Tanya Khovanova's Math Blog

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I am coaching my AMSA students for math competitions. Recently, I gave them the following problem from the 1964 MAML:

The difference of the squares of two odd numbers is always divisible by:
A) 3, B) 5, C) 6, D) 7, E) 8?

The fastest way to solve this problem is to check an example. If we choose 1 and 3 as two odd numbers, we see that the difference of their squares is 8, so the answer must be E. Unfortunately this solution doesn’t provide any useful insight; it is just a trivial calculation.

If we remove the choices, the problem immediately becomes more interesting. We can again plug in numbers 1 and 3 to see that the answer must be a factor of 8. But to really solve the problem, we need to do some reasoning. Suppose 2k + 1 is an odd integer. Its square can be written in the form 4n(n+1) + 1, from which you can see that every odd square has remainder 1 when divided by 8. A solution like this is a more profitable investment of your time. You understand what is going on. You master a method for solving many problems of this type. As a bonus, if students remember the conclusion, they can solve the competition problem above instantaneously.

This is why when I am teaching I often remove multiple choices from problems. To solve them, rough estimates and plugging numbers are not enough. To solve the problems the students really need to understand them. Frankly, some of the problems remain boring even if we remove the multiple choices, like this one from the 2009 AMC 10.

One can holds 12 ounces of soda. What is the minimum number of cans needed to provide a gallon (128 ounces) of soda?

It’s a shame that many math competitions do not reward deep analysis and big-picture understanding. They emphasize speed and accuracy. In such cases, plugging in numbers and rough estimates are useful skills, as I pointed out in my essay Solving Problems with Choices.

In addition, smart guessing can boost the score, but I already wrote about that, too, in How to Boost Your Guessing Accuracy During Tests and To Guess or Not to Guess?, as well as Metasolving AMC 8.

As the AMC 10 fast approaches, I am bracing myself for the necessity to include multiple choices once again, thereby training my students in mindless arithmetic.

Many people ask me when is a good time to teach kids math. In my experience, it can never be too early. You just need to keep some order. Multiplication should be taught after addition, and negative numbers after subtraction. Kids should remember multiplication by heart at the age of seven. They can understand negative numbers as early as four.

In the picture I am explaining Platonic solids to four-month-old Eli, the son of my friends. His homework is to chew on a dodecahedron.

Suppose Romeo is encouraged by love and attention. If Juliet likes him, his feelings for Juliet grow and flourish. If she doesn’t like him, he loses his interest in her.

Juliet, on the other hand, is the opposite. If Romeo doesn’t like her, she needs to win him over and her attraction for him grows. If he likes her, she feels that her task is accomplished and she loses her interest in him. Juliet likes the challenge more than the relationship.

Steven Strogatz used differential equations to model the dynamics of the relationship between Romeo and Juliet. This is a new and fascinating area of applied mathematical research; you can read more about the roller-coaster relationship between Romeo and Juliet in Steven Strogatz’s Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering.

Mathematicians like symmetry: in math literature they switch the roles between Romeo and Juliet randomly. So in some papers they give Romeo the role of preferring a challenge over love and in some papers they give that role to Juliet.

When I teach this subject of love, Alexander Pushkin’s famous quote always pops into my mind. The quote comes from the first lines of Chapter Four of Eugene Onegin, and in Russian it is:

Чем меньше женщину мы любим,
Тем легче нравимся мы ей…

I didn’t like the English translations that I found, so I asked my son Alexey to provide a more literal translation:

The less we love a woman, the more she likes us in return…

I blame Pushkin for my tendency to always pick Juliet as the character who thrives on the challenge, even though men are often assumed to be the chasers. I’d like to ask my readers to comment on these roles: Do you think both genders play these roles equally? If not, then who is more prone to be into the chase?

Let’s return to mathematical models. In the original model, the reactions of Romeo and Juliet are a linear function of feelings towards them. I would like to suggest two other roles, in which people react to the absolute value of feelings towards them. They do not care if it is love or hate: they care about intensity.

First, there is the person, like my friend Connie, who feeds on the emotions of other people. She’s turned on by guys who love her as well as by guys who hate her. If they’re indifferent, she’s turned off.

Second, there is the opposite type, like my colleagues George, Joseph, David and many others. They hate emotion and prefer not to be involved. They lose all interest in people who feel strongly about them and they like people who are distant. I know the name for this role: it’s a mathematician!

My friend Olga Amosova worked as a molecular biologist at Princeton University. Last time I visited her, we talked about her research.

She told me that she and her group designed a repair for a DNA mutation that is highly localized. “What’s the point,” I asked her, “of repairing DNA mutation in one cell?”

I was amazed to learn that not only is there a practical use to her research, but that there is something urgent that I myself must do.

There are many diseases that are caused by localized (so called “point”) mutations. The most famous one is Sickle-cell disease. In Sickle-cell disease, defective hemoglobin causes erythrocytes to adopt a sickle shape that makes it difficult to pass through blood vessels. It is a very painful and debilitating disease. However, it turns out that the results of the research of Olga and her group could make the lives of people with such mutations much easier.

Stem cells have two amazing abilities. They grow fast and they can be turned into any type of cells in the human body. If the mutation is repaired in just one stem-cell, it can be selected and turned into a blood progenitor cell. These progenitor cells produce erythrocytes that actually transport oxygen. If these repaired cells are added to the patient’s blood, they would produce good hemoglobin for half a year. This would improve the patient’s quality of life tremendously.

So what do the rest of us learn from Olga’s research? That we must save all left-over stem-cells that are produced in childbirth, like the umbilical cord and the placenta. It’s not only Sickle-cell, but many other diseases that could benefit from using stem-cells. Research is moving so fast that these frozen stem-cells might become relevant in surprising ways — not only for the child, but also for relatives of the child — like you yourself!

So what’s the urgent thing I must do? My son recently got married, so I must finish this post and send it to my son in case they get pregnant.

Mikhail Zhvanetsky is the most prolific and famous Russian humorist. Here are my own translations of some of his best lines.

• Better a small dollar than a big thank you.
• Better dinner without an appetite than an appetite without dinner.
• Don’t drive faster than your guardian angel can fly.
• I drive too fast to worry about cholesterol.
• Best alibi — be a victim.
• A pedestrian is always right. While he is alive.
• Any car will last you a life-time. If you are hasty enough.
• Better a belly from beer than a hump from hard work.
• A bald patch is a glade trampled by thoughts.
• It is difficult to crawl with your head proudly held high.
• It’s a shame when other people have your dreams come true!
• The lottery is the most accurate measure of the number of optimists.
• A courteous man will not criticize a woman who carries a railroad tie awkwardly.
• The highest degree of embarrassment? Exchanged glances in a keyhole.
• Everything goes well, but past me.
• Let them laugh at you, rather than cry.
• While you measure seven times, others will already make a cut.
• It is not enough to find your place in life, you have to be there first.
• If a person knows what he wants, then he either knows too much or wants too little.
• And then he took a knife and shot himself dead.
• Thinking is too difficult, so most people judge.
• The more I look in the mirror, the more I believe in Darwin.
• Of two evils, I choose the one I haven’t tried before.
• Do not run from a sniper, you’ll die tired.
• You came — thanks; you left — many thanks.
• All great men are long dead, and I am feeling so-so.
• Never exaggerate the stupidity of your enemies and the loyalty of your friends.
• To save a drowning man, it is not enough to lend a hand; it is necessary for him to offer his hand in return.
• What a pity that you are leaving at long last.
• An idea came into his head and now it is desperately trying to find his brain.
• I am infinitely respectful of the terrible choices of my people.
• Some have both hemispheres protected by a skull, others by pants.
• For illusions of grandeur one doesn’t need grandeur; illusions are quite enough.
• Good always wins over evil. Hence, the winner is always good.
• Only on your birthday do you discover how many useless things there are in the world.
• You can recognize a decent man by how difficult it is for him to be nasty.
• Everything in this world is relative. For example, the length of one minute depends on which side of the bathroom door you’re on.
• In the form I filled in before the surgery there was this question: Whom should we call in case of an emergency? I wrote: A more qualified surgeon.

 I love the TV series of Angel and of Buffy the Vampire Slayer. I enjoy the excitement of saving the world every 42 minutes. But as a scientist I keep asking myself a lot of questions.

Where do vampires take their energy from? Usually oxygen is the fuel for the muscles of living organisms, but vampires do not breathe. Vampires are not living organisms, and yet they have to get their energy from somewhere.

When you kill a vampire, it turns to dust. If organisms are 60% water, then a 200-pound vampire should generate 80 pounds of dust. So why, in the series, do you get just a little puff of dust whenever someone plunges a stake into a vampire? Plus 120 pounds of water apparently evaporates instantly during staking. Can someone who is less lazy than me please calculate the energy needed to evaporate 120 pounds of water in one second? Because my first reaction is that you would need an explosion, not just one stab with Buffy’s stake.

All these unscientific elements do not actually bother me that much. What does bother me are inconsistencies in logic. For example, at the end of Season One of Buffy, Angel refuses to give Buffy CPR, claiming that as a vampire he can’t breathe. But then how can Spike and other vampires smoke? If they can smoke that means they are capable of inhaling and exhaling. Not to mention that these vampires talk: wouldn’t they need an airflow through their throats to produce sounds?

It would make more sense for the show to state that vampires do not need to breathe, but are nonetheless capable of inhaling and exhaling. So Angel should have given Buffy CPR. It would have created a great plot twist: Angel saves Buffy at the end of Season One, only for her to send him to the hell dimension at the end of Season Two.

Back to breathing. I remember a scene in “Bring On the Night” in which Spike was tortured by Turok-Han holding his head in water. But if Spike can’t breathe, why is this torture?

Another thing that bothers me in the series is not related to what happens but to what doesn’t happen. For example, vampires do not have reflections. So I don’t understand why every vampire-aware person didn’t install a mirror on the front door of their house to check for reflections before inviting anyone in.

Also, it looks like producers do not care about backwards compatibility. Later in the series we get to know that vampires are cold. Watch the first season of Buffy with that knowledge. In the very first episode, Darla is holding hands with her victim, but he doesn’t notice that she is cold. Later Buffy kisses Angel, before she knows that he is a vampire, and she doesn’t notice that he’s cold either. Unfortunately, the series also isn’t forward compatible. In the second season of Angel in the episode “Disharmony”, when we already know that vampires are cold, Harmony is trying to reconnect with Cordelia. They hug and touch each other. Such an experienced demon fighter as Cordelia should have noticed that Harmony is cold and, therefore, dead.

Finally, let’s look at Spike in the last season of Angel. Spike is non-corporeal for a part of the season; we see him going through walls and standing in the middle of a desk. Yet, one time we see him sitting on a couch talking to Angel. In addition, he can take the stairs. He can go through the elevator wall to ride in an elevator instead of falling down through its floor. And what about floors? Why isn’t he falling through floors? Some friends of mine said that we can assume that floors are made from stronger materials. But, if there is a material that can prevent Spike from penetrating it, they ought to use this material to make a weapon for him.

I’ve never been involved in making a show, but these producers clearly need help. Perhaps they should hire a mathematician like me with an eye for detail to prevent so many goofs.

The jokes are a rough translation from a Russian collection, except the last one I invented myself.

* * *

— Moishe, do you know how many cuckolds are there in Odessa not counting you?
— What? What do you mean by saying “not counting you”?
— Sorry. Okay then, how many counting you?
* * *

At a very prestigious Russian nursery school a teacher talks to a four-year-old applicant.
“Mike, can you count for me?”
Mike counts very fast and with a lot of enthusiasm, “Fifty-nine, fifty-eight, fifty-seven…”
“Super,” says the teacher, “But how did you learn to count backwards?”
Mike replies proudly, “I can heat my own lunch — in the microwave.”

* * *

The curl of the curl equals the gradient of the divergence minus the Laplacian. Why do I remember this shit that I never need, but can’t remember where I put my keys yesterday?

* * *

In a bike store:
Customer: “Can you show me your finest helmet? I’ve already spent $200,000 on my head, so I don’t want to take any risks.”
Clerk, sympathetically: “You had a head trauma?”
Customer: “No, I went to college.”

* * *

A topologist walks into a cafe:
— Can I have a doughnut of coffee, please.

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?