Archive for the ‘Math Competitions’ Category.

## More Gnomes

I recently posted a cute Shapovalov’s puzzle about gnomes. Here is another easier gnome puzzle, also by Alexander Shapovalov.

Puzzle.All gnomes are knights or knaves: knights always tell the truth, and knaves always lie. There is a gnome on every cell of a 4 by 4 chessboard. It is known that both knights and knaves are present in this group. Every gnome states, “Among my neighbors, the number of knaves is the same as the number of knights”. How many knaves are there, if by neighbors they mean orthogonally adjacent gnomes?

The next gnome puzzle has a different author, Alexander Gribalko. Gnomes in this puzzle are not knights or knaves but rather friendly and polite beings.

Puzzle.Nine gnomes repeated the following procedure three times. They arranged themselves on a 3 by 3 chessboard with one gnome per cell and greeted all their orthogonal neighbors. Prove that not all pairs of gnomes greeted each other.

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## Archimedes Helps Again

Below, you can find a lovely problem from the 2016 All-Russian Olympiad suggested by Bogdanov and Knop. I took some liberties translating it.

Problem.King Hiero II of Syracuse has 11 identical-looking metallic ingots. The king knows that the weights of the ingots are equal to 1, 2, …, 11 libras, in some order. He also has a bag, which would be ripped apart if someone were to put more than 11 libras worth of material into it. The king loves the bag and would kill if it was destroyed. Archimedes knows the weights of all the ingots. What is the smallest number of times he needs to use the bag to prove the weight of one of the ingots to the king?

And a bonus question from me.

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Bonus.Add one more weighing to prove the weight of three more ingots.

## My Computational Linguistics Olympiads

Do you know that I participated in Linguistics Olympiads in high school? They are not well-known in the US, but the Soviet Union has been running them since 1965. The first International Linguistics Olympiad was conducted in 2003, and the US joined in 2007. They are called Computational Linguistics because you are expected to discover some phenomenon in an unfamiliar language on the fly instead of knowing a lot of languages already. The problems mostly need logic and are a good fit for a person who likes mathematics. However, a feel for languages is very helpful.

I do not remember why I started attending the Olympiads, but I remember that there were two sets of problems: more difficult for senior and less difficult for non-senior years. I used to be really good at these Olympiads. When I was in 8th grade, I finished my problems before the time ran out and started the senior problems. I got two awards: first place for non-senior years and second place for senior years. In 9th grade, I got two first-place awards. I didn’t know what to do in 10th grade, which was a senior year at that time in the USSR. I couldn’t get two first-place awards, as I could no longer compete in the non-senior category. I felt ashamed that my result could only be worse than in the previous years, so I just didn’t go.

The prizes were terrific: they gave me tons of rare language books. In the picture, a guy from the jury is carrying my prizes for me. I immediately sold the books at used-books stores for a good price. Looking back, I should have gone to the Olympiad in 10th grade: my winter boots had big holes.

Share:## A Splashy Math Problem Solution

I recently wrote a post, A Splashy Math Problem, with an interesting problem from the 2021 Moscow Math Olympiad.

Problem (by Dmitry Krekov).Does there exist a numberAso that for any natural numbern, there exists a square of a natural number that differs from the ceiling ofA^{n}by 2?

The problem is very difficult, but the solution is not long. It starts with a trick. Suppose *A* = *t*^{2}, then *A*^{n} + 1/*A*^{n} = *t*^{2n} + 1/*t*^{2n} = (*t*^{n} + 1/*t*^{n})^{2} − 2. If t < 1, then the ceiling of *A*^{n} differs by 2 from a square as long as *t*^{n} + 1/*t*^{n} is an integer. A trivial induction shows that it is enough for *t* + 1/*t* to be an integer. What is left to do is to pick a suitable quadratic equation with the first and the last term equal to 1, say *x*^{2} – 6*x* + 1, and declare *t* to be its largest root.

## The 41-st Tournament of the Towns

Today I present three problems from the 41-st Tournament of the Towns that I liked: an easy one, one that reminds me of the Collatz conjecture, and a hard one.

Problem 1 (by Aleksey Voropayev).A magician places all the cards from the standard 52-card deck face up in a row. He promises that the card left at the end will be the ace of clubs. At any moment, an audience member tells a numbernthat doesn’t exceed the number of cards left in the row. The magician counts thenth card from the left or right and removes it. Where does the magician need to put the ace of clubs to guarantee the success of his trick?

Problem 2 (by Vladislav Novikov).Numberxon the blackboard can be replaced by either 3x+ 1 or ⌊x/2⌋. Prove that you can use these operations to get to any natural number when starting with 1.

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Problem 3 (by A. Gribalko).There are 2nconsecutive integers written on a blackboard. In one move, you can split all the numbers into pairs and replace every paira,bwith two numbers:a+banda−b. (The numbers can be subtracted in any order, and all pairs have to be replaced simultaneously.) Prove that no 2nconsecutive integers will ever appear on the board after the first move.

## The Anniversary Coin

Konstantin Knop, the world’s top authority on coin-weighing puzzles, suggested the following problem for the 2019 Russian Math Olympiad.

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Puzzle.Eight out of sixteen coins are heavier than the rest and weigh 11 grams each. The other eight coins weigh 10 grams each. We do not know which coin is which, but one coin is conspicuously marked as an “Anniversary” coin. Can you figure out whether the Anniversary coin is heavier or lighter using a balance scale at most three times?