Tanya Khovanova's Math Blog

I gave my students a problem from the 2002 AMC 10-A:

Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, …, 10}. The probability that Sergio’s number is larger than the sum of the two numbers chosen by Tina is: (A) 2/5, (B) 9/20, (C) 1/2, (D) 11/20, (E) 24/25.

Here is a solution that some of my students suggested:

On average Tina gets 6. The probability that Sergio gets more than 6 is 2/5.

This is a flawed solution with the right answer. Time and again I meet a problem at a competition where incorrect reasoning produces the right answer and is much faster, putting students who understand the problem at a disadvantage. This is a design flaw. The designers of multiple-choice problems should anticipate mistaken solutions such as the one above. A good designer would create a problem such that a mistaken solution leads to a wrong answer — one which has been included in the list of choices. Thus, a wrong solution would be punished rather than rewarded.

Readers: here are three challenges. First, to ponder what is the right solution. Second, to change parameters slightly so that the solution above doesn’t work. And lastly, the most interesting challenge is to explain why the solution above yielded the correct result.

I’ve heard many fun problems in which blindfolded parachutists are dropped somewhere and they need to meet up once they’re on the ground. They can’t shout or purposefully leave traces behind. They will recognize each other as soon as they bump into each other. Their goal is to get to the same assembly point. They can design their strategy in advance.

Here is the first problem in a series that gets increasingly difficult:

Two parachutists are dropped at different locations on a straight line at the same time. Both have an excellent sense of direction and a good geographical memory, so both know where they are at any moment with respect to their starting point on the line. What’s their strategy?

The strategy is that the first person stands still and the second one goes forward and back repeatedly, increasing the distance of each leg until they collide.

In the next variation, both are required to execute the same program, that is, if one stands still, then both stand still. To compensate for this increased difficulty, they are allowed to leave their parachutes anywhere. And both of them will recognize the other’s parachute if they bump into it.

In the third variation, the set-up is similar to the previous problem, but they are not allowed to change the direction of their movement. To their advantage, they know which way East is.

I recently heard a 2-D version from my son Sergei in which the parachutists are ghosts. That means that when they bump into each other they go through each other without even recognizing the fact that they met:

Several blindfolded men are sleeping at different locations on a plane. Each wakes up, not necessarily at the same time. At the moment of waking up, each of them receives the locations of all the others in relation to himself at that moment. They are not allowed to interact, nor will they receive any further information as time passes. They need to get together in one place. How can they do that, if they are allowed to decide on their strategy in advance?

They do not know where North is. So they can’t go to the person at the most Northern point. Also they do not know how locations correspond to people, so they can’t all go to where, say, Peter is. Let us consider the case of two men. Suppose they decide to go to the middle of the segment of two locations they receive when they awake. But they get different locations because they wake up at different times. Suppose the first person wakes up and goes to the middle. The fact that he walks while the other is sleeping, means that he changes the middle. So when the second person wakes up, his calculated middle is different from the one calculated by the first person. Consequently, they will never manage to meet. Hence, the solution should be different.

Actually Sergei gave me a more difficult problem:

Not only do they need to meet, but they need to stay together for a predefined finite time period.

Here is as bonus problem.

If there are three parachutists, it is possible to end up in a meeting place and stay there indefinitely. For four people it is often possible too.

In my old essay I presented the following coin problem.

We have N coins that look identical, but we know that exactly one of them is fake. The genuine coins all weigh the same. The fake coin is either lighter or heavier than a real coin. We also have a balance scale. Unlike in classical math problems where you need to find the fake coin, in this problem your task is to figure out whether the fake coin is heavier or lighter than a real coin. Your challenge is that you are only permitted to use the scale twice. Find all numbers N for which this can be done.

Here is my solution to this problem. Let us start with small values of N. For one coin you can’t do anything. For two coins there isn’t much you can do either. I will leave it to the readers to solve this for three coins, while I move on to four coins.

Let us compare two coins against the other two. The weighing has to unbalance. Then put aside the two coins from the right pan and compare one coin from the left pan with the other coin from the left pan. If they balance, then the right pan in the first weighing contained the fake coin. If they are unbalanced then the left pan in the first weighing contained the fake coin. Knowing where the fake coin was in the first weighing gives us the answer.

It is often very useful to go through the easy cases. For this problem we can scale the solution for three and four coins to get a solution for any number of coins that is divisible by three and four by just grouping coins accordingly. Thus we have solutions for 3k and 4k coins.

For any number of coins we can try to merge the solutions above. Divide all coins into three piles of size a, a and b, where a ≤ b ≤ 2a. In the first weighing compare the first two piles. If they balance, then the fake coin must be among the b remaining coins. Now pick any b coins from both pans in the first weighing and compare them to the remaining b coins. If the first weighing is unbalanced, then the remaining coins have to be real. For the second weighing we can pick a coins from the remaining pile and compare them to one of the pans in the first weighing.

The solution I just described doesn’t cover the case of N = 5. I leave it to my readers to explain why and to solve the problem for N = 5.

Among ten given coins, some may be real and some may be fake. All real coins weigh the same. All fake coins weigh the same, but have a different weight than real coins. Can you prove or disprove that all ten coins weigh the same in three weighings on a balance scale?

When I first received this puzzle from Ken Fan I thought that he mistyped the number of coins. The solution for eight coins was so easy and natural that I thought that it should be eight — not ten. It appears that I was not the only one who thought so. I heard about a published paper with the conjecture that the best you can do is to prove uniformity for 2ⁿ coins in n weighings.

I will leave it to the readers to find a solution for eight coins, as well as for any number of coins less than eight. I’ll use my time here to explain the solution for ten coins that my son Sergei Bernstein suggested.

First, in every weighing we need to put the same number of coins in both pans. If the pans are unbalanced, the coins are not uniform; that is, some of them are real and some of them are fake. For this discussion, I will assume that all the weighings are balanced. Let’s number all coins from one to ten.

Consider two sets. The first set contains only the first coin and the second set contains the second and the third coins. Suppose the number of fake coins in the first set is a and a could be zero or one. The number of fake coins in the second set is b where b is zero, one or two. In the first weighing compare the first three coins against coins numbered 4, 5, and 6. As they balance the set of coins 4, 5, and 6 has to have exactly a + b fake coins.

In the second weighing compare the remaining four coins 7, 8, 9, and 10 against coins 1, 4, 5, and 6. As the scale balances we have to conclude that the number of fake coins among the coins 7, 8, 9, and 10 is 2a + b.

For the last weighing we compare coins 1, 7, 8, 9, and 10 against 2, 3, 4, 5, and 6. The balance brings us to the equation 3a + b = a + 2b, which means that 2a = b. This in turn means that either a = b = 0 and all the coins are real, or that a = 1, and b = 2 and all the coins are fake.

Now that you’ve solved the problem for eight and less coins and that I’ve just described a solution for ten coins, can we solve this problem for nine coins? Here is my solution for nine coins. This solution includes ideas of how to use a solution you already know to build a solution for a smaller number of coins.

Take the solution for ten coins and find two coins that are never on the same pan. For example coins 2 and 10. Now everywhere where we need 10, use 2. If we need both of them on different pans, then do not use them at all. The solution becomes:

The first weighing is the same as before with the same conclusion. The set containing the coin 1 has a fake coins, the set containing the coins 2 and 3 has b fake coins and the set containing coins 4, 5, and 6 has to have exactly a + b fake coins.

In the second weighing compare the four coins 7, 8, 9, and 2 against 1, 4, 5, and 6. As the scale balances we have to conclude that the number of fake coins among 7, 8, 9, and 2 is 2a + b.

For the last weighing we compare coins 1, 7, 8, and 9 against 3, 4, 5, and 6. If we virtually add the coin number 2 to both pans, the balance brings us to the equation 3a + b = a + 2b, which means that 2a = b. Which in turn means, similar to above, that either all the coins are real or all of them are fake.

It is known (see Kozlov and Vu, Coins and Cones) that you can solve the same problem for 30 coins in four weighings. I’ve never seen an elementary solution. Can you provide one?

I heard this problem many years ago when I was working for Math Alive.

Three men are fighting in a truel. Andrew is the worst shot; he misses 2/3 of the time. Bob is better; he misses 1/3 of the time. Connor is the best shot; he always hits. Each of the three men have an infinite number of bullets. Each shot is either a kill or a miss. They have to shoot at each other in order until two of them are dead. To make it more fair they decide to start with Andrew, followed by Bob, and then Connor. We assume that they choose their strategies to maximize their probability of survival. At whom should Andrew aim for his first shot?

This is a beautiful probability puzzle, and I will not spoil it for you by writing a solution. Recently, I watched an episode of Numb3rs: The Fifth Season (“Frienemies”) which featured a version of this puzzle. Here is how Dr. Marshall Penfield, a frienemy of the protagonist Charlie Eppes, describes it:

Imagine a duel between three people. I’m the worst shot; I hit the target once every three trials. One of my opponents [Charlie] is better; hits it twice every three shots. The third guy [Colby] is a dead shot; he never misses. Each gets one shot. As the worst I go first, then Charlie, then Colby. Who do I aim for for my one shot?

This is a completely different problem; it’s no longer about the last man standing. Colby doesn’t need to shoot since the other two truelists have already expended their only shots. If Charlie believes that Colby prefers nonviolence, all else being equal, then Charlie doesn’t need to shoot. Finally, the same is true of Marshall. There is no point in shooting at all.

To make things more mathematically interesting, let’s assume that the truelists are bloodthirsty. That is, if shooting doesn’t decrease their survival rate, they will shoot. How do we solve this problem?

If he survives, Colby will kill someone. If Charlie is alive during his turn, he has to shoot Colby because Colby might kill him. What should Marshall do? If Marshall kills Colby, then Charlie misses Marshall (hence Marshall survives) with probability 1/3. If Marshall kills Charlie, then Marshall is guaranteed to be killed by Colby, so Marshall survives with probability 0. If he doesn’t kill anyone, things look much better: with probability 2/3, Colby is killed by Charlie and Marshall survives. Even if Colby is alive, he does not necessarily shoot Marshall, so Marshall survives with probability at least 2/3. Overall, Marshall’s chances of staying alive are much better if he misses. He should shoot in the air!

The sad part of the story is that Charlie Eppes completely missed. That is, he completely missed the solution. In the episode he suggested that Marshall should shoot Charlie. If Marshall shoots Charlie, he will be guaranteed to die.

It is disappointing that a show about math can’t get its math right.

From the 1966 Moscow Math Olympiad:

Prove that you can choose six weights from a set of weights weighing 1, 2, …, 26 grams such that any two subsets of the six have different total weights. Prove that you can’t choose seven weights with this property.

Let us define the sequence a(n) to be the largest size of a subset of the set of weights weighing 1, 2, …, n grams such that any subset of it is uniquely determined by its total weight. I hope that you agree with me that a(1) = 1, a(2) = 2, a(3) = 2, a(4) = 3, and a(5) = 3. The next few terms are more difficult to calculate, but if I am not mistaken, a(6) = 3 and a(7) = 4. Can you compute more terms of this sequence?

Let’s see what can be said about upper and lower bounds for a(n). If we take weights that are different powers of two, we are guaranteed that any subset is uniquely determined by the total weight. Thus a(n) ≥ log₂n. On the other hand, the total weight of a subset has to be a number between 1 and the total weight of all the coins, n(n+1)/2. That means that our set can have no more than n(n+1)/2 subsets. Thus a(n) ≤ log₂(n(n+1)/2).

Returning back to the original problem we see that 5 ≤ a(26) ≤ 8. So to solve the original problem you need to find a more interesting set than powers of two and a more interesting counting argument.

Sergei Bernstein and Nathan Benjamin brought back a variation of the “Rock, Paper, Scissors” game from the Mathcamp. They call it “Rock, Paper, Scissor.” In this variation one of the players is not allowed to play Scissors. The game ends as soon as someone wins a turn.

Can you suggest the best strategy for each player?

They also invented their own variation of the standard “Rock, Paper, Scissors.” In their version, players are not allowed to play the same thing twice in a row.

If there is a draw, then it will remain a draw forever. So the game ends when there is a draw. The winner is the person who has more points.

They didn’t invent a nice name for their game yet, so I am open to suggestions.

Let me describe a variation of Nim that is at the same time a variation of Chomp. Here’s a reminder of what Nim and Chomp are.

In the game of Nim, there are several piles of matches and two players. Each of the players, in turn, can take any number of matches, but those matches must come from the same pile. The person who takes the last match wins. Some people play with a different variation in which the person who takes the last match loses.

Mathematicians do not differentiate between these two versions since the strategy is almost the same in both cases. The classic game of Nim starts with four piles that have 1, 3, 5 and 7 matches. I call this configuration “classic” because it is how Nim was played in one of my favorite movies, “Last Year at Marienbad”. Recently that movie was rated Number One by Time Magazine in their list of the Top 10 Movies That Mess with Your Mind.

In the game of Chomp, also played by two people, there is a rectangular chocolate bar consisting of n by m squares, where the lowest left square is poisoned. Each player in turn chooses a particular square of the chocolate bar, and then eats this square as well as all the squares to the right and above. The player who eats the poisonous square loses.

Here is my Nim-Chomp game. It is the game of Nim with an extra condition: the piles are numbered. With every move a player is allowed to take any number of matches from any pile, with one constraint: after each turn the i-th pile can’t have fewer matches than the j-th pile if i is bigger than j.

That was a definition of the Nim-Chomp game based on the game of Nim, so to be fair, here is a definition based on the game of Chomp. The game follows the rules of Chomp with one additional constraint: the squares a player eats in a single turn must all be from the same row. In other words, the chosen square shouldn’t have a square above it.

The game of Nim is easy and its strategy has been known for many years. On the other hand, the game of Chomp is very difficult. The strategy is only known for 2 by m bars. So I invented the game of Nim-Chomp as a bridge between Nim and Chomp.

I recently published a puzzle about wizards, hats of different colors and rooms. Unfortunately, I was too succinct in my description and didn’t explicitly mention several assumptions. Although such assumptions are usual in this type of puzzle, I realize now, from your responses, that I should have listed them and I apologize.

The Sultan decided to test the wisdom of his wizards. He collected them together and gave them a task. Tomorrow at noon he will put hats of different colors on each of the wizard’s heads.
The wizards have a list of the available colors. There are enough hats of each color for every wizard. The wizards also have a list of rooms. There are enough rooms to assign a different room for every color.
Tomorrow as the Sultan puts hats on the wizards, they will be able to see the colors of the hats of the other wizards, but not the color of their own. Without communicating with each other, each wizard has to choose a room. The challenge comes when two wizards have the same hat color, for they must choose the same room. On the other hand, if they have different hat colors, they must choose different rooms. Wizards have one day to decide on their strategy. If they do not all complete their task, then all of their heads will be chopped off. What strategy would you suggest for the wizards?

In his comment on the first blog about this problem, JBL beautifully described the intended solution for the finite number of wizards, and any potentially-infinite number of colors. I do not want to repeat his full solution here. I would rather describe his solution for two wizards and two colors.

Suppose the colors are red and blue. The wizards will designate one of the rooms as red and another as blue. As soon as each wizard sees the other wizard’s hat color, he chooses the room of the color he sees. The beauty of this solution is that if the colors of hats are different, the color of the rooms will not match the color of the corresponding hats: the blue-hatted wizard will go to the red room and vice versa. But the Sultan’s condition would still be fulfilled.

JBL’s solution doesn’t work if the number of wizards is infinite. I know the solution in that case, but I do not like it because it gives more power to the axiom of choice than I am comfortable with. If you are interested, you can extrapolate the solution from my essay on Countable Wise Men with Hats which offers a similar solution to a slightly different problem.

Sergei just came back from MOP — Mathematical Olympiad Summer Program, where he was a grader. The first question I asked him was, “What was your favorite math problem there?” Here it is:

There are wisemen, hats and rooms. Hats are of different color and there are enough hats of each color for every wisemen. There are enough rooms, so that you can assign a different room for every color. At some moment in time the sultan puts hats on the wisemen’s heads, so as usual they see all other hats, but do not see their own hat color. At the same time, each wiseman has to choose a room to go to. If two wisemen have the same hat color, they should go to the same room. If they have different hat colors, they should go to different rooms. What strategy should the wisemen decide upon before this event takes place?

Oh, I forgot to mention the most interesting part of this problem is that you shouldn’t assume that the number of wisemen or hats or rooms is finite. You should just assume that they have the power of choice.