Repeated Participation at the Mathematical Olympiads: A Comparative Study of the Solutions of Selected Problems

Abstract

The paper analyzes the works of students who have participated in at least three Open Mathematical Olympiads of Latvia (LOMO) in the 6th, 8th and 9th grade. The question of the research discussed in the paper is: What knowledge and problem-solving skills do students demonstrate in the solutions of algebra and number theory problems? Six algebra and number theory problems were selected for the research, as solving them requires sufficiently high levels of abstract thinking, algebraic reasoning, and an accurate use of the mathematical language. The authors developed a special coding system for students’ works and elaborated an assessment tool for assessing each individual student’s levels of problem-solving competence. The implementation of this tool enables comparing the student’s problem-solving success in different grades, and it enables comparing the specific properties of the solutions presented by the group of students as well. While showing that a few students were good problem solvers, the data collected revealed deficiencies of algebra knowledge in a significant part of students’ works.

Appendices

Appendix 1

1.1 Solutions of LOMO Problems Offered by Experts

• Solution of Problem 6.1. (LOMO 39, 2012)

Two numbers are erased and their sum plus 2 is written instead. As the other numbers do not change, their sum stays the same. The common sum of the given numbers increases by 2 after every operation. Finally, the last number equals $S + 2 \cdot n$ where S is the total sum of all given numbers, but n is the number of the executed operations. There were 10 numbers on the board at the beginning, but after each operation the amount of the numbers diminishes by one. Therefore $n = 9$. The number left on the board is:

$$S + 2 \cdot n = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) + 2 \cdot 9 = \frac{(1 + 10) \cdot 10}{2} + 18 = 55 + 18 = 73$$

• Solution of Problem 6.4. (LOMO 39, 2012)

• Case (a) It is possible to write six numbers around the circle as required. See the example in Figure 13.13.

  Fig. 13.13

  

  Arrangement of six numbers around the circle

• Case (b) It is not possible. The sum of two different natural numbers is at least 1 + 2 = 3. All prime numbers except the number 2 are odd. Accordingly, the sum of two adjacent numbers must be an odd number, that is, the parities of any two adjacent numbers must differ. Whatever is the layout of the seven numbers around the circle, there is going to be one spot where two numbers of the same parity with a total sum of an even number bigger than 2 (it is not a prime number) will be next to each other.

• Solution of Problem 8.2. (LOMO 41, 2014)

Express the number in expanded form $x \cdot 10^{k} + Y$ where x is the first digit, but Y is the k–digit number ($1 \le k \le 5$). Then $x \cdot 10^{k} + Y = 15 \cdot Y \Rightarrow x \cdot 10^{k} = 14 \cdot Y \Rightarrow x \cdot 2^{k} \cdot 5^{k} = 2 \cdot 7 \cdot Y$.

We conclude that x is divisible by 7. Taking into account that x is a one-digit number $x = 7$ and $Y = 2^{k - 1} \cdot 5^{k} = 5 \cdot 10^{k - 1}$, $1 \le k \le 5$. There are five natural numbers that satisfy the solution—75, 750, 7500, 75,000, 750,000.

• Solution of Problem 8.5 (LOMO 41, 2014)

The variable x represents the number in the central cell of the given square. The variable y represents the number in the middle cell of the lower row. Then the sum of numbers in every row, column or diagonal is $24 + x + y$. The cells can be filled step by step (see Figs. 13.14 and 13.15).

Fig. 13.14

Filling of the square in case (a)

Fig. 13.15

Filling of the square in the case (b)

Case (a)

We determine that the middle number in the last column must be −10, which is not a natural number. Consequently, the lower left corner cannot contain the number 7.

Case (b) Similarly the cells can be filled in the case (b):

The sum of numbers in the diagonal is 3x. Then $y = 2x - 24$. Choosing $x = 13$ we determine one of the answers (see Fig. 13.16.).

Fig. 13.16

One of the possible answers in the case (b)

• Solution of Problem 9.1. (LOMO 42, 2015)

The letter x and the expression $x + 2015$ are used to represent the given numbers. The function representing their product is $f(x) = x \cdot (x + 2015) = x^{2} + 2015x$. The graph of this function is a parabola that opens upward. The x-coordinate of their vertex is $x_{0} = \frac{ - 2015}{2} = - 1007.5$. The function has the minimal value at this point. Consequently, the two numbers are 1007.5 and −1007.5.

• Solution of Problem 9.3 (LOMO 42, 2015)

The factoring of the expression is

$$\begin{aligned} x^{5} - 5x^{3} + 4x &= x \cdot (x^{4} - 5x^{2} + 4) \\ &= x \cdot (x^{4} - x^{2} - 4x^{2} + 4) = x \cdot (x^{2} (x^{2} - 1) - 4(x^{2} - 1)) \\ & = x \cdot (x^{2} - 1) \cdot (x^{2} - 4) = x \cdot (x - 1) \cdot (x + 1) \cdot (x - 2) \cdot (x + 2) \\ &= (x - 2) \cdot (x - 1) \cdot x \cdot (x + 1) \cdot (x + 2). \\ \end{aligned}$$

We determined that the given expression is a product of five consecutive whole numbers. At least two of the numbers are divisible by 2, while one of them is also divisible by 4. At least one number is divisible by 3, and at least one number is divisible by 5. Consequently, the product is divisible by $2 \cdot 3 \cdot 4 \cdot 5 = 120$.

Appendix 2

2.1 Solutions of Problems, Presented by Students

Comment. We tried to keep in translation the writing style presented by students. Usually students did not use correct mathematical language and failed to construct grammatically correct sentences. Sometimes they omitted the subject or the verb.

• Solution of Problem 6.1, presented by student Emils

Taking into account that we reach every next number in the equation $a + b + 2 = c$, we can predict:

d = the number that left the last,

there are 10 numbers, therefore 5 of them will be A, but others 5—will be b.

In one’s turn $d = a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + b_{1} + b_{2} + b_{3} + b_{4} + b_{5} + (n \cdot 2)$

n = the number of attempts of the operation $a + b + 2 = c$

n = 9 because after every attempt left per 1 number less on the blackboard, so only one number left after 9 attempts.

Not important, which number is $a_{1}$, which number is $a_{2}$, etc., because all given numbers will be used sooner or later. The common sum of given numbers is $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55$, therefore $d = 55 + (9 \cdot 2) = 55 + 18 = 73$.

• Solution of Problem 8.5, presented by student Janis

Problem 5, version (a). If we insert number 7 in the left lower corner, the square is impossible to complete. The square in the picture is almost completed (see Fig. 13.17). However, number 7 does not allow the highlighted diagonal (all sums are 35). If number 7 could be turned into number 17 at least, the square would be completed as I am going to demonstrate in version (b).

Fig. 13.17

Filling of the square in case (a)

The square in Problem 5 version (b) will be possible to complete because of the lack of a small number that disturbs completion. At the beginning the square is drawn up so that the last digit of the sum in each row, column and diagonal is 7. Then a number of tens is added so that the sums would be 57 (see Fig. 13.18).

Fig. 13.18.

Solution of case (b), presented by Janis

Answer: it is impossible to complete the square in version (a), but it is possible to complete it in version (b).

• Solution of Problem 9.1, presented by student Arvis

Numbers where the difference is 2015 will have the smallest product when the second number is negative, because if a negative number is subtracted, it is added. For example, $2 - ( - 4) = 2 + 4 = 6$. Therefore the smallest product for whole numbers is 1008 and −1007. But it can be slightly increased if we take numbers 1007.5 and −1007.5.

• Solution of Problem 9.1, presented by student Alex

We assume that 2015 and 0 are such numbers. If it is not true, then one of the numbers is less than 0 (for the product of two numbers to be less than 0, one number is <0, but the other is bigger). We assume that the positive (P ₁) is subtracted from the negative (−P ₂):

$- P_{2} - P_{1} < 0 < 2015$ is not good.

Second version:

$P_{1} - ( - P_{2} ) = P_{1} + P_{2}$—can be completely equal to 2015.

As we have to determine the smallest $P{}_{1} \cdot ( - P_{2} )$, let’s determine the biggest $P{}_{1} \cdot P_{2}$, so that $P{}_{1} + P_{2} = 2015$.

According to the rule of the square where the perimeter is equal to the perimeter of other rectangles (not squares), a square will always have the biggest area.

That is, $P{}_{1} \cdot P_{2}$ is going to be the biggest if $P{}_{1} = P_{2}$.

As $P{}_{1} + P_{2} = 2015$, $P{}_{1} = P_{2} = 1007.5$.

Consequently, the smallest product is going to be $- (1007.5^{2} )$. Truly:

$1007.5 - ( - 1007.5) = 2015$, so these are the numbers we need. Answer 1007.5 and −1007.5.

• Solution of Problem 9.3, presented by student Elena

If this number is divisible by 120, it contains prime factors 2, 2, 3, 2, 5. This number is divisible by 5, because subtrahend $( - 5x^{3} )$ is divisible by 5, and the transformed expression $x(x^{4} + 4)$ is also divisible by 5. In order to prove it, let’s determine how many different numbers can be obtained in the case of each digit (see Fig. 13.19).

Fig. 13.19

Last digits of the terms x and $x^{4}$

If the last digit of number $x^{4}$ is 1, then, if we add 4, it is divisible by 5. Accordingly, every product is divisible by 5. It is likewise with the digit 6 at the end of a number. If the last digit of number $x^{4}$ is 5, then x contains 5. But all addends contain 5, therefore $\vdots$ 5.

We see that the sum of the last digits is 0 (see Fig. 13.20). 0 is divisible by $2 \cdot 2 \cdot 2 = 8$ therefore the given expression is divisible by 8. The expression is divisible by 3 too if x is a whole number. The expression $x^{5} - 5x^{3} + 4x$ is divisible by 120 if x is a whole number.

Fig. 13.20

Research on the last digits of given terms

Appendix 3

3.1 Example of Coding System for Problem 6.1 (LOMO 39)

We selected all solution steps contained in the works of the students of the selected focus group. They are as follows:

• 1. Try one or some examples to see how the given algorithm works.

• 2. Describe one step of the process algebraically.

• 3. Systematically investigate several examples to detect common properties of the given numbers during the process.

• 4. Visualize the algorithm—construct the tree, draw the arcs, or use colored pencils—to visualize the calculation of an example.

• 5. Calculate the sum of the given numbers (S = 55).

• 6. Express the sum of the given numbers S algebraically.

Consider and explain what happens at every step of the process:

• 7. the sum of the given numbers does not change,

• 8. the total sum of the numbers on the black-board increases by 2,

• 9. the amount of the numbers on the black-board diminishes by 1.

Calculate the number of steps in the process:

• 10. the use of an algebraic formula (10 – n = 1),

• 11. calculate (10–1 = 9).

Calculate the increase of the sum S:

• 12. the use of an algebraic formula (2n),

• 13. numerically (2·9).

Write the calculation formula:

• 14. algebraically (S + 2n),

• 15. numerically ((1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) + 2·9).

• 16. Calculate the result (73).

• 17. Justify the invariance of the result obtained.

It is not necessary to execute all of these steps to complete the solution of the given problem. The best students’ solutions were similar to the expert solving path, which contains the following steps: 5, 7, 8, 9, 10, 12, 14, 16, and 17.

All the steps mentioned were coded corresponding to the competencies needed for problem solving. Some of the steps are attributed to different groups of competencies. For example, the creation of an algebraic formula to describe the whole process corresponds to modelling skills, technical skills, and problem-solving skills. Problem-solving skills applied in the solutions represent such heuristic strategies as the method of trial and error (step 1), the visualization of the process (step 4), systematic investigation (step 3), the detection of relevant properties of the process (steps 7–13)—here explanations and reasoning are coded as argumentation skills; the creation of algebraic formula (step 14), justifying the invariance of the process (step 17). Students’ decisions to complete calculations or to apply some algebraic terms are coded as technical skills. If a student additionally formulated the established definitions and described the applied operations, we coded these argumentation skills as knowledge reproduction and description (see Table 13.3).

Table 13.5 contains codes for each solution step of Problem 6.1. We added the number of the step to the code. In the case of Problem solving some steps are coded by the same code because the meaning of these steps is the same. For example, step 12 and step 13 are coded by ap12, where the second letter is used to differentiate it from the code am12 used for coding the same step as a case of modelling.

Table 13.5. Coding system of solution steps for Problem 6.1 LOMO 39

Student Emils used 10 steps in the draft and fair copy for solving this problem. He tried to calculate the example (step 1) and made one mistake (this step was evaluated with average quality). Then he systematically investigated the examples (step 3) in the draft (he scored average quality because he called an example “proof”). He calculated the sum S (step 5) and the increase of the sum (step 13). In the fair copy, Emils used an algebraic description (step 2), created and explained the formula (step 14), transferred this into a numerical form (step 15) and calculated the result (step 16). He calculated and reasoned the number of steps in the process (step 11) and justified the invariance of the process (step 17). All the steps completed were of a high quality, except two. The quality map GMC of the steps completed is as follows: 210 110 400 100.

Emils made 3 steps of the first level of competencies and two of the second level, where two of these steps were considered to be of an average quality. Other five steps were completed very well. Emils presented exhaustive explanations and a correct justification, as demonstrated in the argumentation section in Table 13.6 that characterized Emils work in details. His solution obtained the highest score.

Table 13.6. Expanded map of Emils’ mathematical skills