one is also basic math, you just have to think a bit.

2 :D

deep Wrote:2 :D

Sorry, but you're wrong. It's 3. I said the half of 2 + 2, so, the half of 2 is 1 and + 2 means it's 3.

what's the half of (2+2),.. you mean,.!?

3rd one is way to difficult for me,.. i quess you start of with half of the coins on each plate,.!?

u BIIIIIIIIIITTTTTTTTTTTTTTCCCCCCCCCCHHHHHH I HATE UUUUUUU!!!! NOW I WILL KILL YOU AND UR FAMALAMAS IN UR SLEEEEPS!!!

lol deep is no sport,.. ^^^^

3rd one is easy.



Let  x | y represent putting x coins in the left and y coins in the right.



1st: 25 | 25 needs no explaination

2nd: 9 | 9 ok, if the left side is heavier, then the heavy coin is in left. if right heavier, then in right. if left and right the same, then heavy coin in remaining 7 coins that are not tested.

3rd: 3 | 3 same reasoning as second step. it is regadless of whether wee narrowed down to 9 or 7 coins previously, it works both way.

4th: 1 | 1 same reasoning as 2nd step. This step is not needed in the case that the heavy coin passes to the group of 7 in test 2, and test 3 yeilds equal weight.


it's possible with more than 50 coins. u can still solve this (in 4 tries) with at max 3^4 = 81 coins.

deep Wrote:u BIIIIIIIIIITTTTTTTTTTTTTTCCCCCCCCCCHHHHHH I HATE UUUUUUU!!!! NOW I WILL KILL YOU AND UR FAMALAMAS IN UR SLEEEEPS!!!

ROFL

Ahh, Assassinator should have a new title - "Forum Math Whizz" or something.

For the third one though, the method used to solve it is somewhat similar to a binary search (a programming algorithm).  Assassinator should learn programming sometime - some algorithms are quite interesting :P

I couldn't get the 2nd one though...  It also partially depends on how the buttons are arranged, but I'm not an outside thinker >_>