Tanya Khovanova's Math Blog

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.

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.

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.

A year ago I posted a chessboard puzzle. Recently I stumbled on a September 2008 issue of “Math Horizons” where it was presented as a magic trick.

When the magician leaves the room, the trickees lay out eight coins in a row deciding which side is turned up according to their whim. They also think of a number between 1 and 8 inclusive. The magician’s assistant then flips exactly one of the coins, before inviting the magician back in. The magician looks at the coins and guesses the number that the trickees thought of.

The magician’s strategy can be derived from the solution to the chessboard puzzle. The assistant numbers the coins from zero to seven from left to right. Then s/he flips the coin so that the parity addition (XORing) of all the numbers corresponding to heads is the number that the magician needs to guess. For this trick to work, the number of coins needs to be a power of 2.

Andrey Zelevinsky posted (in Russian) a cool variation of this trick with two decks of cards.

The magician has two identical card decks and he is out of the room for now. A random person from the audience thinks of a card. Next, the audience chooses several cards from the first deck. Then the assistant adds one card from the second deck to the set of chosen cards, lays them on a table, and then invites the magician back. The magician looks at the cards on the table and guesses the card that was thought of.

Unlike in the coin trick above, the number of cards in the deck doesn’t need to be a power of 2. This flexibility is due to the fact that the magician has two decks of cards, as opposed to one set of coins. Having the second deck is equivalent to the assistant in the coin trick being allowed to flip one or ZERO coins.

Nikolay Konstantinov, the creator and the organizer of the Tournament of the Towns, discussed some of his favorite tournament problems in a recent Russian interview. He mentioned two beautiful geometry problems by Sergey Markelov that I particularly loved. The first one is from the 2003 tournament.

An ant is sitting on the corner of a brick. A brick means a solid rectangular parallelepiped. The ant has a math degree and knows the shortest way to crawl to any point on the surface of the brick. Is it true that the farthest point from the ant is the opposite corner?

The other one is from 1995.

There are six pine trees on the shore of a circular lake. A treasure is submerged on the bottom of the lake. The directions to the treasure say that you need to divide the pine trees into two groups of three. Each group forms a triangle, and the treasure is at the midpoint between the two triangles’ orthocenters. Unfortunately, the directions do not explain how exactly to divide the trees into the groups. How many times do you need to dive in order to guarantee finding the treasure?

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.