Problems with Count-Down Clues

PROBLEMS plus COUNT-DOWN CLUES and ANSWERS

For each problem-page there are several clues that count down (... 3, 2, 1) — so you can do the maximum amount of "figuring it out on your own" — until the answer is revealed at "0".
Eventually I'll find a higher-tech method (using pop-up windows or in some other way) that makes it easy for you to see each clue by itself, one at a time, so you won't also see the next clue, but for now you'll just have to cover up the clues with a piece of paper, or by controlling the scrolling.

A. What has more mass, a pound of feathers or a pound of gold? clues and answer

B. You have 9 silver dollars — 8 genuine and 1 fake (it is heavier than the others) — and an accurate "two pan balance" (with the fulcrum in the middle of a meter stick, with everything symmetric) that lets you distinguish between real and fake coins. What is the least number of weighings that guarantees you'll be able to find the fake? What is your strategy for doing this? clues and answer

C. You have 9 stacks of coins, each containing 20 coins; one stack is entirely counterfeit. A real coin weighs 2.0000 grams, while a bogus coin is 2.0100 grams. You have a balance that accurately weighs a stack to within .0001 gram. What is the least number of weighings that will guarantee you can find the counterfeit stack? What is your strategy for doing this? clues and answer

Problem A: What has more mass, a pound of feathers or a pound of gold?

4-A Clue: This problem is not trivial (like asking "What color was Napoleon's white horse?") because a pound is a measure of force, not mass. If I asked "What has more weight,..." the answer would be "they both have the same weight." But for the actual question, "Which has more mass,...", usually (except under very special circumstances of weighing) the masses will be different. An object's weight is defined as the downward force of gravity acting on its mass.

3-A Clue: Two special circumstances would be if you did the weighings on the moon, or in a vacuum tube with (almost) all of the air removed.

2-A Clue: The masses must be different, for the same reason that a helium balloon rises in the air. (Archimedes' Principle)

1-A Clue: Draw a force diagram (if you've had a physics course, you'll know what this is and how to do it) that includes the upward "buoyancy force" acting on each object, the feathers and the gold. Archimedes' Principle, which is a theory describing what we observe, states that an object's buoyancy force is equal to the weight of fluid (air, water,...) displaced by the object, which is proportional to its volume and thus depends on its density. {On the moon, or in a vacuum, there is no air-fluid so there is no buoyant force.}

0-A Answer: The high-density gold has a small volume, so it displaces a small amount of air, and thus has a small upward buoyant force. But the lower-density feathers have a larger volume, and thus a larger upward buoyant force, so they require a larger downward gravity force (due to gravity acting on its mass) in order to get a downward net force of one pound. The feathers need a larger mass, to overcome their larger buoyant force, so (except in a special "no air" situation) a pound of feathers has more mass than a pound of gold.

A Numerical Example: I don't know what the density of feathers (like goose down) is, but it's much smaller than the density of gold. But feathers are not the lowest-density material. What is? By googling [lowest density solid] you will discover aerogel (1 2 3) with a density (in the Guiness Book of Records) of 1.9 mg/cm3, compared with air (1.2 mg/cm3) and gold (19.3 g/cm3). For aerogel, 1 cubic centimeter has a downward weight-force proportional to 1.9 (its own mass) and upward buoyant-force of 1.2 (mass of air displaced), for a net weight-force of .7, so to find the "effective weight" of an aerogel object you would have to multiply by .7/1.9, and to get a weight of 1 pound you would need (1 pound)(1.9/.7) = 2.7 pounds. This is a very significant buoyancy effect. But for gold (19.3 g/cc) the multiplying factor is only 1.000062, so a pound of aerogel would have a mass that is larger by a factor of 2.7 because "2.7/1.000062" = 2.7.

Problem B: You have 9 silver dollars — 8 genuine and 1 fake (it is heavier than the others) — and an accurate "two pan balance" (with the fulcrum in the middle of a meter stick, with everything symmetric) that lets you distinguish between real and fake coins. What is the least number of weighings that guarantees you'll be able to find the fake? What is your strategy for doing this?

2-B Clue: A comparison of two stacks (we'll call them L & R, for Left & Right) can give you three results — L is heavier, R is heavier, or they're the same weight.

1-B Clue: What is a systematic way to logically use this "3 results" principle in a step-by-step strategy for getting more and more knowledge about the coins.

0-B Answer #1 (for "What is the least number of weighings?"): You can always find the fake coin with 2 weighings. { With luck, you could do it in 1 weighing, but I asked for "the least number of weighings that guarantees you'll be able to find the fake." } What is the strategy?

0-B Answer #2 (for "What is your strategy?): Split the coins into 3 groups, each with 3 coins. If you weigh two of these stacks, you'll know whether either is heavier, or neither, so you'll know which stack has the fake. Using similar logic, a second weighing, using 2 coins in this stack, will tell you which coin is the fake.

Problem BB: In three weighings, what is the largest number of coins for which you can always find the fake? {and what about four weighings?}

1-BB Clue: How can you get a "group of 9" so you can do the two-step strategy described above?

0-BB Answer: You can split 27 coins into 3 groups of 9, use one weighing to choose the fake-group, and then use the two steps above. Similarly, with four weighings you can find the fake in 28 to 81 coins.

Problem BBB: How can you find the fake, in two weighings, in a group of 4 coins? or in a group of 5, 6, 7, or 8 coins? What about three weighings for a group of 10, or 11, 12,... 25, 26?

1-BBB Clue: What will a comparison of two groups-of-2 (L with 2 coins, R with 2 coins) tell you?

0-BBB Answer: With 4 coins, first compare one 2-group with another 2-group, then a second weighing (to compare the two coins in the fake-group) will show you the fake. How can you use this strategy with 5, 6, 7, or 8 coins?

With 5 coins, do 2-versus-2, and (if necessary) a second weighing. With 6 coins, do 2-versus-2 with 2 leftovers. With 7 coins, do 2-versus-2 with 3 leftovers. With 8 coins, do 3-versus-3 with 2 leftovers.

and for the second part of the problem — What is your strategy for 10 coins? and for 11, 12,... 25, 26?

For 10 coins, you could do 3-versus-3, and then you'll know the coin is in a 3-group (requiring one more weighing) or a 2-group (requiring two more weighings, as in the "4 coin strategy" above). For 11 coins, after a 3-versus-3 you'll need either one more weighing or (for the group of 5) two more. A similar strategy works for 12, 13, 14, or 15, but not for 16? So how can you find the fake in a group of16 coins? or 17,...?

For 16, do 8-versus-8, and then the "strategy for 8" above. For 17, do 8-versus-8, and then (if necessary) the 8-strategy. 8-versus-8 also works for 18 thru 24, and (with a 9-strategy for the leftovers) for 25. But what can you do for 26 or 27?

For 26, compare 9 with 9, followed by a 9-strategy or 8-strategy. And for 27, it's 9-versus-9 plus the usual strategy (as in the original B-Problem) for the group of 9 that contains the fake.

Problem C: You have 9 stacks of coins, each containing 20 coins; one stack is entirely counterfeit. A real coin weighs 2.0000 grams, while a bogus coin is 2.0100 grams. You have a balance that accurately weighs a stack to within .0001 gram. What is the least number of weighings that will guarantee you can find the counterfeit stack? What is your strategy for doing this?

4-C Clue: This balance is different than in the B-Problem (it can give you more information) so you should use a different strategy.

3-C Clue: This balance can tell you HOW MUCH HEAVIER a fake coin is. and vice versa?

2-C Clue: Compared with two real coins, how much extra weight is in two fake coins? If you detect this much extra weight (.0200 g extra) how many fake coins do you have?

1-C Clue: Can you use this strategy to know how many fake coins are on the scale? How can you use this ability to find the fake stack in the least number of weighings?

0-C Answer #1 (for "What is the least number of weighings?"): You can always find the fake coin with 1 weighing. How? {answer is below}

0-C Answer #2 (for "What is your strategy?): Weigh 1 coin from Stack #1, and 2 coins from Stack #2, and so on through 9 coins for Stack #9. If the 45 coins were all real, the weight would be 90.000 grams. If the weight is 90.0400 g (which is .0400 g too much) you know there are 4 fake coins, so they must have come from Stack #4.

Problem CC: Using this strategy, how many 20-coin stacks could be evaluated?

1-CC Clue: The correct answer is not 20. Why?

0-CC Answer: You can evaluate 21 stacks by taking 1 coin from Stack #1,... and 20 coins from Stack #20. If there is no extra weight, then you know that Stack #21 has all of the fakes. { Of course, in the C-Problem you could take any number of coins from each stack, as long as they're all different numbers and you keep good records. For example, you could take 5 coins from Stack #1, and 19 from Stack #2,... and 14 from Stack #9. }

Later, this page will end with a link to another page that describes my efforts to make a high-tech version of this "count-down clues plus answer" strategy for education, and sharing ideas with other teachers about how this kind of teaching strategy (with creative variations and extensions) can be used to help students learn a wide variety of ideas and thinking skills.