Official material: complete proof · OBMEP solutions · collection of the year.

The resolutions below follow the numbering of the official notebook. Open the test to follow the statements, alternatives and diagrams; The explanations here are copyright. Attempt each question before checking the answer.

Question 1 — Wagons, colors and flags

a) The white cars have numbers $1,4,7,10,\ldots$; with a flag, they must also be pairs. The first one after 8 is 10.

b) $2023=3\cdot674+1$. The remainder 1 identifies a white wagon.

c) The even blanks form $4,10,16,\ldots,2020$. Its general term is $4+6k$. The condition $4+6k\le2023$ gives $0\le k\le336$, that is, 337 flags. The carriage 2023 is white, but has no flag.

Question 2 — Perimeter of hexagons

Call the central hexagon C and number the outer six 1 to 6, in circular order. For $n$ painted hexagons and $e$ shared sides, the perimeter is $P=6n-2e$.

a) For 12 cm, color C, 1 and 2: there are three contacts and $P=18-6=12$. For 18 cm, paint 1, 3 and 5: there are no common sides and $P=18$.

b) Paint the exteriors 1, 2, 4 and 5. There are two contacts and $P=24-4=\boxed{20\text{ cm}}$. This is the maximum: including C, there are at least three contacts; Excluding C, four positions of a cycle of six have at least two contacts, as the two empty positions separate at most two blocks of occupied positions.

c) In the formula $P=6n-2e$, both installments are even. Thus every possible perimeter is par.

Question 3 — Cards and multiples

a) One possibility is $6,2,4,8,1,9,3$. Checking the pairs, we have divisibility in all of them: 6 by 2, 4 by 2, 8 by 4, 8 by 1, 9 by 1 and 9 by 3.

b) With eight cards: $9,3,6,2,4,8,1,5$. No number is repeated and each neighboring pair satisfies the rule.

c) Use $5,1,7$ or its reverse order. Among the available cards, only 1 can be a neighbor of 5 and only 1 can be a neighbor of 7.

d) If we used the nine cards, 5 and 7 would have to be at the ends, both connected to 1. But this would form the entire section $5,1,7$, without the possibility of attaching another card: 1 would already have its two neighbors. Therefore, there is no sequence with the nine.

Question 4 — Additions with equal digits

Write each installment as $111a$, with $1\le a\le9$. So just work with the sum of three different digits in ascending order.

a) As $1332=111\cdot12$, we can choose $3+4+5=12$: $333+444+555=1332$.

b) As $2109=111\cdot19$, we need $a

  • $222+888+999$;
  • $333+777+999$;
  • $444+666+999$;
  • $444+777+888$;
  • $555+666+888$.

c) Be $s=a+b+c$, with $6\le s\le24$. The carry from units to tens is $u=\lfloor s/10\rfloor$. The next transport is $v=\lfloor(s+u)/10\rfloor$. The tens and hundreds digits come from $s+u$ and $s+v$, reduced modulo 10. For $s=6,\ldots,9$, we have $u=v=0$; for $10\le s\le18$, $u=v=1$; to $20\le s\le24$, $u=v=2$. The only exception is $s=19$, with $u=1$ and $v=2$. The corresponding result is $111\cdot19=2109$.

Question 5 — Coded paths

a) For the code 1001001100, scroll over the squares 1, 4, 7 and 8 and under the others. Starting at A, go up before square 1, go down after it, go up before 4, go down after 4, go up before 7, go down after 8 and, after 10, go up to B. This description determines the entire design.

b) Each of the ten squares offers two choices, up or down. Vertical passages are determined by changes between choices, so each binary word corresponds to a path. Total: $2^{10}=1024$.

c) In the two squares that meet at C, three of the four local combinations pass through the point. The other eight squares are free. So there are $3\cdot2^8=768$ paths. Equivalently, we subtract the $2^8$ paths that avoid C from the 1024 totals.

Question 6 — A parity invariant

a) Number the sectors from 1 to 6 clockwise, starting with the top gray sector. In each step, perform the two movements indicated simultaneously:

Step Counterclockwise movement Clockwise movement Quantities in sectors 1 to 6
Home — — $(1,1,1,1,1,1)$
1 $2\to1$ $6\to1$ $(3,0,1,1,1,0)$
2 $3\to2$ $5\to6$ $(3,1,0,1,0,1)$
3 $2\to1$ $6\to1$ $(5,0,0,1,0,0)$

b) Each ball moved changes sector color. If both leave white sectors, the blank quantity decreases 2; if ash comes out, 2 increases; if they come out in different colors, it doesn’t change. As there are initially three blank balls, this quantity remains odd.

c) All in a single sector would produce zero or six marbles in the set of white sectors. Both numbers are even, contradicting the invariant. Therefore concentration is impossible.

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