Math problem :)

Pre-College Issues High School Life
1

<p>Given 12 balls; where one ball is lighter/heavier than other 11. Find which one differs from other for a minimum amount of weighings.
I know solution for 3 weighings but won't tell ya :P</p>

2

<p>Haha, I saw this exact problem on Cyberchase.</p>

3

<p>split into groups of 6 and 6
of the one that lighter split into groups of 3 and 3
of the three lighter balls remove on and do 1 and 1
if one is lighter than you found it,
if they are equal, then removed one is lighters.</p>

<p>Yeah I just figured it out!</p>

4

<p>What if the "special" ball is HEAVIER?</p>

5

<p>It still works,
split into groups of 6 and 6
of the one that heavier split into groups of 3 and 3
of the three heavier balls remove on and do 1 and 1
if one is heavier than you found it,
if they are equal, then removed one is heavier.
GO LOGIC!</p>

6

<p>Check out the original problem. It says the ball is either heavier or lighter but you don't know which one it is -- you know that it is different.</p>

7

<p>lol mr. chipset, clearly, you have to use inductive combined with deductive reasoning here so he is right,</p>

8

<p>Ok well you have got me stumped. W/e</p>

9

<p>
[quote]
lol mr. chipset, clearly, you have to use inductive combined with deductive reasoning here so he is right,

[/quote]

Can you use profanity like normal people do?</p>

10

<p>Hmmm, let me take a stab at it. Weigh two sets of four, if equal, then set one from the original eight ball X and compare it to ball Y and Z. If X and the original is lighter, then compare Y and Z, if equal, X is light, if Y>Z Y is heavy, if Z>Y, then Z is heavy. If X and the original is heavier, compare Y and Z and use the same mentality only opposite [heavier is now lighter], if the two sets are equal then leftover ball Q is either heavier or lighter depending on when it's compared against others, please don't make me go on, it gets much more complicated</p>

11

<p>
[quote]
if equal, then set one from the original eight ball X and compare it to ball Y and Z.

[/quote]

Can you elaborate on this step? What do you mean under "original eight ball X"?</p>

12

<p>you know Profanity literally means In Front Of from the Pro. I'm guessing fanity would be conscience.</p>

13

<p>sorry one of the originally eight weighed balls and a left over ball X weighed against left over balls Y and Z.</p>

14

<p>would u rather me explain in letters or something to be more specific</p>

15

<p>How do you choose "one of the originally eight weighted balls"?</p>

16

<p>
[quote]
lol mr. chipset, clearly, you have to use inductive combined with deductive reasoning here so he is right,

[/quote]
</p>

<p>Is it really necessary to state the obvious? Am I just retarded, or does this apply to nearly every situation?</p>

17

<p>Tony88, how many weightins to maximum are you using by the way?
Psycho, you can ignore Masterus :)</p>

18

<p>Okay fine, but you wouldn't want to see me when I'm angry:</p>

<p>Okay twelve objects labeled A to L</p>

<p>ABCD = EFGH</p>

<p>then
if HI = JK then L is the odd one and can be easily found whether to be lighter or heavier when weighed against any other object</p>

<p>or if HI > JK then if J = K; H is heavy, if J > K, then K is light if K > J then J is light</p>

<p>or if HI < JK then if J = K; H is light, if J > K, then J is heavy, if K > J then K is heavy.</p>

<p>There's obviously more which use similar logic but I'm really not up for explaining much further then this though obviously this doesn't describe every possible scenario but you get the picture.</p>

19

<p>How many steps in maximum are you using when ABCD > EFGH or ABCD < EFGH. I can trace your logic, but in the case I mentioned it's not obvious whether ABCD will equal IJKL or EFGH. So will it use one more weightin and therefore make maximum of 4 for your method?</p>

20

<p>Fine, here goes nothing:</p>

<p>ABCD > EFGH </p>

<p>then if ABE = CFL, then if G = H, then D is heavy, if G>H then H is light; if H>G then G is light</p>

<p>then if ABE > CFL, if A=B then F is light, if A>B then A is heavy, if B>A then B is heavy</p>

<p>then if ABE < CFL, if C=L then E is light, if C > L then C is heavy and L > C can't exist</p>

<hr>

<p>and if ABCD < EFGH, then you follow each independent procedure and what was once found to be heavy is now light and vice versa. I hope this all made sense and any scenario can be found in three weighs</p>