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# Varsity Math Week 108

Photo: Luci Gutierrez

The coach loves to pose problems involving weight to refine the team’s logical and analytical skills.

Nine Coins
The coach provides the team with nine coins that are identical in appearance, telling them that three of the coins weigh 3 units each, three weigh 2 units each, and three weigh 1 unit each. Hannah knows what each coin weighs, but needs to convince the coach that she knows.

How many weighings on a simple two-pan balance does she need to make to convince the coach that she knows which coins are which?

Five Questionable Coins
Ethan is holding a true coin and is told that among five identical looking coins on the table, one is counterfeit. The counterfeit is heavier or lighter than a true coin.

How can he find the counterfeit in two weighings using a simple two-pan balance?

*****

See answers to last week’s puzzles below.

Come back next week for answers and more puzzles.

The National Museum of Mathematics (MoMath) is North America’s only museum devoted to mathematics. Learn more at momath.org.

The answer for Ten from Two involves thinking outside the box (or more properly, thinking outside the star; see the diagram below). The answer for Triangular Boundary is 18.

Photo: National Museum of Mathematics

See MoMath’s website for extended solutions.

D B Miller user

I like to look at the Nine Coins problem this way:

The coins are actually colored. The 3-unit coins are red, the 2-unit coins are brown, and the 1-unit coins are green. But the coach is red/green color blind, and can't tell them apart. He needs "proof" that Hannah can tell the difference, and the 333 v 222111 and 3 v 111 (Or 33 v 222) weighings provides such proof.

Dorit Hochbaum subscriber

Nine Coins - I agree with the majority that the 2-weightings procedure (2WP) suggested by DBM, or its variants, is the answer the puzzle-master expects, but I have a different take on the solution. My starting point is: H cannot strictly prove her prior knowledge of the full weights assignment (FWA). Even if she simply tells the FWA, there is 1/1680 chance that she has 0-knowledge and just made a lucky guess. So rather than aim at disclosing the FWA (not a necessary requirement), she should devise a series of operations that minimize P(p|NK), the probability of passing (p) successfully the procedure while not knowing (NK) the FWA. A lower P(p|NK) procedure makes the proposition that H knew the FWA more plausible. TBC.

Dorit Hochbaum subscriber

“Not knowing (NK) the FWA” consists of 8 possibilities: knowing the weights of 0, or 1, …or 7 coins. Clearly, P(p|NK) > P(p|K0). For the 2WP I got: P(p|K0) = 1/1680; P(p|K1) = 1/504; and P(p|K7)=5/6. Beside the difficulty in calculating the rest of the terms, the puzzle does not provide enough information to actually calculate P(p|NK). Instead, I attempted to use P(p|K0) as a criterion for the degree of plausibility provided by any given procedure.

Unfortunately, the 2WP and a 3WP (3-21, three times) both gave P(p|K0) = 1/1680.  P(p|K0) is probably too crude to tell if there is a procedure that makes H’s proposition more plausible than if she simply tells the FWA.

Dorit Hochbaum subscriber

Nine Coins – All the 8 proposed 2WP below have equal degree of plausibility as measured by P(p|K0) = 1/1680. This is the same as if H just tells the FWA. All procedures, except JL’s, end in disclosing the FWA.

Consider an extension of JL’s 2WP: 333-111222, 22-112, 2-11. it provides a better plausibility: 1/(1680*3) and the FWA. An even better result is a 4WP: 12-3, 12-3, 12-3, 22/111, which provides the FWA and P(p|K0) = 1/(1680*20). Using the P(p|K0) criterion, this 4WP makes H’s claim 20 times more plausible than the benchmark.

Note: Once the FWA is known, additional operations do not increase plausibility. The 22/111 step was borrowed from ST. It can be replaced by: 11*-2, 11*-2 (*- using same coin), or by: 1-2, 1-2, 1-2, all give the same increase factor of 20.

Stuart Tucker subscriber

I believe she can't prove that she KNEW which were which (she could have picked randomly and lucked out), but only that she does AFTER her demonstration.

Stuart Tucker subscriber

Nine Coins.  There seem to be several 2 weighing solutions.  Mine is: 111222 vs 333. Which proves 333 are known.  Then 111 vs 22.  Which proves 111 are known.  So the other must be a 2.

I still considering Droit's 1 weighing solution.

D B Miller user

@Stuart Tucker Interesting second weighing. Even though 111 vs 22 does not balance, the 111 will be lighter than the 22, and with the 3's already marked as known, this works as well. Nice use of the balance scale and an mismatched weighing.

Hitesh Shah subscriber

NINE COINS: First, Hannah puts three 1-unit coins on one side of the balance, and one 3-unit coin on the other side. By doing this, she proves that she knows all three 1-unit coins and one 3-unit coin since the largest weight one coin can be is 3-units and only three 1-unit coins can match that weight. Then she removes the 1-unit coins (because she has already proved that she knows their weight) and adds all three 2-unit coins to that side. To match this 6-unit weight, she adds one 3-unit coin to the other side of the balance, which still has the one 3-unit coin added in the first step. This proves that she knows all three 2-unit coins and one more 3-unit coin, and by elimination, the last coin must be a 3-unit coin. Thus with only two weighings on a simple two-pan balance, Hannah proves to her coach that she knows which coins are which. - Amil & Hitesh Shah

June Ewell subscriber

I think to "prove" she knows the IDs of the 9 coins (granted there's a small probability of a lucky guess), she puts the coins into their respective 3, 2, 1 piles of 3 each. One 3 opposite three 1s balances, showing that she knows the three 1s and one of the 3s. Then she puts the two remaining 3s opposite the three 2s. When they balance, she's shown that she knows them all. -Bob Ewell

Fallaw Sowell subscriber

It seems that she must perform two measurements.

First place the three 1 unit coins on one side of the scale and a single 3 unit coin on the other side.  If this balances then the three 1 unit coins have been identified.   No other combination of three coins equalling a single coin.

The other test is to place the three 2 unit coins on top of the three 1 unit coins on one side of the scale and the three 3 unit coins on the other side of the scale.    This will balance only if the three coins on top of the three 1 unit coins are the three 2 unit coins and on the other side are the three 3 unit coins.

Werner Weingartner user

Hi math friends, A friend of mine looked @ my solution and said I have no idea whatyou did so here goes Jay.   weighing #1   she has to prove she knows weight of each coin     place on left side 333 on the right side 222111 bot =9 and balance now the probality that she knows   is 1/3x1/4x1/7=1/84    1-1/84 =83/84.     on the second weighing place 222 on left side and 1113 on the right side which balances proving she knows (not certainty but  1/2x1/3x1/4=1/24 so probabilty of her getting this by chance is 1/84x1/24=1/2016   so the prob that she knew is 2015/2016 very high is that Ok Jay

Dorit Hochbaum subscriber

Nine Coins - In my opinion, none of the suggested weighing procedures constitutes a tight proof that Hannah knew, IN ADVANCE to weighing, the correct weights assignment. They only make it plausible, to different degrees of probability, that she did know. It is possible that she knew nothing and still pass JL’s procedure as if she knew.

For example, Ignorant-Hannah can follow JL’s procedure and randomly choose the correct three 3-coins to balance the other 6 with probability P = 1/84. Ignorant-Hannah has a P=1/7560 chance to twice balance the balance and pass as Knowledgeable-Hannah. So passing successfully JL’s test is not strictly a proof but rather an indication of high degree of plausibility for prior knowledge. For a traditional proof one needs to demonstrate a procedure where random choices have P=0 to be the correct choices and that, I think, is impossible. The coach after all only asked to be “convinced”, not “proved”.

D B Miller user

Sure looks like two weighings for absolute proof in the Nine Coins problem. Doing any weighing where the scale does NOT balance the first time proves nothing. So the initial weighing must balance. As the total weight of the 9 coins is 18 units, each side on the first weighing will have 9 units on each side. This can be done via:

333 = 222111

3321 = 32211

3222 = 33111

The 333 vs 222111 weight proves she knows the three 3-unit coins, but doesn't prove she can tell the difference between the 2-unit and 1-unit coins. So for a second weighing, take one of the known 3-unit coins and put it against the three 1-unit coins.

The other two weighings above are ambiguous, you don't know which of the two scenario's it is, so it would take at least another weighing to prove anything. Since we already have a solution with two weighings, there is no need to proceed any further.

Werner Weingartner user

Hi  prob#1    to prove she knows   weigh   333/222111 so she can state left side is 333   on the next weighing   weigh   222/1113 clear proof she knows which is which

Richard Reiter subscriber

@Werner Weingartner agree with that and also it is similar to the 2nd weighing of 2-3s verses 3-2s then when equal the remaining 3 must be 1s

Richard Reiter subscriber

Part can be done in 2 weighing's   T 1,2,3,4,5

Weigh T&1 verses 2&3 if same it is either 4 or 5 so put 4 verses T and see if they are the same then it is 5 or if different was 4

If T&1 different from 2&3( say T&1  less than 2&3 but does not matter) but now 4 &5 are known to be real

For the 2nd weighing take T&4 vs 1&2   if T&4 still less than 1&2 then 2 is the heavy one and counterfeit,  if T&4 is now heavier then 1& 2 then 1 is the counterfeit. If they are the same then then 3 was the counterfeit.

Dorit Hochbaum subscriber

Nine Coins - 1 weighing. There are two combinations of balancing 4 coins with 5 coins: 3321-32211 and 3222-33111. Both require knowing to which group each coin belongs.

Richard Reiter subscriber

@Dorit Hochbaum I wrote the same explanation as James just posted on needing 2 weighings to positively prove each coin identity

3- 3s vs the other 6 if they are equal the 3s are proved

then 2-3s and 3-2s if they are equal then she identifies the 3-2s leaving the remaining 3 as  3-1s identitified

Your method does not prove she can identify each coin just lumping them together could be a lucky lump for each of your 2 groups not identifying which coin is which

James Lewis subscriber

@Dorit Hochbaum Suppose she can identify two of the 3's, one of the 2's and one of the 1's. Then she can easily make the first measurement, and she knows there is one 3, two 2's and two 1's on the other side, but how does she identify which ones exactly are 3's, 2's or 1's without further measurement? If she can't, then she has only proven to the coach that she knows the weight of 4 of the coins

Werner Weingartner user

part b is easy- call coins abcde  weighing1    ab /cd if they balance ans E say cd goes down     now weighing 2     bc/de  now its easy tell the odd coin.    prob a  need some time to solve but it looks like long road

James Lewis subscriber

Five Questionable Coins - Label the true coin t, and the other five coins a, b, c, d, e. Put t and a in the left pan and b and c in the right pan. If there is no deflection, remove a and put td in one pan and bc in the other pan. Since b and c are true, if there is no deflection the counterfeit is e; if there is, the counterfeit is d. Now go back to the first measurement. If there was a deflection on the first measurement, put td in the left pan and ac in the right pan and perform the second measurement. If there is no deflection, the counterfeit coin is the one removed, b. Otherwise, it is either a or e, and the counterfeit can be lighter or heavier. On the second measurement, if there is a deflection and it is in the opposite direction to the deflection on the first measurement, the coin that was moved, a, is counterfeit. If the same deflection occurs, coin c is counterfeit.

James Lewis subscriber

Nine Coins - Two measurements. 1) Put the three 3's in one side and the other 6 coins in the other side. The scale balances, proving that she knows which three coins have weight 3, because no other arrangement of six on one side and three on the other balances. Put these three aside. 2) Put two of the 2's in one side, the other 2 and two of the ones in the other side. Put the third 1 aside. The scale balances, which proves she knows which coins have weight two, because no other arrangement of the six she coins, with two on one side and three on the other, balances. Since she has proven she knows the weights of the three coins that weight three, and the three coins that weigh two, she must know the weights of the three others that weigh one, so she knows the weights of each of the nine coins.

Richard Reiter subscriber

@James Lewis After you put aside the 3 -3s you know that there were 3 -2unit and 3 -1

unit coins  remaining. if you place 2 coins on one side and 3 coins on the other side and they are equal you have only proven that the 2 coins on one side were each 2 unit coins but you haven't proven which of the three on the other side is the 2. It could be argued that it was any of the three coins on the other side.

But instead, after identifying the 3 -3 unit coins in the first weighing,  if you then take 2 of them and weigh them against the three 2 unit coins and it is equal the 3 -2 unit coins are now identified leaving the other 3 as definitely 1 unit coins.

James Lewis subscriber

@Richard Reiter

Hannah has to prove to the coach that she knows the weight of each coin, but she does not have to prove to the coach which individual coin has which weight. In my second measurement she is able to select the three two-unit coins and place two of the three in one side and the other one in the other side, so she obviously knows the weights of the three two-unit coins. Since she’s already identified the three-weight coins, she must also know the identities of the one-unit coins. This proves to the coach she knows the weights of all nine coins.

What if she only had known the weights of two of the 2-unit coins? She could put those in side 1. But then how does she select the remaining 2-unit coin to put in side two? She could end up putting three 1-unit coins in side 2. The fact that she selected the correct 2-unit coin to put in side 2 shows that she did in fact know the identity of the third 2-unit coin. This proves that she knew the weights of each of the coins, which is what was required that she prove to the coach.

I agree there are several ways of dividing up the coins to conduct the weighings. Both our methods prove that she knows the weights of each of the 9 coins. However, she still can’t in general prove to the coach which coin is which, even though she herself knows which is which. For example, in your method, how would she prove to the coach which of the 4 coins remaining after the second weighing is the 3-unit coin and which are the 1-unit coins, without another measurement?
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