Tanya Khovanova's Math Blog

My blog is getting more famous. Now I don’t need to look around for nice problems, for my readers often send them to me. In response to my blog about him, Sergey Markelov’s Best, Markelov sent me more of his problems. Here is a cute tetrahedron problem that he designed:

Six segments are such that you can make a triangle out of any three of them. Is it true that you can build a tetrahedron out of all six of them?

Another reader, Alexander Shen, sent me a different tetrahedron problem from a competition after reading my post on Problem Design for Multiple Choice Questions:

Imagine the union of a pyramid based on a square whose faces are equilateral triangles and a regular tetrahedron that is glued to one of these faces. How many faces will this figure have?

Shen wrote that the right answer to this problem had been rumored to have a negative correlation with the result of the entire test.

86 is conjectured to be the largest power of 2 not containing a zero. This simply stated conjecture has proven itself to be proof-resistant. Let us see why.

What is the probability that the nth power of two will not have any zeroes? The first and the last digits are non-zeroes; suppose that other digits become zeroes randomly and independently of each other. This supposition allows us to estimate the probability of 2ⁿ not having zeroes as (9/10)^k-2, where k is the number of digits of 2ⁿ. The number of digits can be estimated as n log₁₀2. Thus, the probability is about cxⁿ, where c = (10/9)² ≈ 1.2 and x = (9/10)^log₁₀2 ≈ 0.97. The expected number of powers of 2 without zeroes starting from the power N is cx^N/(1-x) ≈ 40 ⋅ 0.97^N.

Let us look at A007377, the sequence of numbers such that their powers of 2 do not contain zeros: 1, 2, 3, 4, 5, 6, 7, 8, 9, 13, 14, 15, 16, 18, 19, 24, 25, 27, 28, 31, 32, 33, 34, 35, 36, 37, 39, 49, 51, 67, 72, 76, 77, 81, 86. Our estimates predicts 32 members of this sequence starting from 6. In fact, the sequence has 30 conjectured members. Similarly, our estimate predicts 2.5 members starting from 86. It is easy to check that the sequence doesn’t contain any more numbers below 200 and our estimate predicts 0.07 members after 200. As we continue checking larger numbers and see that they do not belong to the sequence, the probability that the sequence contains more elements vanishes. With time we check more numbers and become more convinced that the conjecture is true. Currently, it has been checked up to the power 4.6 ⋅ 10⁷. The probability of finding something after that is about 1.764342396 ⋅10^-633620.

Let us try to approach the conjecture from another angle. Let us check the last K digits of powers of two. As the number of possibilities is finite, these last digits eventually will start cycling. If we can show that all the elements inside the period contain zeroes, then we need to check the finite number of powers of two until this period starts. If we can find such K, we can prove the conjecture.

Let us look at the last two digits of powers of two. The sequence starts as: 01, 02, 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 04. As we would anticipate, it starts cycling. The cycle length is 20, and 90% of numbers in the cycle don’t have zeroes.

Now let’s continue to the last three digits. The period length is 100, and 19 of them either start with zero or contain zero. The percentage of elements in the cycle that do not contain zero is 81%.

The cycle length for the last n digits is known. It is 4 ⋅ 5^n-1. In particular the cycle length grows by 5 every time. The number of zero-free elements in these cycles form a sequence A181610: 4, 18, 81, 364, 1638, 7371, 33170. If we continue with our supposition that the digits are random, and study the new digits that appear when we move from the cycle of the last n digits to the next cycle of the last n+1 digits, we can expect that 9/10 of those digits will be non-zero. Indeed, if we check the ratio of how many numbers do not contain zero in the next cycle compared to the previous cycle, we get: 4.5, 4.5, 4.49383, 4.5, 4.5, 4.50007. All of these numbers are quite close to our estimation of 4.5. If this trend continues the portion of the numbers in the cycle that don’t have zeroes tends to zero; however, the total of such numbers grows exponentially. We can even estimate that the expected growth is 4 ⋅ 4.5^n-1. From this estimation, we can derive the conjecture:

Conjecture. For any number N, there exists a power of two such that its last N digits are zero-free.

Indeed, the last N digits of powers of two cycle, and there are an increasing number of members inside that cycle that do not contain zeroes. The corresponding powers of two don’t have zeroes among N rightmost digits.

So, how do we combine the two results? First, the expected probability of finding the power of two larger than 86 that doesn’t contain zero is minuscule. And second, we most certainly can find a power of two that has as many zeroless digits at the end as we want.

To combine the two results, let us look at the sequences A031140 and A031141. We can deduce from them that for the power 103233492954 the first zero from the right occupies the 250th spot. The total number of digits of that power is 31076377936. So 250 is a tiny portion of the digits.

As time goes by we grow more and more convinced that 86 is the largest power of two without zeroes, but it is not at all clear how we can prove the conjecture or whether it can be proven at all.

My son, Sergei, suggested that I claim that I have a proof of this conjecture, but do not have enough space in the margin to fit my proof in. The probability that I will ever be shamed and disproven is lower than the probability of me winning a billion dollars in the lottery. Though then, if I do win the big bucks, I will still care about being shamed and disproven.

Janet Mertz encouraged me to find IMO girls and compare their careers to that of their teammates. I had always wanted to learn more about the legendary Lida Goncharova — who in 1962 was the first girl to win an IMO gold medal. So I located her, and after an interview, wrote about her. Only 14 years later, in 1976, did the next girl get a gold medal. That was me. I was ranked overall second and had 39 points out of 40.

As I did in the article about Lida, I would like to compare my math career to that of my teammates.

I got my PhD in 1988 and moved to the US in 1990. My postdoc at MIT in 1993 was followed by a postdoc at Bar-Ilan University. In 1996 I got a non-paying visiting position at Princeton University. In 1998 I gave up academia and moved to industry, accepting an offer from Bellcore. There were many reasons for that change: family, financial, geographical, medical and so on.

On the practical level, I had had two children and raising them was my first priority. But there was also a psychological element to this change: my low self-esteem. I believed that I wasn’t good enough and wouldn’t stand a chance of finding a job in academia. Looking back, I have no regrets about putting my kids first, but I do regret that I wasn’t confident enough in my abilities to persist.

I continued working in industry until I resigned in January 2008, due to my feeling that I wasn’t doing what I was meant to do: mathematics. Besides, my children were grown, giving me the freedom to leave a job I did not like and return to the work I love. Now I am a struggling freelance mathematician affiliated with MIT. Although my math blog is quite popular and I have been publishing research papers, I am not sure that I will ever be able to find an academic job because of my non-traditional curriculum vitae.

The year 1976 was very successful for the Soviet team. Out of nine gold medals our team took four. My result was the best for our team with 39 points followed by Sergey Finashin and Alexander Goncharov with 37 points and by Nikita Netsvetaev with 34 points.

Alexander Goncharov became a full professor at Brown University in 1999 and now is a full professor at Yale University. His research is in Arithmetic Algebraic Geometry, Teichmuller Theory and Integral Geometry. He has received multiple awards including the 1992 European Math Society prize. Sergey Finashin is very active in the fields of Low Dimensional Topology and Topology of Real Algebraic Varieties. He became a full professor at Middle East Technical University in Ankara, Turkey in 1998. Nikita Netsvetaev is an expert in Differential Topology. He is a professor at Saint Petersburg State University and the Head of the High Geometry Department.

Comparing my story to that of Lida, I already see a pattern emerging. Now I’m curious to hear the stories of other gold-winning women. I believe that the next gold girl, in 1984, was Karin Gröger from the German Democratic Republic. I haven’t yet managed to find her, so can my readers help?

I wrote a series of essays about AMC competitions:

• Metasolving AMC 8
• AMC, AIME, USAMO Contradiction
• Multiple Choice Proofs
• Should You Check Your Answers During Tests?
• To Guess or Not to Guess?
• Solving Problems with Choices
• How to Boost Your Guessing Accuracy During Tests
• Choices or No Choices
• Problem Design for Multiple Choice Questions

This essay is next in the series. Although it is not strictly about AMC, it should be useful during any test when you need to check your answers. There are several important rules which are helpful.

Rule 0. Checking is important. If wrong answers are punished, then correcting a mistake brings more points than solving a new problem. In addition, problems that were solved are often easier than problems yet to be solved, so finding a mistake might be faster than solving a new problem.

Rule 1. Your checking methods must be fast. The tests are generally timed. This means that in order to check your answers, you need to sacrifice your work on the next problem.

Rule 2. Customize how you check according to your strengths and weaknesses. For example, if you tend to jump to conclusions about what the question is going to be, and as a result answer your anticipated question instead of the one that is actually on the test, then when you are checking you should start reading the problem from the question. Or, if you usually make mistakes in geometry problems, you should allocate more time to geometry problems when you are checking. If you never make mistakes in arithmetic problems then you do not need to check those.

Rule 3. Mark problems that might need checking. If you do not have enough time to check all the problems, check only those you are not sure about.

Rule 4. Do not repeat your solution when you check. While solving the problem your brain often creates a pathway from start to finish. If on this pathway your brain decided to believe that two plus two is five, very often during checking, your brain will make the same mistake again. Because of that it is crucial to use other methods for checking than repeating your reasoning. In case you can’t find a way to check your answers using a different method and have to repeat your reasoning, you should repeat it in a different order.

This rule is so important, that I am providing some methods to change your brain pathway when you are checking your answers.

Plug in. Plugging in the answer you found is much faster than finding it. Use this method whenever possible. It is perfect for problems like this one below from 2004 AMC10-A:

What is the value of x if |x – 1| = |x – 2|?

Plug in an intermediate result. Sometimes you can’t plug in the answer, but you can plug in an intermediate result. In the following problem from 2004 AMC10-B you can plug in the number of nickels and dimes:

Patty has 20 coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have 70 cents more. How much are her coins worth?

Calculate something else related to your answer. For example a negation. Here is a problem from 2004 AMC10-B:

How many two-digit positive integers have at least one 7 as a digit?

If you calculated the answer directly, to check it you may want to calculate the number of two-digit positive integers that do not contain 7.

Create an example. Sometimes you solve a problem by reasoning, but to check it you might create a particular example. Here is a problem from 2001 AMC10:

Let P(n) and S(n) denote the product and the sum, respectively, of the digits of the integer n. For example, P(23) = 6 and S(23) = 5. Suppose N is a two-digit number such that N = P(N) + S(N). What is the units digit of N?

If we denote the tens digit by a and the units digit by b, then N = 10a + b, P(N) = a*b, and S(N) = a + b. We get an equation a(b+1) = 10a, from which the answer is 9. To check the answer we do not need to repeat the reasoning. It is enough to check that 19 is the sum of the product of its digits plus the digits.

Here is another problem from 2001 AMC10:

Suppose that n is the product of three consecutive integers and that n is divisible by 7. Which of the following is not necessarily a divisor of n?

The list of choices is: 6, 14, 21, 28, 42. Your solution might go like this: the product of three consecutive numbers is divisible by 6. Hence, n is divisible by 42. So, the answer must be 28. To check you might consider a product of three consecutive numbers: 5*6*7=210 and see that it is not divisible by 4, hence it is not divisible by 28.

Rule 5. Embrace the partial check. It is very important to check your answers fast. Sometimes you can gain speed if you do not check the problem completely, but check it partially. For example, you can check that your answer is one of the two correct answers. There are many methods for partial checking.

Try an example. Sometimes an example doesn’t guarantee that your choice is correct, but it increases your confidence in your answer. Here is another problem from 2001 AMC10:

The sum of two numbers is S. Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?

The choices are: 2S + 3, 3S + 2, 3S + 6, 2S + 6, 2S + 12. You can reason that increasing each summand by 3, increases the sum by 6. After that doubling each summand increases the resulting sum twice, so the answer is 2S + 12. To check the answer you can use an example. Usually an example doesn’t guarantee the confirmation of your answer, but it might help you eliminate some of the wrong answers. For example, if you choose zero and zero as your initial two numbers, then S = 0, and your transformation brings the result to 12, which confirms your answer 2S + 12. In this particular case, a very easy specific example excluded all the wrong answers.

Divisibility. Sometimes it is faster to calculate the remainder of the answer by some number.

For example, look at the following problem from 2003 AMC10:

What is the units digit of 13²⁰⁰³?

The choices are 1, 3, 7, 8, 9. We can immediately say that the answer must be an odd number.

Approximation check. One important example of a partial check is an approximation check. By estimating an approximate answer you might exclude most of the wrong answers. Consider this problem from 2001 AMC12:

How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?

The divisibilities by 3, 4 or 5 shouldn’t correlate with each other. Approximately one third of those number are multiples of 3 and one quarter are multiples of 4. Let’s say that one twelfth are multiples of both 3 and 4. Hence, we estimate the portion of numbers that are multiples of 3 or 4 as 1/3 + 1/4 – 1/12 = 1/2. We have about 1,000 such numbers. The number of numbers that are, in addition, not divisible by 5, are less than that. So out of the given choice of (A) 768, (B) 801, (C) 934, (D) 1067, (E) 1167, we can immediately confirm that the answer is among the first three.

The methods above can be useful even if you do not have multiple choices. But if you do…

Rule 6. Use given choices as extra information. In the previous examples you saw how to use a partial check to exclude some of the choices. Here is a specific example from 2006 AMC10-A of how to exclude choices:

What non-zero real value for x satisfies (7x)¹⁴ = (14x)⁷?

The choices are: 1/7, 2/7, 1, 7, 14. If you solved the problem directly, to check it you can reason why other choices do not work. In this particular case it can be done very fast. 1/7 doesn’t work because the left part of the equation becomes 1 when the right is clearly not. 1 and 7 do not work because the left part is odd and the right is even; 14 doesn’t work because the left is clearly bigger than the right.

Rule 7. Use meta considerations. If you get into the mind of the designers you can better anticipate when you should check more thoroughly. Consider this problem from 2006 AMC10-A:

A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?

The most common mistake would be to assume that 12:59 supplies the largest sum, which is 17. But look at the choices: 17, 19, 21, 22, 23. When the designers are asking to find the largest number with some property, they assume that some students will make a mistake and chose a smaller number over a larger one. That means the designers would include this potential mistake among the choices. So the answer is extremely unlikely to be the smallest number on the list of choices. Thus, if you think the answer is 17, understanding how these problems are constructed should alert you to thoroughly check your answer. Indeed, the correct answer is 23 which corresponds to 9:59. Not surprisingly, it is the largest on the list of choices.

AMC 10/12 is coming on February 8 and HMMT on February 12. Happy checking.

I’ve been celebrating my 29th birthday for many years. Once, when I was actually 45 and wanted to have a big party, I invited everyone to the 5th anniversary of my 29th birthday.

Last week my son turned 29 and I realized that it is time to drop this beautiful, prime, evil, deficient, lazy-caterer number, that in addition is the largest power of two to have all different digits. No more celebrating 29.

For my next age, I picked 42. Not because it is the smallest abundant odious number, but rather because it is the answer to life, the universe and everything.

Thank you everyone who congratulated me on my birthday two days ago. For your information, from now on I am 42.

As I did for 2010 and for previous years, here are math-related puzzles from the MIT Mystery Hunt 2011.

• Another Small Collection of Immense Integers — Large integers.
• Two Heads are Better than One — A logic puzzle.
• Worthy of Picasso — Paint-by-Numbers followed by a mathematical extraction.
• Unfair Cryptogram — A cryptogram.
• The Eternal Struggle — Something happens at every time step.
• N-tris — A tetris variation.
• Scrambling Attributes Yields Conundrum — A hidden logic puzzle.
• Basic Knowledge — An integer circuit diagram.
• Nik-holey — A set of Nikoli puzzles with a twist, including Nurikabe, Slitherlink, Hitori, Heyawake, Fillomino, Hashiwokakero.
• Games — Related to games.
• You Shall Understand What Hath Befallen — An Othello game.
• Pointillisme — Paint-by-Numbers.
• The Crypt — A cryptogram.
• Technological Crisis at Shikakuro Farms — My favorite puzzle, a blend of shikaku, kakuro, and sudoku.
• Showcase — This one looks the most mathematical, but you need to physically visit MIT’s math department to solve it.
• Advanced Maths — My second favorite, this looks like binary operations, but not quite.
• Efficiency — A Scheme code.
• Losing My Nerve — A changing cipher cryptogram with extra math at the extraction stage.
• Clues and Coordinates — 3D geometry.

Two more puzzles deserve a special mention for their nerdiness. My teammates loved them.

• Sufficiently Advanced Technology — It is difficult to believe that this puzzle is not over-constrained.
• Unlikely Situations — This puzzle needed cooperation from a very special person in advance.

Tired of the same old sudoku? Here’s an opportunity to try many variations of it. Thomas Snyder and Wei-Hwa Huang wrote a book called Mutant Sudoku. The authors are both Sudoku champions. I like the book because the authors are trying to bring everyone up to their level, rather than dumbing down their puzzles. So the book is not at all boring as are most Sudoku books.

The book contains about 180 fun puzzles. Look at the variety:

• Tight Fit Sudoku
• Extra Space Sudoku
• Tile Sudoku
• 3-D Sudoku
• Outside Sudoku
• Shape Sudoku
• Target Sum Sudoku
• Thermo-Sudoku
• Consecutive Sudoku
• Surplus Sudoku
• Deficit Sudoku
• Chimeric Sudoku

Wei-Hwa Huang kindly sent me this sample Thermo Sudoku puzzle from the book to use on my blog. The grey areas represent thermometers. Every particular thermometer has to have numbers in increasing order (not necessarily consecutive) starting from the bulb.

The second book by the same authors Sudoku Masterpieces: Elegant Challenges for Sudoku Lovers, is itself a masterpiece. With about 100 puzzles, there are fewer than in the first book, but there are more types of puzzles. As a consequence, you’ll have less practice for each particular type, but more variety. In addition, as you can see from the cover, the second book is elegantly designed.

I bought both books and immediately started scribbling in the first one. My bad handwriting would seem so out of place in the beautiful second book that I have not even started working in it yet. Maybe I will give it as a gift to someone with better penmanship.

Two days ago I threw at my readers the following problem:

A plane takes off and goes east at a rate of 350 mph. At the same time, a second plane takes off from the same place and goes west at a rate of 400 mph. When will they be 2000 miles apart?

The purpose of throwing this problem was to discuss the nature of the implicit assumptions that we are asked to make when solving math problems, and the implicit assumptions we teach our children to make when we teach them to solve math problems. This is especially important for problems like this, that are phrased in terms of a situation in the real world. The real world is too complex to model all of; the great power of mathematics is that sufficiently idealized situations are predictable. But which idealizations are appropriate? How does one choose? How does one teach youngsters what to choose?

Before I get to the actual discussion, however, I want to re-throw this problem at my readers, in an effort to highlight what originally jumped out at me as being wrong with it.

Neglecting the effects of altitude, differential wind, acceleration, relativity, measurement error, finite size and non-superimposability of the planes, and the Earth’s deviations from perfect sphericity,

1. Find how much time it takes them to become 2000 miles apart, assuming that the planes are starting from Boston and the distance is measured as
   1. a straight line in 3-space.
   2. the shortest surface distance.
2. How far from the closest pole may the starting point be located, so that the answer to the problem is “never”? Solve separately for
   1. the 3D distance.
   2. the shortest surface distance.
3. What portion of the Earth’s surface do the “never”-locations of the previous question occupy?
   1. under the 3D distance?
   2. under the shortest surface distance?

Hint: The easiest question is 2b.

My Jewish ex-father-in-law, Naum Bernstein, is 96 years old and is full of life. He has a joke for every situation. In the last decade he wrote several volumes of memoirs in Russian. One of the books was a collection of his favorite jokes and his explanations of them. I decided to retell some of the jokes from his selection.

Arithmetic

An arithmetic teacher calls the student Rabinovich to the blackboard. “It is known that from 1 kilogram of sour cream you can make 200 grams of butter. Imagine, Rabinovich, that your father bought 2 kilograms of sour cream. How much butter can he make?”

“Five hundred grams,” Rabinovich replies.

The teacher frowns, “Rabinovich, you do not know arithmetic!”

Rabinovich answers, “Sir, you do not know my father.”

Billions

An astronomy teacher explains that in the future the Earth will lose its heat energy, continents will collide, and solar radiation will increase. In six billion years life will be extinct. A student looking really scared raises his hand and asks, “In how many years will life become extinct?”

“In about six billion years,” the teacher repeats.

“Whew,” says the student, “you got me so scared. I thought you said six million.”

Soccer Player

Two professors are chatting while watching a soccer game. The first one says, “They say that soccer players have their brains in their legs. So their heads are really empty.”

“Not quite,” the second professor replies. “The player on the right passed my exam yesterday.”

The first professor expresses interest, so the second one elaborates. “As a rule, I ask two questions. If the student gives a correct answer to one of them, he passes.”

“So, what did you ask that guy?”

“My first question was ‘What color are red blood cells?’ He answered ‘Yellow.’ That was an incorrect answer. The second question was ‘How is sulfuric acid produced?’ To this he replied, ‘I do not know,’ which was absolutely true, so he passed.”

Pushkin’s “Eugene Onegin”

A Russian literature teacher asks a pupil, “Who wrote Eugene Onegin?” The pupil gets scared that he is being blamed for something and replies, “No, not me! I swear I didn’t write it!” Everyone laughs. The teacher decides that the pupil disrupted the class on purpose and asks for his father to come by.

The father arrives and after the teacher explains what happened, the father says, “Maybe he is not guilty; maybe he really didn’t write it. I doubt that he is capable of writing anything.”

The teacher is stunned and later tells the whole story in the teachers’ lounge to her colleagues. An astronomy teacher comes home and retells the story to her husband who works for the KGB. The husband comments, “Do not worry, we are on it. Three people already confessed to writing it.”

Death of an Anti-Semite

A hardcore anti-Semite was dying. As he got weaker he made a last request. He wanted to convert to Judaism. Everyone was extremely surprised, but decided not to interfere. After the conversion, his wife summoned the courage to ask him what was going on. “Do you think you were mistaken, hating Jews all your life?”

“No,” he replied happily, “But now with my death, the world will get rid of one more Jew.”

Shaving

An old Jew comes to a Rabbi and asks if he can shave his beard off, because his children think that he is old-fashioned. The Rabbi tells him that by Jewish law he is not allowed to shave. The old man turns to go home when he realizes that the Rabbi himself doesn’t have a beard. He stops and asks, “Dear Rabbi, you just forbade me to shave my beard, but how come you are clean-shaven yourself?”

The Rabbi replies, “I didn’t ask anyone’s permission.”

A Bureaucrat

When Rabinovich came to a bureaucrat with a request, the bureaucrat replied, “Come back tomorrow.” Rabinovich returned the next day and received the same reply. Rabinovich was very persistent and returned day after day.

Finally, the bureaucrat lost his patience and attacked Rabinovich, “This is outrageous! Don’t you understand simple language? I keep telling you to come tomorrow and you keep coming today.”

Bathroom Tissue

The communist committee of a supermarket in the USSR received a lot of complaints about the rudeness of their salespeople. The committee decided to improve the quality of service and provided special training in which salespeople were taught politeness. The training emphasized what to do in case a particular item was unavailable. The salespeople were supposed to politely explain that the item is temporarily unavailable and to offer a substitute.

The next thing one of their salespeople said to a customer was “I am very sorry, we are temporarily out of toilet paper. May I offer some sandpaper?”

13th Floor

There is panic in an apartment on the 13th floor. The wife recognizes the sound of her husband’s approach, even though he was supposed to be on a business trip. The lover asks, “What should I do, honey?”

“What do people do in such cases? Jump out the window!”

“But we are on the 13th floor!”

“This is no time for superstition!”

Smelly Socks

A young man had smelly feet, plus he always forgot to change his socks. His girlfriend got tired of it and asked him to promise that he would always change his socks before coming to see her.

Next visit the young man smelled as bad as ever. Outraged, the girl said, “But you promised to change your socks!”

The young man answered, “I did as I promised.”

“I don’t believe you, you smell awful.”

“I was sure you wouldn’t believe me. Good thing I brought my dirty socks with me as proof.”

A Recipe for a Happy Marriage

At the 50th anniversary of a very happy couple, someone asked the husband for their secret. He said that right before the wedding they agreed that the husband would decide all the crucial and very important things, and the wife would be responsible for all minor decisions. “For example,” he continued, “yesterday I decided that the US should withdraw their troops from Iraq, and my wife decided where to buy our vacation house.”

Coffee in Bed

Two long-time girlfriends meet after several years without being in touch. “How are your children?” asks one of them.

The other replies, “My daughter is fine, she married a nice young man, who is providing for her. He also helps her with chores and even brings her coffee in bed every morning.”

“What about your son?”

“It’s a disaster. I don’t know what to do. He married a really lazy woman. Even though she’s not working, she wants him to help her with the chores. Can you imagine that? She even dared to ask him to bring her coffee in bed every morning.”

Debt

Two friends are walking along a street very late at night. Robbers attack them with guns, demanding their wallets. One of the friends asks the robbers, “Can you give me 30 seconds?” The robbers agree. He takes out $100 from his wallet and gives it to his friend, “Remember I owed you $100? I am paying back my debt in front of witnesses.”

Struggle

Life is a struggle. Before lunch with hunger, after lunch with sleepiness.

Window

A mother says to her son, “Please, close the window — it’s cold outside.”

The son replies, “Do you think it will get warmer outside if I close the window?”

Pessimist and Optimist

How are pessimists and optimists different from normal people?

A pessimist uses both a belt and suspenders, an optimist uses neither.

Ads

In a cemetery there is a beautiful monument with a picture of a bald, wrinkled old man. He is smiling, showing his perfect white teeth. His epitaph says:

Here lies Mr. X, who lived more than 100 years, lost his hair, became all wrinkled, but kept his perfect teeth. That is because he always used our company’s toothpaste.

A nearby monument has a picture of an old toothless woman with beautiful, voluminous hair. The inscription explains which brand of shampoo she used.

Many other tombstones with ads are scattered throughout the cemetery. But in the middle there is a huge mausoleum with an inscription reading:

No one is buried here and no one ever will be, because his or her parents used condoms made by our company.

Shower

A Russian man marries an American woman. After a while he writes a letter home.

My wife must be very dirty. She showers every day.

Last

Rabinovich was asked why he didn’t attend the last committee meeting. He replied, “If I knew it was the last, I would certainly have come.”

I stumbled upon the following problem in Mathematics Teacher v.73 (September 1980):

A plane takes off and goes east at a rate of 350 mph. At the same time, a second plane takes off from the same place and goes west at a rate of 400 mph. When will they be 2000 miles apart?

Ooh, boy!

Question for my readers: explain my reaction.