SAMO – Mathemafrica http://www.mathemafrica.org All about maths in Africa Mon, 20 Apr 2020 06:44:50 +0000 en-US hourly 1 https://wordpress.org/?v=4.4.2 The 2018 South African Mathematics Olympiad — Problem 6 http://www.mathemafrica.org/?p=15416 http://www.mathemafrica.org/?p=15416#comments Tue, 23 Jul 2019 14:29:33 +0000 http://www.mathemafrica.org/?p=15416 The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. I have been writing about some of the problems from the senior paper from 2018. A list of all of the problems can be found here.

Today we will look at the sixth and final problem from the 2018 South African Mathematics Olympiad:

Let n be a positive integer, and let x_1, x_2, \dots, x_n be distinct positive integers with x_1 = 1. Construct an n \times 3 table where the entries of the k-th row are x_k, 2x_k, 3x_k for k = 1, 2, \dots, n. Now follow a procedure where, in each step, two identical entries are removed from the table. This continues until there are no more identical entries in the table.

  1. Prove that at least three entries remain at the end of the procedure.
  2. Prove that there are infinitely many possible choices for n and x_1, x_2, \dots, x_n such that only three entries remain,

There are some heuristics that are often helpful when solving a problem, such as

  • Looking at small cases:

    This helps us to understand the problem and how the various pieces in the problem relate to each other.

…]]> The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. I have been writing about some of the problems from the senior paper from 2018. A list of all of the problems can be found here.

Today we will look at the sixth and final problem from the 2018 South African Mathematics Olympiad:

Let n be a positive integer, and let x_1, x_2, \dots, x_n be distinct positive integers with x_1 = 1. Construct an n \times 3 table where the entries of the k-th row are x_k, 2x_k, 3x_k for k = 1, 2, \dots, n. Now follow a procedure where, in each step, two identical entries are removed from the table. This continues until there are no more identical entries in the table.

  1. Prove that at least three entries remain at the end of the procedure.
  2. Prove that there are infinitely many possible choices for n and x_1, x_2, \dots, x_n such that only three entries remain,

There are some heuristics that are often helpful when solving a problem, such as

  • Looking at small cases:

    This helps us to understand the problem and how the various pieces in the problem relate to each other.

…]]> http://www.mathemafrica.org/?feed=rss2&p=15416 1 The 2018 South African Mathematics Olympiad — Problem 5 http://www.mathemafrica.org/?p=15351 http://www.mathemafrica.org/?p=15351#comments Sun, 21 Jul 2019 12:03:57 +0000 http://www.mathemafrica.org/?p=15351 The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. I have been writing about some of the problems from the senior paper from 2018. A list of all of the problems can be found here.

Today we will look at the fifth problem from the 2018 South African Mathematics Olympiad:

Determine all sequences a_1, a_2, a_3, \ldots of nonnegative integers such that a_1 < a_2 < a_3 < \ldots, and a_n divides a_{n - 1} + n for all n \geq 2.

Since the sequence a_1, a_2, \ldots is strictly increasing, we know that a_n \geq n - 1 for all positive integers n. (We could prove this rigorously by induction.) This means that a_{n - 1} + n \leq (a_n - 1) + (a_n + 1) = 2a_n for all n, and so we know that a_{n - 1} + n is equal to either a_n, or to 2a_n for all positive integers n. Perhaps we should try to figure out exactly when it is equal to a_n, and when it is equal to 2a_n. If we knew, for example, that we always have that a_{n - 1} + n = a_n, then we have reduced the problem to solving this recurrence relation.…

]]> The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. I have been writing about some of the problems from the senior paper from 2018. A list of all of the problems can be found here.

Today we will look at the fifth problem from the 2018 South African Mathematics Olympiad:

Determine all sequences a_1, a_2, a_3, \ldots of nonnegative integers such that a_1 < a_2 < a_3 < \ldots, and a_n divides a_{n - 1} + n for all n \geq 2.

Since the sequence a_1, a_2, \ldots is strictly increasing, we know that a_n \geq n - 1 for all positive integers n. (We could prove this rigorously by induction.) This means that a_{n - 1} + n \leq (a_n - 1) + (a_n + 1) = 2a_n for all n, and so we know that a_{n - 1} + n is equal to either a_n, or to 2a_n for all positive integers n. Perhaps we should try to figure out exactly when it is equal to a_n, and when it is equal to 2a_n. If we knew, for example, that we always have that a_{n - 1} + n = a_n, then we have reduced the problem to solving this recurrence relation.…

]]> http://www.mathemafrica.org/?feed=rss2&p=15351 1 The 2018 South African Mathematics Olympiad — Problem 2 http://www.mathemafrica.org/?p=15219 http://www.mathemafrica.org/?p=15219#comments Mon, 15 Jul 2019 13:10:50 +0000 http://www.mathemafrica.org/?p=15219 The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. In the week and half leading up the the contest, I plan to take a look at some of the problems from the senior paper in 2018, and have already written about the first problem

The second problem from the 2018 South African Mathematics Olympiad was

In triangle ABC, AB = AC, and D is on BC. A point E is chosen on AC, and a point F is chosen on AB, such that DE = DC and DF = DB. It is given that \frac{DC}{BD} = 2 and \frac{AF}{AE} = 5. Determine the value of \frac{AB}{BC}.

The first step of solving any geometry problem should always be to draw a sketch. This helps you to understand how different parts of the figure relate to each other, and an accurate sketch may help you to form conjectures. Sometimes having a deliberately inaccurate sketch on hand is also helpful as it may help to avoid circular reasoning.…

]]> The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. In the week and half leading up the the contest, I plan to take a look at some of the problems from the senior paper in 2018, and have already written about the first problem

The second problem from the 2018 South African Mathematics Olympiad was

In triangle ABC, AB = AC, and D is on BC. A point E is chosen on AC, and a point F is chosen on AB, such that DE = DC and DF = DB. It is given that \frac{DC}{BD} = 2 and \frac{AF}{AE} = 5. Determine the value of \frac{AB}{BC}.

The first step of solving any geometry problem should always be to draw a sketch. This helps you to understand how different parts of the figure relate to each other, and an accurate sketch may help you to form conjectures. Sometimes having a deliberately inaccurate sketch on hand is also helpful as it may help to avoid circular reasoning.…

]]> http://www.mathemafrica.org/?feed=rss2&p=15219 1 The 2018 South African Mathematics Olympiad — Problem 1 http://www.mathemafrica.org/?p=15210 http://www.mathemafrica.org/?p=15210#comments Sat, 13 Jul 2019 10:28:18 +0000 http://www.mathemafrica.org/?p=15210

The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. In the two weeks leading up to the contest, I plan to take a look at some of the problems from the senior paper from 2018.

The first problem from the 2018 South African Mathematics Olympiad was

One hundred empty glasses are arranged in a 10 \times 10 array. Now we pick a of the rows and pour blue liquid into all glasses in these rows, so that they are half full. The remaining rows are filled halfway with yellow liquid. Afterwards, we pick b of the columns and fill them up with blue liquid. The remaining columns are filled with yellow liquid. The mixture of blue and yellow liquid turns green. If both halves have the same colour, then that colour remains as is.

  1. Determine all possible combinations of values for a and b so that exactly half of the glasses contain green liquid at the end.
…]]>

The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. In the two weeks leading up to the contest, I plan to take a look at some of the problems from the senior paper from 2018.

The first problem from the 2018 South African Mathematics Olympiad was

One hundred empty glasses are arranged in a 10 \times 10 array. Now we pick a of the rows and pour blue liquid into all glasses in these rows, so that they are half full. The remaining rows are filled halfway with yellow liquid. Afterwards, we pick b of the columns and fill them up with blue liquid. The remaining columns are filled with yellow liquid. The mixture of blue and yellow liquid turns green. If both halves have the same colour, then that colour remains as is.

  1. Determine all possible combinations of values for a and b so that exactly half of the glasses contain green liquid at the end.
…]]> http://www.mathemafrica.org/?feed=rss2&p=15210 1