OBMEP 2023 – 1st phase – Level 1: 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.

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Question 1 — Cans that support other cans

Start from the blue can and mark the two that depend on it. Repeat the marking on the following layers: two cans, then two and, finally, the top can. These dependencies need to be removed from the top down; Removing cans from outside this set does not help release the blue one. There are $2+2+2+1=7$ cans before it. Alternative C.

Question 2 — Shopping comparison

José pays $80+65=145$ reais; Luiz pays $90+60=150$ reais. Subtract the totals, not just the prices of one of the pieces: $150-145=5$. Luiz spends R$ 5,00 more — A.

Question 3 — A single necklace

A closed collar must form a single circuit. In each alternative, start at one end of the visible part and follow the thread, including at the intersections, until you return to the beginning. If another separate circuit remains, the figure represents two collars. Only the composition D closes a circuit that runs through the entire wire; crossing two stretches does not mean joining them.

Question 4 — Card order

Solve backwards. And it covers up the others and was the last one. Apart from E, A is over B, so A was fourth. B is over C and D, which occupy the first two positions, in some order. Therefore B was the third card, corresponding to alternative B.

Question 5 — Identification of stamps

Only compare sections that are outside the lane, as the interior is hidden. In the first line, the visible outline selects 1B. In the second, the top and bottom tips select 2B. In the third, the stamp needs to hide one of its protrusions within the strip, leaving the others in the indicated positions: 3C. The set is 1B, 2B and 3C — C.

Question 6 — Fractions and decimals

Write all alternatives as decimals and align the places:

The third value is $0{,}166$ — E. The decimal $1/6$ is a little larger, as it continues with digits 6 after the third place.

Question 7 — Colored area

Divide the rectangle into four strips of the same area. The lower yellow band takes up $1/4$ of the total. In the upper bands, the yellow triangular parts can be gathered in a region of equal area to another band. Thus, the yellow fraction is $1/4+1/4=\boxed{1/2}$. Alternative C. The slope of one side does not change the formula base times height divided by 2.

Question 8 — Division of expenses

Pedro spent $4\cdot5{,}50+5\cdot2{,}20=33$ reais. Paulo spent $1{,}4\cdot5=7$ reais. Each person must pay $(33+7)/2=20$ reais, so Paulo must transfer $20-7=13$ reais to Pedro. The expression is $(33-7)/2$, alternative A. Transferring the 26 real difference would only change the expenses.

Question 9 — Coverage by pieces

Measure the area in squares: the eight square pieces add up to 8, and the nine triangles add up to $9/2$. The total area is $12{,}5$. This eliminates the different area board. It is also necessary to accommodate eight squares: A, C and D only accommodate seven in the indicated division. The board B meets the area and allows you to fit the eight squares and the nine half squares. Alternative B.

Question 10 — Sticker collections

Taking Benedito’s collection as a unit, the quantities are $1$, $4/5=0{,}80$ and $3/4=0{,}75$. Logo $1>0{,}80>0{,}75$: Benedito has more and Carlos has less — C.

Question 11 — Complement of a piece

The scalloped edge must fit without gaps or overlap. Note the protrusion in two steps and compare the complementary cutout in the options, allowing you to rotate the piece. In A, B, C and E, an incompatible step or cavity remains. The piece D completes the external rectangular contour and exactly fills the internal cutout. Alternative D.

Question 12 — Even Sundays

Successive Sundays differ by 7, so their dates alternate between odd and even. Three even Sundays require five Sundays, starting with an even date. The first can only be day 2: the dates are 2, 9, 16, 23 and 30. Therefore, day 9 is Sunday and day 10 is Monday — A.

Question 13 — Pizzas and divisibility

If $k$ of the five pizzas have eight slices, the rest have six. The total is $8k+6(5-k)=30+2k$, with $0\le k\le5$. Among 30, 32, 34, 36, 38 and 40, only 36 is divisible by 9. Each friend received $36/9=4$ slices — A. This does not require the slices to be the same size.

Question 14 — Number of birthdays

The five months already specified add up to $2\cdot6+3\cdot2=18$ students. In the other seven months, each quantity is between 3 and 5, as there cannot be more months with 2 or 6. Therefore,

Of the options, only 45 is in this range. It is achievable: in the remaining seven months, six groups of 4 and one of 3 add up to 27. Answer: 45 — B.

Question 15 — Hidden faces of cubes

A cube with some visible white face cannot be entirely blue. Count the cubes in this situation once, even when they have two white faces: there are nine. Of the 14 cubes, there remain $14-9=5$ which can have all blue faces. The figure does not require any hidden face to be painted white. Alternative E.

Question 16 — Fraction of a fraction

The population with bicycles corresponds to $1/4$ of the total. Within this group, $2/3$ has only one. Therefore, the desired portion is

Alternative B. Adding the fractions would mix two different reference groups.

Question 17 — Reading the hour hand

The hour hand moves through five divisions in 60 minutes; each division represents 12 minutes. The departure was at 6:24 am, and the return was at 2:12 pm. To subtract, write 2:12 pm as 1:72 pm:

Alternative D. The marks on the dial do not represent minutes when we are reading the hour hand.

Question 18 — Route length

Separate horizontal and vertical segments. The verticals add up to two heights of 8 cm: 16 cm. On the horizontals, a recess of size $d$ decreases one section and increases another; the sum is $5+(5-d)+8+8+d=26$ cm. Therefore the total length is $16+26=42$ cm — E.

Question 19 — The child who lies

The 8 year old child is the only liar. If she said she didn’t have 7 or didn’t have 9, she would be telling the truth. Therefore, it is necessarily whoever claims not to have 8. Of the remaining two, whoever says they don’t have 9 has 7; whoever says they don’t have 7 has 9. This is how the youngest and oldest in the figure can be identified. Alternative B.

Question 20 — Path count

Separate the paths by the mandatory point of the halfway line they cross. Through the two extreme points, there is a path for each one. For each internal point, there are six ways to get there and six ways to continue, totaling $6\cdot6=36$. The classes are disjoint, therefore

Alternative E. We multiply successive choices and add the alternative cases.

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