Need help with math extra credit.?

Question

My teacher gave us a problem as extra credit and the problem is:
You are given 12 coins and 1 of those coins weights more or lesser than other 11 coins. So, if you were given a balance and you were to find out the the coin that was heavier or lighter and could use the balance only three times, how would you get to you answer for finding the counter coin

Answer 1

The complete procedure is explained in my reference, which includes a very nicely formatted chart.

For simplicity, I will refer to the coin of different weight as "fake" and the other eleven as "real." Start by dividing the twelve coins into three groups of four. Weigh two of the groups against each other (1). If they are equal, one of the remaining four is the fake one, and all eight that you weighed are real. So of the four that might be fake, take two of them and weigh them against a third possible fake and a known real coin from the first weighing (2a). If they are equal, it is the fourth possible fake that is in fact the fake, and you can weigh it against any of the known real coins to see if it is heavy or light (3a). If weighing (2a) showed that they were unequal, weigh the two possible fakes that were on the same pan against each other (3b). If they are the same, they are both real and the possible fake from the other pan is fake; if that pan was heavy in weighing (2a), the fake is heavy, and if that pan was light, the fake is light. If they are unequal, it is the heavier one if that pan was heavy in weighing (2a), and the lighter one if that pan was light.

If the first weighing shows that the two groups of four were unequal, you know that the third group contains four real coins, and one of the eight you weighed is fake. Once you figure out which coin is the fake, you'll also know if it was heavy or light, depending on if the pan it was in was heavy or light in that first weighing. Now weigh two coins from the first group and one from the second (the two groups that might contain the fake), against one from each of the three groups (2b). If they are equal, the fake must be either the one from the first group that you didn't weigh, or one of the two from the second group that you didn't weigh. Take the two from the second group and weigh them against each other (3c). If they are equal, it is the one from the other group that is fake. If they are unequal, it is whichever one matches the status of the pan the second group was in during the first weighing (heavy or light).

If the mixed groups from weighing (2b) were unequal, then either the two coins from the same original group were on a pan that was heavier both times or lighter both times, or was heavier once and lighter once. If they had the same weight both times, one of two things is true: a) One of them is the fake, and you know whether it is the heavy one or the light one; weigh them against each other (3d) to find out. Or b) they are both real, and the same weighing (3d) will tell you that, in which case it is the one from the second group that was on the other pan in both weighings that is fake. If, on the other hand, the two coins I mentioned were heavier once and lighter the other time, the fake is either the one from the first group that was weighed with coins from both other groups in weighing (2b), or the one from the second group that was weighed with two from the first. Take this two candidates and weigh one against a known real coin (3e). Either it will be real, too, in which case the other candidate is the fake, or it will be proven to be the fake itself.

Answer 2

To solve this problem, all you have to do is take 10 of those 12 coins.

CASE 1: The heavy coin will NOT be included among the 10 coins.

In this case, all we have to do is test the remaining two coins, determine which one is heavier, and we would have found our coin in two tries.

CASE 2: The heavy coin IS included among the 10 coins.

This means one side of the balance will be heavier than the other. Gather these 5 coins.

For your second try, use the balance to weigh 2 and 2 coins.

CASE 2A: They weigh the same. In this case, it's obvious; the unweighed coin must be it.

CASE2B: They don't weigh the same. Take the heavier side, and the weigh 1 and 1 coins. On your third try, you should have gotten the heaviest coin.

Answer 3

For this you have to know if the coin weighs more or less than the others.
Put 6 on each side of the scale, the one that weighs more/less (depending on what the odd coin weighs) has the odd coin in it. Split those 6 into two groups of 3 and compare them. Find the one with the odd coin. Now compare 2 of the 3 in the group with the odd coin. If one weighs more/less then that is the coin. If they weigh the same then it is the 3rd coin you did not weigh.

Answer 4

It depends on what kind of balance it is, and what your teacher counts as "using the balance." If the balance has 2 trays, like the old-fashioned ones, I would place 6 coins on each tray, then take off one coin from each tray simultaneously, repeating this until the trays were level. In a sense you'd only be putting items on the balance once, and when the trays were level, you'd know that the one coin you just pulled off from the once heavier end was the heavier coin.

Answer 5

Split the coins in two groups (A + B) with six coins on each.

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Assumption 1 (the coin is lighter)

Weight groups A and B

keep the lighter of the two groups and split it into two new groups ( C and D) with 3 coins on each

Weight groups C and D

keep the lighter of the two groups and split it into three new groups ( E, F, G) with 1 coins on each

Weight groups E and F

Keep the lighter of the two and that is your coin if they are both equal then G group has your coin.
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Assumption 2 (the coin is heavier)

Weight groups A and B

keep the heavier of the two groups and split it into two new groups ( C and D) with 3 coins on each

Weight groups C and D

keep the heavier of the two groups and split it into three new groups ( E, F, G) with 1 coins on each

Weight groups E and F

Keep the heavier of the two and that is your coin if they are both equal then G group contains your coin.
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Answer 6

okay, there are 3 8 hour shifts throughout the time of the day and no elf is permitted to artwork better than one shift. If there are 9,000 elves operating collectively, this suggests they are all element of the same shift, fantastic? If it really is genuine than those 9,000 elves won't be able to artwork again for something of the day because they are in effortless words required one shift. So if there are 3 shifts and 9,000 are continuously operating collectively, you should mulitply 9,000 by technique of three to get the entire type of elves operating each day: 27,000 elves. Now if each and each elf needs cookies for the time of his smash, you in effortless words multiply 27,000 by technique of two: fifty 4,000 cookies. So very last solutions: 27,000 man or woman elves artwork each day. fifty 4,000 cookies are necessary each day

Answer 7

is this the problem or are their actual numbers.... cuz if this is how the question is actually wrote then without numbers or a balance by theory you couldn't give the answer....