You have 12 objects that r identical to each other, but one object weighs either less or more then the others, U have a balancing scale to use, but u only get to use it 3 times. how do u find which object is the one that weighs either less or more then ther others.
2007-05-02 14:59:10 · 5 answers · asked by sir_septamus 1 in Science & Mathematics ➔ Other - Science
This is a famous old problem with a very clever solution that someone will probably Google up while I'm typing this. Not being that clever, I'll just describe the general, brute force method.
Each of the three weighings can have one of three results - left pan heavy, right pan heavy, or balanced. That means there are 3x3x3 = 27 possible outcomes. The case where the two pans are always balanced tells us nothing, so eliminate it. That leaves 26 distinct results. The number of allowed variations in the coins is 24; twelve in which one coin is heavier, and 12 in which one coin is lighter. So, you should be able to pick out the odd coin of up to 13 with three weighings.
For that brute force solution, just write out all the possible permutations of outcomes - L, 0, 0; L, R, 0; etc. Now, assign one of the permutations to each possible solution. For example, if you assign "coin 2 heavy" to L, R, 0, you put coin 2 on the left in the first weighing, on the right in the second weighing, and leave it out of the third weighing. Once you have created a table of weighings, do them, and from the result pick the solution that matches the result.
2007-05-02 17:54:11 · answer #1 · answered by injanier 7 · 0⤊ 0⤋
Balance 6 on one side and 6 on the other side.
Take the side that weighs more/less than the other, and put 3 on one side and 3 on the other.
Take the side that weights more/less than the other, and put 1 on one side and 1 on the other, and leave one out.
You will see that one side weights more/less than the other. If the sides weigh the same, the one you have left out weighs more/less.
2007-05-02 22:03:45 · answer #2 · answered by Mathlady 6 · 1⤊ 0⤋
ok i've thought about this and...i honestly have f*ing idea! head! its hurts. :( ummm...oh oh oh. ok i'm gonna take a guess. put 6 on each side. then seperate the 6 that weigh less to 3 on each side. then ummm. i don't know. i'm stuck.
2007-05-02 22:15:18 · answer #3 · answered by OTHcutie 3 · 0⤊ 0⤋
6/6
3/3
1/1
1 out
2007-05-02 22:17:31 · answer #4 · answered by MARIA 2 · 0⤊ 0⤋
um choose wich ever one is either bigger or smaller?
lol i don't know
2007-05-02 22:05:08 · answer #5 · answered by brendz 2 · 0⤊ 0⤋