Here are some problems that I liked from the YuMSh (Youths Math School in St. Petersburg) Olympiad.

Problem for 6th grade. Twenty people from an island of knights and knaves have a party. Knights always tell the truth, and knaves always lie. Each party-goer got a card with a different number from 1 to 20. When they were asked about their numbers, each answered with a number from 1 to 20. The sum of all the answers is 156. What is the minimum possible number of liars that have to be at the party?

Problem for 7th grade. Alice and Bob bought a deck of playing cards (52 cards total) and took turns gluing the cards on the wall one at a time. Alice was first. The game is lost if, after a move, the wall has 4 cards of the same suit or 4 cards of consecutive values (for example, 8-9-10-jack). Can Alice or Bob guarantee themselves a win, regardless of their opponent’s moves?

Problem for 7th grade. Buddhist monks gather in an infinite cave where a finite number of prime numbers are written on the wall. The numbers might not be distinct. Every second, one of the monks performs one of the following operations.

1. Adds to one of the numbers one of its digits.
2. Shuffles the digits of one of the numbers.

Every time they do it, they erase the old number and write the new one. The rule is that the new number has to be greater than the old one. If a composite number gets written on the wall of this cave, then the world collapses into nothingness. Can the monks save the world for eternity?

Problem for 8th grade. The incenter of a triangle is equidistant from the midpoints of the sides of the triangle. Prove that the triangle is equilateral.

Problem for 9th grade. Bob was given 30 distinct natural numbers. He wrote down all the 435 pairwise sums. It appears that among those sums, 230 are divisible by 3. How many of the original 30 numbers are divisible by 3?

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1. #### Sanjay Godse:

Minimum possible number of liars is 3. Let the persons having numbers 1 to 17 be all truth tellers. Their sum is 153. The remaining 3 persons with numbers 18, 19, 20 will lie and tell their numbers as 1, 1 and 1. The total sum will be 156.

2. #### Sanjay Godse:

21 numbers of the original 30 numbers are divisible by 3.

3. #### Andrew:

6: Three. Numbers 18, 19, and 20 all changed their number to 1.

7a: Bob can win by symmetrical play. Define a mapping of suits (e.g. spadeshearts, clubsdiamonds), and always play the card of same rank, and corresponding suit, as Alice’s last move.

9: Two things contribute to the 230 sums: pairs of multiples of three, and pairs of a 3n+1 and a 3n+2. That works out to n0(n0-1)/2 + n1n2 = 230. The problem is symmetric in n1 and n2, and we only have to figure out n0, so let’s start with n1n2 = 230 – n0(n0-1)/2, and n1+n2 = 30-n0. Requiring n1n2 to be positive leads to a quadratic equation implying n020, so the number is 21 (specifically, 21, 5, and 4; or 21, 4, and 5).

4. #### Andrew:

6: Three. Numbers 18, 19, and 20 all changed their number to 1.

7a: Bob can win by symmetrical play. Define a mapping of suits (e.g. spadeshearts, clubsdiamonds), and always play the card of same rank, and corresponding suit, as Alice’s last move.

9: Two things contribute to the 230 sums: pairs of multiples of three, and pairs of a 3n+1 and a 3n+2. That works out to n0(n0-1)/2 + n1n2 = 230. The problem is symmetric in n1 and n2, and we only have to figure out n0, so let’s start with n1n2 = 230 – n0(n0-1)/2, and n1+n2 = 30-n0. Requiring n1n2 to be positive leads to a quadratic equation implying n0<22, and more algebra and playing with bounds gives n0>20, so the number is 21 (specifically, 21, 5, and 4; or 21, 4, and 5).

5. #### Gennardo:

7a) Alice wins. She starts with an 8. Then after Bob taking a 2, 3, 4, 5, 6, 7, 9, 10, J, Q, K or A she tooks an A, K, Q, J, 10, 9, 7, 6, 5, 4, 3 or 2 respectively. if Alice would have had 4 in a row, Bob would have that one move earlier. If Bob takes the second 7, Alice takes the third 7 preventing Bob from taking the last 7.

6. #### Gennardo:

Sorry, in the last post I meant: If Bob takes the second 8, Alice takes the third 8 preventing Bob from taking the last 8.

7. #### Gennardo:

7b) seems to be hard. An approach: Assume that the first number is a large prime of the form
64800000…0000031 if such a number exists. This number is +1 (mod 2, 3 and 5) and the digts 6, 4 and 8 are available for operation 1 for a long time.
Then add either a 6 or a pair of a 4 and an 8 in such a way that you add a 6 if the current number is +1 or +3 (mod 5)
and a 4 and an 8 if the current number is +4 (mod 5) to avoid a new number divisible by 5

So you can start with 6, 6, 6 (4+8 ist now forced), 4+8 (6 is now forced), 6, 6
and get the numbers 64800000…0000031, 64800000…0000037, ……0000043, …….0000049, …….0000053,
…..000061, …….00000067, …… These numbers are never divisible by 2, 3 or 5

Of course these numbers could be divisible by 7, but maybe you can also avoid reaching such a number ?

Then there is the second operation which is very limited (once the digits are sorted you have to wait for the next overflow to reuse it), but still can help when you are stucked.

I think I’m missing something

8. #### Gennardo:

7b) Start the process with a prime which must be between 10^(k-1) and 10^k for a certain k. Since the primes can only grow
during the process the sequence will exceed 10^k at some point. Since the second operation cannot
lead to an additional digit this must happen after a step of the first operation. So the prime
must be 100…001, 100…003, 100…007 or 100…009 at this point. Now another first operation would lead to
an even number, i.e. a non-prime. 100…001 cant be the subject for the second operation without
giving an non-prime, the others can if 300…001, 700…001, 900….001 are primes. But thats the
end neither the fist nor the second operation again can give another prime.
So the process is finite for every prime and since you start with a finite numbers of primes the whole process is finite.