Euler's Eyes
2026-09-26
OBMEP 2023 – 2nd phase – Level 3: complete resolution
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 — 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 2 — Coloring rules
Use both rules: adding 5 preserves the color; a white product must have at least one white factor.
a) $38-3=35$ is a multiple of 5. Going back seven times from 5 to 5, we conclude that 3 is white.
b) If $4=2\cdot2$ is white, one of the factors 2 must be white; as they are the same, 2 is white.
c) If 1 is white, 16 is also, by the first rule. As $16=4^2$, 4 is white, by the same argument as in item b.
d) Let us show the two implications. If 3 is white, 8 is also white. In $8=2\cdot4$, either 2 is white directly, or 4 is white and item b forces 2 white. Conversely, if 2 is white, 12 is also white. In $12=3\cdot4$, either 3 is white, or 4 is white. In the last case 9 is white, as it differs from 4 by 5, and $9=3^2$ makes 3 white. Therefore 2 is white if, and only if, 3 is white; otherwise, both are black.
Question 3 — Nearest antenna
Consider the road as the horizontal axis. The squared distances to the antennas are $d_C^2=(x-2)^2+1$ and $d_D^2=(x-5)^2+4$.
a) In $x=3$, these values are 2 and 8. The function chooses the smallest: $f(3)=2$.
b) To equalize the distances, equalize their squares: $(x-2)^2+1=(x-5)^2+4$. Canceling $x^2$, we obtain $6x=24$, therefore $x=4$.
c) The difference $d_C^2-d_D^2=6x-24$ changes sign in 4. Logo
$$f(x)=\begin{cases}(x-2)^2+1,&0\le x\le4,\\(x-5)^2+4,&4Question 4 — Elevator choices
Let us adopt destinations on the four floors above the ground floor, independent and equiprobable. The three people can choose from $4^3=64$ modes.
a) For a single common target, there are four choices: $4/64=1/16$.
b) It is simpler to count the absence of stops on the first floor: each person has three possible destinations, totaling 27 cases. So the chance of stopping there is $1-27/64=37/64$.
c) The three consecutive stops are $\{1,2,3\}$ or $\{2,3,4\}$. In each set, the three people can choose different destinations in $3!=6$ orders. The chance is $2\cdot6/64=3/16$.
The official guide also agrees to consider the ground floor as a destination. In this interpretation there are $5^3=125$ choices; the responses become, respectively, $1/25$, $61/125$ and $18/125$. It is necessary to declare which sample space is being used.
Question 5 — Triangular sticker on the cylinder
Open the side surface into a rectangle of width 12 cm and height 15 cm. With each complete turn, the sticker advances 12 cm on its horizontal base.
a) For base 10, there is no complete turn or overlap. The surface measures $12\cdot15=180$ and the triangle, $10\cdot15/2=75$. The discovered area is $105\text{ cm}^2$.
b) For base 60, the height of the sticker decreases $15\cdot12/60=3$ cm with each turn. The first round already covers everything that will be covered; the rest overlap below it. What remains is a triangle with base 12 and height 3, with area $18\text{ cm}^2$.
c) For base 90, the height reduction per turn is $15\cdot12/90=2$ cm. The boundaries between two, three and four layers are parallel. The strip with exactly three layers is a parallelogram with a horizontal base 12 and a vertical height 2. Its area is $24\text{ cm}^2$.
Question 6 — Boards and parity
Represent white by 0 and black by 1. The requirement is that each subboard $3\times3$ has an odd sum.
a) For houses K, L, M and N in the last line, the four possibilities are BBPP, BPBP, PBBB and PPPB. Choosing K and L freely fixes M by the parity of the left block and fixes N by the right one.
b) On the $3\times4$ board, choose the eight squares in the first two rows and the first two squares in the third. The remaining two are forced, one for each condition $3\times3$. There are $2^{10}=1024$ paintings.
c) In $4\times4$, choose the first two lines and the first two columns of the other lines: they are $8+4=12$ binary choices. Complete each line from left to right, imposing the parity of the block that ends in the new house. The total is $2^{12}=4096$.
d) In $2023\times2023$, the same procedure leaves $2\cdot2023+2\cdot2021=8088$ spaces free. Each remaining square is determined by the block $3\times3$ that has the lower right corner of it. No condition goes unapplied and no box is determined twice. Therefore the total is $2^{8088}$.