Beam Balance

Let the weights of the 6 coins be $A,B,C,D,E,F$ . Pair them off sequentially onto the beam balance, making a total of 3 weighings. Without loss of generality suppose we find that $A \lt B, C \lt D$ and $E \lt F$ . We then know that the lightest coin will be among $A,C$ and $E$ , and that the heaviest coin will be among $B, D$ and $F$ .

Now weigh $A$ against $C$ . Whichever is the lightest, say $A$ , we then just need to weigh $A$ against $E$ to determine the lightest of all the coins. Similarly, to find the heaviest coin we start by weighing $B$ against $D$ , and then weighing the heaviest of these against $F$ . We have made 4 additional weighing in the process, giving us a total of 7 weighings to determine both the lightest and heaviest coins. The answer to the question is therefore $\boxed{\text{True}}$ .

1. Weigh any 2 coins. Put each into either a heavy pile or a light pile.
2. Do the same with the next 2 coins. You now have 2 heavy and 2 light and 2 unknown after 2 weigh ins.
3. Weigh the 2 heavy coins and discard the loser.
4. Weigh the 2 light coins and discard the loser.
5. Weigh the 2 unknowns and put each into the correct pile, either heavy or light.
6. Weigh the 2 heavy coins and discard the loser.
7. Weigh the 2 light coins and discard the loser. You have now identified both the heaviest and the lightest coins using 7 weighings.

Imagine a tournament of coins to find the heaviest. It takes one round (3 matches) to eliminate half of the coins (A B C and D E F). Of the remaining three coins (D E F), two matches determine the heaviest (F). Now, to find the lightest, you know it lost the first round, so you use the other half of the coins (A B C) and have another round of two matches to discover it (A).

A variant of this question is to ask how many matches you need to find the second heaviest. Clue: It lost against the heaviest.

if you set this up as a tournament with five "matches" you'll find the heaviest. to find the lightest take all the losers (3) from the first round and instead advancing the winer (heaviest) you'll advance the loser (lightest). in two matches you should have your answer

Label all coins 1,2,3,4,5,6. Now make pair as (1,2),(3,4),(5,6) The three above pairs are now sent for a dead match ...suppose 1,3,5 wins in the allotted pairs,then (1,3) are sent to dead match...and suppose 3 wins.. Then (3,5) sent for dead match .Finally say,5 wins(Heavy weight champion) So overall we had 5 matches so far... Let's give the failures their "second chance" (1,4) are sent for a dead match ...say 1 wins..now (1 ,6) are the final challengers ..who will the winer,will be the lightest-weight champion... So total seven matches are done..

Let the weights of the 6 coins be A,B,C,D,E,F. Choose a coin in random to be a constant. Now, weigh the other 5 coins against it. The heaviest coin will be the one which causes the biggest tilt and the lightest coin will be the one which causes the smallest tilt.

First, put 3 coins in each pan of the balance. The set with heavier coins will go down. Now we have three coins which contain the heaviest and other 3 which contain the lightest. You now have two options, to weigh the heaviest first or the lightest. It is simple. You just have to first weigh two coins against each other(this activity will be separate for both the sets) and determine the heaviest among them and then repeat this activity for the remaining coin. You still have 4 chances left of using the balance. So repeat this for the other set and you will get the heaviest and the lightest coins!

Example list with 6 weights: a = [8,4,3,9,1,2]

Find the lowest with 5 comparisons:

cmp(8,4) low = 4, hi = 8

cmp(4,3) low = 3, hi = 8

cmp(3,9) low = 3, hi = 8

* remember number 9, the first non-lower number *

cmp(3,1) low = 1, hi = 8

cmp(1,2) low = 1, hi = 8

Found the lowest number=1, with 5 comparisons.

Find the greatest with 2 additional comparisions:

* start from previous number 9, because all the left is lower or equal to 8 *

cmp(8,9) low = 1, hi = 9

* no need to compare to 1, as it's already proved the lowest *

cmp(9,2) low = 1, hi = 9

Found the highest number=9, with 2 additional comparisons.

This can be done with 7 weighings.

1) Randomly create three pairs of coins.

2) Compare each coin pair using the scale. Separate coins that are heavier into a 'heavy' pile and the lighter coins into a 'light' pile (this will require 3 weighings)

3) Select two coins from the 'heavy' pile at random and compare with the scale. Remove the lighter coin from the scale and place the remaining coin from the 'heavy' pile on the scale. The heaviest coin from this weighing is the heaviest of all the original coins (this will require 2 weighings)

4) Select two coins from the 'light' pile at random and compare with the scale. Remove the heavier coin from the scale and place the remaining coin from the 'light' pile on the scale. The lightest coin from this weighing is the lightest of all the original coins (this will require 2 weighings)

Let the weights be $A<B<C<D<E<F$ and the coins be $c_1, c_2, ... c_6$ . Since we do not know the relationship between the weights of the coins, weighing with more than one coin on a side gives no information. For example, when trying two coins on one side and one on the other, weighing $C, E$ with $F$ may result in either side being heavier than the other because we do not know if $C+E$ is greater or less than $F$ . This will be seen with any other situation where more than one coin is on one side.

Now we weigh two coins, $c_1$ and $c_2$ , against each other. Suppose $c_1<c_2$ . Now we know $c_1$ does not weigh $F$ and $c_2$ does not weigh $A$ . If we weigh one of these two, $c_1$ for example, with another coin, $c_3$ , we will find that one cannot be $A$ and the other might be. If we weigh two unrelated coins $c_3$ and $c_4$ , we will learn this information, but we will also learn that one cannot be $F$ . Therefore, weighing two new coins gives the most information. We will do the same for the last two coins.

WLOG, we have $c_1<c_2$ , $c_3<c_4$ , and $c_5<c_6$ , and we took $3$ separate measurements so far. We know one of $c_1$ , $c_3$ , and $c_5$ weigh $A$ , and one of $c_2$ , $c_4$ , and $c_6$ weigh $F$ . Cross comparing the two groups gives no information about $A$ or $F$ since we already eliminated these weights from each of the groups. We will weigh two of the lighter coins, $c_1$ and $c_3$ , and we find that $c_1<c_3$ . Now $c_3$ cannot weigh $A$ . A similar result would occur if we weighed any two of the lighter coins. This alone does not tell us the lightest coin, so at least one more weight is needed. This can be accomplished by weighing the last two coins that might weigh $A$ , $c_1$ and $c_5$ . With $2$ additional weighings, we found the lightest coin. A very similar method will locate the heaviest coin in the second group with $2$ weighings. Therefore, the minimum number of weighings needed to find the lightest and heaviest coins is $3+2+2=\boxed{7}$ .