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 — 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 3 — Carpets and overlap

a) The room measures $7\times7$, and the two rugs each have an area of ​​14. Its intersection measures $2\times2=4$. By inclusion-exclusion, the discovered area is $49-(14+14-4)=25\text{ m}^2$.

b) In the second arrangement, use the MN diagonal common to mats. It divides each mat into two triangles of area 7. As the mats are congruent, the distances from their vertices to MN are equal. If this distance is $h$, then $MN\cdot h/2=7$. Compare with the two halves of the room, with the same base MN and height $h$: each half has area $MN\cdot h=14$. So the room has $28\text{ m}^2$.

c) Denote the overlapping area by $I$. The union of the mats has area $14+14-I=28-I$. As the room measures 28, its uncovered part measures $28-(28-I)=I$. Thus, the uncovered and overlapping areas are equal.

Question 4 — Bracelets

Write P for black and B for white; the words are read in a circle, identifying turns and inversions.

a) The two classes are PPBB and PBPB: black together or separately.

b) There are three classes, represented by PPPBBB, PPBPBB and PBPBPB. We classify by the largest number of consecutive blacks: three, two or one.

c) Fix the four white ones and record the amounts of black in the four intervals. Short of spins and reflections, all possibilities are

$$(4,0,0,0),\ (3,1,0,0),\ (3,0,1,0),\ (2,2,0,0),$$

$$(2,0,2,0),\ (2,1,1,0),\ (2,1,0,1),\ (1,1,1,1).$$

The sorting comes from the partitions of 4: four together; three and one, at neighboring or opposite intervals; two pairs, also neighbors or opposites; one pair and two isolated ones, with the empty interval neighboring or opposite the pair; or all separately. So there are eight bracelets.

Question 5 — 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 6 — L-shaped pieces

In the tables below, each letter represents a three-square piece; matching letters form an L. Read the rows from top to bottom and the columns from left to right.

a) A complete coverage, respecting the five initial pieces A to E, is:

A A F F G G
B A F H H G
B B E H I I
C C E E J I
C D K J J L
D D K K L L

There are 12 pieces, each with three spaces, without overlapping: $12\cdot3=36$.

b) Keep A and B in the given positions and add C, D, E and F:

. A A B B .
. A . . B .
. C C D D .
. C . . D .
. E E F F .
. E . . F .

The dot represents a free house. To place an L, some block $2\times2$ would need to have three free spaces. In this configuration, each block $2\times2$ has at least two occupied blocks; therefore does not fit a seventh piece.

c) Divide the board into nine disjoint blocks $2\times2$. Five pieces occupy only 15 spaces. If all blocks had two occupied houses, at least 18 would be required. Therefore, any block has at most one occupied and at least three free. Any three squares in a block $2\times2$ form an L; so one more piece always fits.

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