diff --git a/readme.md b/README.md similarity index 100% rename from readme.md rename to README.md diff --git a/maths-club-app/src/app/data/weeklyProblems.ts b/maths-club-app/src/app/data/weeklyProblems.ts index d0e56c11..72215112 100644 --- a/maths-club-app/src/app/data/weeklyProblems.ts +++ b/maths-club-app/src/app/data/weeklyProblems.ts @@ -50,153 +50,153 @@ export const WEEKLY_PROBLEMS = { "2021-08-01": "letter-matching", "2021-08-08": "countdown", "2021-08-15": "15-game", - "2021-08-22": "pell-numbers", - "2021-08-29": "paths", - "2021-09-05": "number-challenge-1", - "2021-09-12": "making-squares", - "2021-09-19": "logic-puzzles-1", - "2021-09-26": "iq-challenge-1", - "2021-10-03": "find-the-path", - "2021-10-10": "counting-squares", - "2021-10-17": "buffons-needle", - "2021-10-24": "balls-and-books", - "2021-10-31": "addition-squares", - "2021-11-07": "adding-to-15", - "2021-11-14": "9-dots", - "2021-11-21": "treasure-hunt", - "2021-11-28": "russian-multiplication", - "2021-12-05": "river-crossing", - "2021-12-12": "poisoned-wine-puzzle", - "2021-12-19": "matchstick-puzzles", - "2021-12-26": "magic-cards", - "2022-01-02": "langfords-problem", - "2022-01-09": "handshake-puzzles", - "2022-01-16": "geometry-puzzle-1", - "2022-01-23": "gabriels-problem", - "2022-01-30": "fence-around-a-field", - "2022-02-06": "domino-tilings", - "2022-02-13": "collatz-conjecture", - "2022-02-20": "coin-game", - "2022-02-27": "coconut-trader", - "2022-03-06": "circumference", - "2022-03-13": "apple-teaser", - "2022-03-20": "two-eggs", - "2022-03-27": "special-numbers", - "2022-04-03": "scales-and-weights", - "2022-04-10": "picture-puzzles-1", - "2022-04-17": "monkey-business", - "2022-04-24": "locks-and-keys", - "2022-05-01": "frog-party", - "2022-05-08": "counting-chickens", - "2022-05-15": "tic-tac-toe", - "2022-05-22": "tic-tac-toe-with-levels", - "2022-05-29": "tangrams", - "2022-06-05": "secret-santa", - "2022-06-12": "patience", - "2022-06-19": "paper-pieces", - "2022-06-26": "nim-related-games", - "2022-07-03": "nim-21", - "2022-07-10": "mastermind", - "2022-07-17": "make-many", - "2022-07-24": "letter-matching", - "2022-07-31": "countdown", - "2022-08-07": "15-game", - "2022-08-14": "pell-numbers", - "2022-08-21": "paths", - "2022-08-28": "number-challenge-1", - "2022-09-04": "making-squares", - "2022-09-11": "logic-puzzles-1", - "2022-09-18": "iq-challenge-1", - "2022-09-25": "find-the-path", - "2022-10-02": "counting-squares", - "2022-10-09": "buffons-needle", - "2022-10-16": "balls-and-books", - "2022-10-23": "addition-squares", - "2022-10-30": "adding-to-15", - "2022-11-06": "9-dots", - "2022-11-13": "treasure-hunt", - "2022-11-20": "russian-multiplication", - "2022-11-27": "river-crossing", - "2022-12-04": "poisoned-wine-puzzle", - "2022-12-11": "matchstick-puzzles", - "2022-12-18": "magic-cards", - "2022-12-25": "langfords-problem", - "2023-01-01": "handshake-puzzles", - "2023-01-08": "geometry-puzzle-1", - "2023-01-15": "gabriels-problem", - "2023-01-22": "fence-around-a-field", - "2023-01-29": "domino-tilings", - "2023-02-05": "collatz-conjecture", - "2023-02-12": "coin-game", - "2023-02-19": "coconut-trader", - "2023-02-26": "circumference", - "2023-03-05": "apple-teaser", - "2023-03-12": "two-eggs", - "2023-03-19": "special-numbers", - "2023-03-26": "scales-and-weights", - "2023-04-02": "picture-puzzles-1", - "2023-04-09": "monkey-business", - "2023-04-16": "locks-and-keys", - "2023-04-23": "frog-party", - "2023-04-30": "counting-chickens", - "2023-05-07": "tic-tac-toe", - "2023-05-14": "tic-tac-toe-with-levels", - "2023-05-21": "tangrams", - "2023-05-28": "secret-santa", - "2023-06-04": "patience", - "2023-06-11": "paper-pieces", - "2023-06-18": "nim-related-games", - "2023-06-25": "nim-21", - "2023-07-02": "mastermind", - "2023-07-09": "make-many", - "2023-07-16": "letter-matching", - "2023-07-23": "countdown", - "2023-07-30": "15-game", - "2023-08-06": "pell-numbers", - "2023-08-13": "paths", - "2023-08-20": "number-challenge-1", - "2023-08-27": "making-squares", - "2023-09-03": "logic-puzzles-1", - "2023-09-10": "iq-challenge-1", - "2023-09-17": "find-the-path", - "2023-09-24": "counting-squares", - "2023-10-01": "buffons-needle", - "2023-10-08": "balls-and-books", - "2023-10-15": "addition-squares", - "2023-10-22": "adding-to-15", - "2023-10-29": "9-dots", - "2023-11-05": "treasure-hunt", - "2023-11-12": "russian-multiplication", - "2023-11-19": "river-crossing", - "2023-11-26": "poisoned-wine-puzzle", - "2023-12-03": "matchstick-puzzles", - "2023-12-10": "magic-cards", - "2023-12-17": "langfords-problem", - "2023-12-24": "handshake-puzzles", - "2023-12-31": "geometry-puzzle-1", - "2024-01-07": "gabriels-problem", - "2024-01-14": "fence-around-a-field", - "2024-01-21": "domino-tilings", - "2024-01-28": "collatz-conjecture", - "2024-02-04": "coin-game", - "2024-02-11": "coconut-trader", - "2024-02-18": "circumference", - "2024-02-25": "apple-teaser", - "2024-03-03": "two-eggs", - "2024-03-10": "special-numbers", - "2024-03-17": "scales-and-weights", - "2024-03-24": "picture-puzzles-1", - "2024-03-31": "monkey-business", - "2024-04-07": "locks-and-keys", - "2024-04-14": "frog-party", - "2024-04-21": "counting-chickens", - "2024-04-28": "tic-tac-toe", - "2024-05-05": "tic-tac-toe-with-levels", - "2024-05-12": "tangrams", - "2024-05-19": "secret-santa", - "2024-05-26": "patience", - "2024-06-02": "paper-pieces", - "2024-06-09": "nim-related-games", - "2024-06-16": "nim-21", - "2024-06-23": "mastermind" + "2021-08-22": "tetris-shapes", + "2021-08-29": "konigsberg-bridges", + "2021-09-05": "kaprekars-number", + "2021-09-12": "diffy-squares", + "2021-09-19": "pell-numbers", + "2021-09-26": "paths", + "2021-10-03": "number-challenge-1", + "2021-10-10": "making-squares", + "2021-10-17": "logic-puzzles-1", + "2021-10-24": "iq-challenge-1", + "2021-10-31": "find-the-path", + "2021-11-07": "counting-squares", + "2021-11-14": "buffons-needle", + "2021-11-21": "balls-and-books", + "2021-11-28": "addition-squares", + "2021-12-05": "adding-to-15", + "2021-12-12": "9-dots", + "2021-12-19": "treasure-hunt", + "2021-12-26": "russian-multiplication", + "2022-01-02": "river-crossing", + "2022-01-09": "poisoned-wine-puzzle", + "2022-01-16": "matchstick-puzzles", + "2022-01-23": "magic-cards", + "2022-01-30": "langfords-problem", + "2022-02-06": "handshake-puzzles", + "2022-02-13": "geometry-puzzle-1", + "2022-02-20": "gabriels-problem", + "2022-02-27": "fence-around-a-field", + "2022-03-06": "domino-tilings", + "2022-03-13": "collatz-conjecture", + "2022-03-20": "coin-game", + "2022-03-27": "coconut-trader", + "2022-04-03": "circumference", + "2022-04-10": "apple-teaser", + "2022-04-17": "two-eggs", + "2022-04-24": "special-numbers", + "2022-05-01": "scales-and-weights", + "2022-05-08": "picture-puzzles-1", + "2022-05-15": "monkey-business", + "2022-05-22": "locks-and-keys", + "2022-05-29": "frog-party", + "2022-06-05": "counting-chickens", + "2022-06-12": "tic-tac-toe", + "2022-06-19": "tic-tac-toe-with-levels", + "2022-06-26": "tangrams", + "2022-07-03": "secret-santa", + "2022-07-10": "patience", + "2022-07-17": "paper-pieces", + "2022-07-24": "nim-related-games", + "2022-07-31": "nim-21", + "2022-08-07": "mastermind", + "2022-08-14": "make-many", + "2022-08-21": "letter-matching", + "2022-08-28": "countdown", + "2022-09-04": "15-game", + "2022-09-11": "tetris-shapes", + "2022-09-18": "konigsberg-bridges", + "2022-09-25": "kaprekars-number", + "2022-10-02": "diffy-squares", + "2022-10-09": "pell-numbers", + "2022-10-16": "paths", + "2022-10-23": "number-challenge-1", + "2022-10-30": "making-squares", + "2022-11-06": "logic-puzzles-1", + "2022-11-13": "iq-challenge-1", + "2022-11-20": "find-the-path", + "2022-11-27": "counting-squares", + "2022-12-04": "buffons-needle", + "2022-12-11": "balls-and-books", + "2022-12-18": "addition-squares", + "2022-12-25": "adding-to-15", + "2023-01-01": "9-dots", + "2023-01-08": "treasure-hunt", + "2023-01-15": "russian-multiplication", + "2023-01-22": "river-crossing", + "2023-01-29": "poisoned-wine-puzzle", + "2023-02-05": "matchstick-puzzles", + "2023-02-12": "magic-cards", + "2023-02-19": "langfords-problem", + "2023-02-26": "handshake-puzzles", + "2023-03-05": "geometry-puzzle-1", + "2023-03-12": "gabriels-problem", + "2023-03-19": "fence-around-a-field", + "2023-03-26": "domino-tilings", + "2023-04-02": "collatz-conjecture", + "2023-04-09": "coin-game", + "2023-04-16": "coconut-trader", + "2023-04-23": "circumference", + "2023-04-30": "apple-teaser", + "2023-05-07": "two-eggs", + "2023-05-14": "special-numbers", + "2023-05-21": "scales-and-weights", + "2023-05-28": "picture-puzzles-1", + "2023-06-04": "monkey-business", + "2023-06-11": "locks-and-keys", + "2023-06-18": "frog-party", + "2023-06-25": "counting-chickens", + "2023-07-02": "tic-tac-toe", + "2023-07-09": "tic-tac-toe-with-levels", + "2023-07-16": "tangrams", + "2023-07-23": "secret-santa", + "2023-07-30": "patience", + "2023-08-06": "paper-pieces", + "2023-08-13": "nim-related-games", + "2023-08-20": "nim-21", + "2023-08-27": "mastermind", + "2023-09-03": "make-many", + "2023-09-10": "letter-matching", + "2023-09-17": "countdown", + "2023-09-24": "15-game", + "2023-10-01": "tetris-shapes", + "2023-10-08": "konigsberg-bridges", + "2023-10-15": "kaprekars-number", + "2023-10-22": "diffy-squares", + "2023-10-29": "pell-numbers", + "2023-11-05": "paths", + "2023-11-12": "number-challenge-1", + "2023-11-19": "making-squares", + "2023-11-26": "logic-puzzles-1", + "2023-12-03": "iq-challenge-1", + "2023-12-10": "find-the-path", + "2023-12-17": "counting-squares", + "2023-12-24": "buffons-needle", + "2023-12-31": "balls-and-books", + "2024-01-07": "addition-squares", + "2024-01-14": "adding-to-15", + "2024-01-21": "9-dots", + "2024-01-28": "treasure-hunt", + "2024-02-04": "russian-multiplication", + "2024-02-11": "river-crossing", + "2024-02-18": "poisoned-wine-puzzle", + "2024-02-25": "matchstick-puzzles", + "2024-03-03": "magic-cards", + "2024-03-10": "langfords-problem", + "2024-03-17": "handshake-puzzles", + "2024-03-24": "geometry-puzzle-1", + "2024-03-31": "gabriels-problem", + "2024-04-07": "fence-around-a-field", + "2024-04-14": "domino-tilings", + "2024-04-21": "collatz-conjecture", + "2024-04-28": "coin-game", + "2024-05-05": "coconut-trader", + "2024-05-12": "circumference", + "2024-05-19": "apple-teaser", + "2024-05-26": "two-eggs", + "2024-06-02": "special-numbers", + "2024-06-09": "scales-and-weights", + "2024-06-16": "picture-puzzles-1", + "2024-06-23": "monkey-business" } \ No newline at end of file diff --git a/maths-club-pack/content/facilitator/diffy-squares.md b/maths-club-pack/content/facilitator/diffy-squares.md new file mode 100644 index 00000000..10a1ee2d --- /dev/null +++ b/maths-club-pack/content/facilitator/diffy-squares.md @@ -0,0 +1,125 @@ +# Diffy Squares + +## Introduction + +This problem, at the simplest level, is a great way to practice subtraction and, for higher level students, can reveal some interesting mathematics. + +With regards to the question asked, it is quite difficult to find particularly long seqences and not at all obvious when they will occur. In the solutions section, you fill find an example of the longest sequence possible under certain assumptions for the starting numbers. However, allowing any starting numbers to be used, this challenge can be as difficult as you want it to be. + +It is technically possible to choose four numbers that make an infinite sequence where you never end up at `0, 0, 0, 0`. However, demonstrating this is tricky. [Here](https://mathforlove.com/2011/10/squares-of-difference-iii-a-surprising-solution/) is a link to a blog discussing the problem. + +## Solution + +Using just single-digit integers, the longest sequence length possible is 7. +Here is an example of such a sequence (using the column notation): + +``` +0, 1, 4, 9 +1, 3, 5, 9 +2, 2, 4, 8 +0, 2, 4, 6 +2, 2, 2, 6 +0, 0, 4, 4 +0, 4, 0, 4 +4, 4, 4, 4 +0, 0, 0, 0 +``` + +It is interesting to point out that the same four starting numbers will not always give the same length of sequence. Taking `0, 4, 1, 9` as the starting line for example: + +``` +0, 4, 1, 9 +4, 3, 8, 9 +1, 5, 1, 5 +4, 4, 4, 4 +0, 0, 0, 0 +``` + +We managed to make this sequence quite a bit shorter just by swapping the position of the $1$ and the $4$. You might like to think about how many different lengths of sequence it is possible to get from the same four starting numbers (see Extention section for solution). + +If we allow any integers less that 100 to be used, the longest sequence length possible is 12. +Here is an example of such a sequence: + +``` +0, 7, 20, 44 +7, 13, 24, 44 +6, 11, 20, 37 +5, 9, 17, 31 +4, 8, 14, 26 +4, 6, 12, 22 +2, 6, 10, 18 +4, 4, 8, 16 +0, 4, 8, 12 +4, 4, 4, 12 +0, 0, 8, 8 +0, 8, 0, 0 +8, 8, 8, 8 +0, 0, 0, 0 +``` + +Introducing fractions is an effective way to increase the length of sequences which we can obtain. Using fractions we can find sequences with numbers less that 10 which are longer than the longest sequence possible using integers up to 100. +Here is an example: + +``` +0, $\frac{17}{12}$, 4, $\frac{35}{4}$ + +$\frac{17}{12}$, $\frac{31}{12}$, $\frac{19}{4}$, $\frac{35}{4}$ + +$\frac{7}{6}$, $\frac{13}{6}$, 4, $\frac{22}{3}$ + +1, $\frac{11}{6}$, $\frac{10}{3}$, $\frac{37}{6}$ + +$\frac{5}{6}$, $\frac{3}{2}$, $\frac{17}{6}$, $\frac{31}{6}$ + +$\frac{2}{3}$, $\frac{4}{3}$, $\frac{7}{3}$, $\frac{13}{3}$ + +$\frac{1}{3}$, 1, 2, $\frac{11}{3}$ + +$\frac{1}{3}$, 1, $\frac{5}{3}$, 3 + +$\frac{2}{3}$, $\frac{2}{3}$, $\frac{4}{3}$, $\frac{8}{3}$ + +0, $\frac{2}{3}$, $\frac{4}{3}$, 2 + +$\frac{2}{3}$, $\frac{2}{3}$, $\frac{2}{3}$, 2 + +0, 0, $\frac{4}{3}$, $\frac{4}{3}$ + +0, $\frac{4}{3}$, 0, $\frac{4}{3}$ + +$\frac{4}{3}$, $\frac{4}{3}$, $\frac{4}{3}$, $\frac{4}{3}$ + +0, 0, 0, 0 +``` + +(In particular, 13 is the longest a sequence can be using numbers which are integer multiples of $\frac{1}{12}$ and less than 10, above is such an example) + +## Extension + +How many sequences of different length are possible from the same four starting numbers? + +If we take a general starting square, using the numbers `a, b, c, d`. + +Say that if we start with line `(a, b, c, d)`, this means that `a` is in the top-left corner, `b` is in the top-right, `c` the bottom-right and `d` the bottom-left. + +By rotating the square, we will not change the length of the sequence generated, and we can see therefore that the starting lines `(b, c, d, a)`, `(c, d, a, b)` and `(d, a, b, c)` will give sequences of the same length as that given by `(a, b, c, d)`. + + + +We could also reflect the square in any of its four lines of symmetry without changing the length of the resulting square. So the starting lines `(a, d, c, b)`, `(d, c, b, a)`, `(c, b, a, d)` and `(b, a, d, c)` will all give sequences of the same length as `(a, b, c, d)`. + + + +So for each starting line `(a, b, c, d)` there are 8 lines in total using the numbers `a`, `b`, `c` and `d` which give sequences of the same length. + +But how many starting lines are possible in total from the numbers `a`, `b`, `c` and `d`? + +If we choose the number in the top-left corner first, we have four choices `a`, `b`, `c` or `d`. Next, if we choose the number in the top-right we have a choice of the three remaining number. Likewise, for the bottom-right, we have a choice from two and the bottom-left must be the one remaining number. + +So, the total number of possible lines is `4 x 3 x 2 x 1 = 24` + +But since we know that these lines come in groups of 8, all of which have the same length. The largest number of different lengths possible must be $\frac{24}{3}$ which is 3. + +Another way to see this would be to fix the position of `a` to the top-left, which we can do since rotation does not change the length of the sequence as we have seen. We can then ask, of the remaining numbers `b`, `c` and `d`, how many different ways can we choose the two numbers which `a` will share an edge with? + +The answer is 3. Either `b` and `c`, `b` and `d` or `c` and `d`. diff --git a/maths-club-pack/content/facilitator/kaprekars-number.md b/maths-club-pack/content/facilitator/kaprekars-number.md new file mode 100644 index 00000000..29820dd7 --- /dev/null +++ b/maths-club-pack/content/facilitator/kaprekars-number.md @@ -0,0 +1,54 @@ +# Kaprekar's Number + +## Introduction + +This problem is a good way to practice subtraction, it also demonstrates an interesting result. + +After trying a few examples, it becomes quite apparent that all four-digit numbers seem to lead to the same number under repeated application of this opperation, namely 6174 or *Kaprekar's Number*. + +Students will notice a similar pattern for three-digit numbers, with the resulting number being 495. They may also be able to demonstrate that this will happen for any three-digit number (see Extension section), although this is a difficult task for the four-digit problem. + +## Solution + +Starting with 4252, as in the question, and applying the operation defined in the question (known as the *Kaprekar Mapping*) the process would be as follows: + +$5422 - 2245 = 3177$ +$7731 - 1377 = 6354$ +$6543 - 3456 = 3087$ +$8730 - 0378 = 8352$ +$8532 - 2358 = 6174$ +$7641 - 1467 = 6174$ +... + +So we can see that the sequence eventually reaches 6174 at which point, if we continued, it would repeat 6174 indefinately. + +## Extension + +For three-digit numbers, we can demonstrate that repeated application of the Kaprekar mapping will always lead to 495. + +Consider a general three-digit number $*abc* = 100a + 10b + c$ +where $9 \geq a \geq b \geq c \geq 0$ + +Applying the Kaprekar mapping gives the second number in our sequence: + +$(100a + 10b + c) - (100c + 10b + a) = 100(a - c) + (c - a)$ + +We can simplify this to: $99(a - c)$ + +So, after one application of the Kaprekar mapping, the only possible results are the multiples of 99: + +099, 198, 297, 396, 495, 594, 693, 792, 891, 990 + +In fact, each of these numbers uses the same digits as exactly one other, and will give the same result under the Kaprekar mapping. So, we only need to consider how half of them will be changed by a second application. + +$990 - 099 = 891$ +$981 - 189 = 792$ +$972 - 279 = 693$ +$963 - 369 = 594$ +$954 - 459 = 495$ + +We could express this information as a flow diagram: + + + +After one application of the Kaprekar mapping, the sequence will be one of the numbers on the flow diagram and then will follow it to reach 495. In total, any number will reach 495 in a maximum of 5 applications of the Kaprekar mapping. For example, 545 will require 5 applications. diff --git a/maths-club-pack/content/facilitator/konigsberg-bridges.md b/maths-club-pack/content/facilitator/konigsberg-bridges.md new file mode 100644 index 00000000..3cc8a384 --- /dev/null +++ b/maths-club-pack/content/facilitator/konigsberg-bridges.md @@ -0,0 +1,41 @@ +# Bridges of Konigsberg + +## Introduction + +This is a classic mathematics problem. Its resolution in 1736 by Leonhard Euler is considered a foundation of the mathematical field of graph theory. + +Without knowing the problem, it is initially quite enjoyable to try any find a solution. However, there is in fact no way to cross every bridge once, as Euler showed (see Solution section). + +You could say that the ‘solution’ to this problem is to wait for the second world war, where two of the bridges were destroyed, making it possible to find a route crossing each remaining bridge once. + +## Solution + +As mentioned, there is no solution to the problem as it is outlined in the question. To demonstrate this, it is helpful to reduce the situation to a mathematical structure called a *graph* (a different type of graph to what you are probably used to). + + + +We have replaced the sections of land with circles and the bridges with connecting lines. We usually refer to these circles as *vertices*, and the lines as *edges*. + +Say we start our walk on the middle vertex and walk to the top one, and then to the vertex on the right. Our path might look like this: + + + +Notice how, since we have walked both to and from the top vertex, we have used two of the edges connected to it. Similarly, once we leave the rightmost vertex, we will use another edge and so will have removed two of its connected edges: + + + +We can see that, as we walk, we always use pairs of edges connected to a particular vertex. The only time that this is not true is for the vertices that we begin and end on. + +Because of this, after walking we will always have used an even number of edges connected to a given vertex (unless they are one that we start or end on). So, if we want to remove all the edges, the vertices in the graph must have an even number of edges connected to them. The only two vertices that can have an odd number of edges are the starting and ending ones. + +This was Euler's discovery, that to be able to pass through all the edges in a graph exactly once, there can be no more than two vertices which have an odd number of edges connected to them. + +In the case of Konigsberg, three of the vertices have 3 edges and one (the central island) has 5. Since all four of the vertices have an odd number of edges, it is impossible to find a path which passes over each edge exactly once. + +## Extension + +Test this result by creating other graphs and trying to find a route passing over each edge once, so-called *Eulerian paths*. See if you can predict whether a given graph will have a Eulerian path. + +Where a Eulerian path is possible, do you notice anything else about the number of vertices which can have an odd number of edges? Try finding a path for a graph which only has one vertex with an odd number of edges, if you cannot, can you see why? + +*Hint: we said that only the starting and ending vertices can have an odd number of connected edges, what happens in the two possible cases where these are different vertices and where these are the same vertex?* diff --git a/maths-club-pack/content/facilitator/tetris-shapes.md b/maths-club-pack/content/facilitator/tetris-shapes.md new file mode 100644 index 00000000..cf471038 --- /dev/null +++ b/maths-club-pack/content/facilitator/tetris-shapes.md @@ -0,0 +1,50 @@ +# Tetris Shapes + +## Introduction + +This creative puzzle is a good exercise in spatial visualtisation and deduction. For most of the +shapes, there is a clear starting point where a certain space can only be filled by one of the +blocks, and a solution can then be logically worked through. + +It should be noted that the choice of blocks is arbitrary and only done in reference to the +game ‘Tetris’. The essence of this puzzle can be maintained using blocks of smaller or larger +size, depending on the age and ability of students. + +## Solution + +The shape which cannot be made by fitting together the blocks given is the Star. It may be +interesting for students to try and reason this logically (see Extension). + +The other four solutions, for the Tree, Boat, Spaceship, and Flower, are as follows: + + + +Note that there may be other solutions not given here, any which uses each block once is equally valid. + +## Extension + +In the case of the star, we can reason as to why is cannot be created from the blocks given: + + + +Firstly, looking at the bottom two points, there is no way to fill this space other than to use +the **S** and **Z** blocks. + +If we fill these bottom points, we can consider the remaining three, which have a similar +three-square line structure. We cannot fit the **O** or **T** blocks in these spaces, so we need to +use the **I**, **J** and **L** blocks to fill these three points. + +We could attempt to do this in a few ways: + +Firstly, we could fill the top point with the **I** block and use the **J** and **L** blocks in the two side points (left on the diagram). At this point, we will not be able to fit the **O** block in the remaining space, so this way will not work. + +If we instead fill the top point with either the **J** or **L** block and use the remaining one to fill the side points, along with the **I** block, we will get a situation like on the right of the +diagram. Notice that, had we used the **L** block at the top, rather that the **J**, this would have +given a situation which is a reflection of the case shown in the diagram. In either case, there +is only one space where we can fit the **O** block and after this the remaining space will not fit +the **T** block, so this way will also not work. + +Therefore, we have exhausted all possibilities and can conclude that the star shape is +impossible to make from the blocks given. + + diff --git a/maths-club-pack/content/student/diffy-squares.md b/maths-club-pack/content/student/diffy-squares.md new file mode 100644 index 00000000..b2ba9289 --- /dev/null +++ b/maths-club-pack/content/student/diffy-squares.md @@ -0,0 +1,37 @@ +--- +title: Diffy Squares +type: problem +printOrder: 999 +added: 2021-01 +--- + +# Diffy Squares + +Create a Diffy Square by following this process: + +1. Draw a square and choose four numbers to place on each vertex. + +2. Along each edge, write the positive difference of the numbers at the vertices that the edge connects. Join these four numbers up to form a new square. + +3. Now repeat this process with the new square. Find the positive difference between each pair of numbers which share an edge, write these differences on the edge and connect up the new numbers. + +4. At some point, you will reach a square with four zeros on the edges. At this point, you should stop. + +5. That's it. You should now count the number of squares which you drew, not including the largest square on the outside. This is the *length* of the sequence of squares. + +Here is an example, starting with the numbers 7, 15, 9 and 13: + + + +We drew two smaller squares, so this sequence has length 2. + +Try following this process with different starting numbers. + +What is the longest sequence of squares that you can find? + + +Tip: For longer sequences, it may be easier to write your sequences as a column of numbers, where each row represents a square. + +For our example of length 2, we could have written: + + diff --git a/maths-club-pack/content/student/kaprekars-number.md b/maths-club-pack/content/student/kaprekars-number.md new file mode 100644 index 00000000..e3c64ba8 --- /dev/null +++ b/maths-club-pack/content/student/kaprekars-number.md @@ -0,0 +1,26 @@ +--- +title: Kaprekar's Number +type: problem +printOrder: 999 +added: 2021-01 +--- + +# Kaprekar's Number + +Choose a four digit number where the digits are not all the same (so not 1111, 2222, ...). + +Rearrange the digits to get the largest and smallest numbers that these digits can make. + +Subtract the smallest number from the largest to get a new number, and carry on repeating this operation for each new number. + +For example, starting with 4252: + +5422 - 2245 = 3177 +7731 - 1377 = 6354 +6543 - 3456 = 3087 +... + +What do you notice as you continue this process? +What happens when you use different starting numbers? + +Does something similar happen for three digit numbers? diff --git a/maths-club-pack/content/student/konigsberg-bridges.md b/maths-club-pack/content/student/konigsberg-bridges.md new file mode 100644 index 00000000..d50a41c5 --- /dev/null +++ b/maths-club-pack/content/student/konigsberg-bridges.md @@ -0,0 +1,20 @@ +--- +title: Konigsberg Bridges +type: problem +printOrder: 999 +added: 2021-01 +--- + +# Bridges of Konigsberg + +The Prussian city of Konigsberg (now Kaliningrad, Russia) is divided into four landmasses by the river Pregel. + + + +You can see that there are seven bridges connecting these sections of the city. + +Can you find a way to walk through the city crossing each bridge exactly once? + +Do you think that it is possible? + +Assume that there are no other ways to cross the river than by walking over one of these bridges. You may begin and end anywhere in the city. diff --git a/maths-club-pack/content/student/tetris-shapes.md b/maths-club-pack/content/student/tetris-shapes.md new file mode 100644 index 00000000..d806525e --- /dev/null +++ b/maths-club-pack/content/student/tetris-shapes.md @@ -0,0 +1,23 @@ +--- +title: Tetris Shapes +type: problem +printOrder: 999 +added: 2021-01 +--- + +# Tetris Shapes + +Using these seven blocks (from the arcade game *Tetris*), can you create the shapes at the bottom of this page? + + + +You are allowed to rotate the blocks but not to reflect them, so you cannot change the 'L-block' into a 'J-block' by flipping it over. + +Four of these five shapes can be created using exactly one of each block. +(left to right: Tree, Boat, Spaceship, Star, Flower) + + + +Can you spot which one is not possible? + +What other shapes can you make using these blocks? diff --git a/maths-club-pack/cover_images/diffy-squares.svg b/maths-club-pack/cover_images/diffy-squares.svg new file mode 100644 index 00000000..624a45ab --- /dev/null +++ b/maths-club-pack/cover_images/diffy-squares.svg @@ -0,0 +1,95 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/maths-club-pack/cover_images/kaprekars-number.svg b/maths-club-pack/cover_images/kaprekars-number.svg new file mode 100644 index 00000000..e54dc601 --- /dev/null +++ b/maths-club-pack/cover_images/kaprekars-number.svg @@ -0,0 +1 @@ + \ No newline at end of file diff --git a/maths-club-pack/cover_images/konigsberg-bridges.svg b/maths-club-pack/cover_images/konigsberg-bridges.svg new file mode 100644 index 00000000..02cf40a8 --- /dev/null +++ b/maths-club-pack/cover_images/konigsberg-bridges.svg @@ -0,0 +1,81 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/maths-club-pack/cover_images/tetris-shapes.svg b/maths-club-pack/cover_images/tetris-shapes.svg new file mode 100644 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a/maths-club-pack/images/tetris-shapes-4.png b/maths-club-pack/images/tetris-shapes-4.png new file mode 100644 index 00000000..fe428b14 Binary files /dev/null and b/maths-club-pack/images/tetris-shapes-4.png differ diff --git a/translations/en/facilitator/diffy-squares.md b/translations/en/facilitator/diffy-squares.md new file mode 100644 index 00000000..3125a658 --- /dev/null +++ b/translations/en/facilitator/diffy-squares.md @@ -0,0 +1,125 @@ +# Diffy Squares + +## Introduction + +This problem, at the simplest level, is a great way to practice subtraction and, for higher level students, can reveal some interesting mathematics. + +With regards to the question asked, it is quite difficult to find particularly long seqences and not at all obvious when they will occur. In the solutions section, you fill find an example of the longest sequence possible under certain assumptions for the starting numbers. However, allowing any starting numbers to be used, this challenge can be as difficult as you want it to be. + +It is technically possible to choose four numbers that make an infinite sequence where you never end up at `0, 0, 0, 0`. However, demonstrating this is tricky. [Here](https://mathforlove.com/2011/10/squares-of-difference-iii-a-surprising-solution/) is a link to a blog discussing the problem. + +## Solution + +Using just single-digit integers, the longest sequence length possible is 7. +Here is an example of such a sequence (using the column notation): + +``` +0, 1, 4, 9 +1, 3, 5, 9 +2, 2, 4, 8 +0, 2, 4, 6 +2, 2, 2, 6 +0, 0, 4, 4 +0, 4, 0, 4 +4, 4, 4, 4 +0, 0, 0, 0 +``` + +It is interesting to point out that the same four starting numbers will not always give the same length of sequence. Taking `0, 4, 1, 9` as the starting line for example: + +``` +0, 4, 1, 9 +4, 3, 8, 9 +1, 5, 1, 5 +4, 4, 4, 4 +0, 0, 0, 0 +``` + +We managed to make this sequence quite a bit shorter just by swapping the position of the $1$ and the $4$. You might like to think about how many different lengths of sequence it is possible to get from the same four starting numbers (see Extention section for solution). + +If we allow any integers less that 100 to be used, the longest sequence length possible is 12. +Here is an example of such a sequence: + +``` +0, 7, 20, 44 +7, 13, 24, 44 +6, 11, 20, 37 +5, 9, 17, 31 +4, 8, 14, 26 +4, 6, 12, 22 +2, 6, 10, 18 +4, 4, 8, 16 +0, 4, 8, 12 +4, 4, 4, 12 +0, 0, 8, 8 +0, 8, 0, 0 +8, 8, 8, 8 +0, 0, 0, 0 +``` + +Introducing fractions is an effective way to increase the length of sequences which we can obtain. Using fractions we can find sequences with numbers less that 10 which are longer than the longest sequence possible using integers up to 100. +Here is an example: + +``` +0, $\frac{17}{12}$, 4, $\frac{35}{4}$ + +$\frac{17}{12}$, $\frac{31}{12}$, $\frac{19}{4}$, $\frac{35}{4}$ + +$\frac{7}{6}$, $\frac{13}{6}$, 4, $\frac{22}{3}$ + +1, $\frac{11}{6}$, $\frac{10}{3}$, $\frac{37}{6}$ + +$\frac{5}{6}$, $\frac{3}{2}$, $\frac{17}{6}$, $\frac{31}{6}$ + +$\frac{2}{3}$, $\frac{4}{3}$, $\frac{7}{3}$, $\frac{13}{3}$ + +$\frac{1}{3}$, 1, 2, $\frac{11}{3}$ + +$\frac{1}{3}$, 1, $\frac{5}{3}$, 3 + +$\frac{2}{3}$, $\frac{2}{3}$, $\frac{4}{3}$, $\frac{8}{3}$ + +0, $\frac{2}{3}$, $\frac{4}{3}$, 2 + +$\frac{2}{3}$, $\frac{2}{3}$, $\frac{2}{3}$, 2 + +0, 0, $\frac{4}{3}$, $\frac{4}{3}$ + +0, $\frac{4}{3}$, 0, $\frac{4}{3}$ + +$\frac{4}{3}$, $\frac{4}{3}$, $\frac{4}{3}$, $\frac{4}{3}$ + +0, 0, 0, 0 +``` + +(In particular, 13 is the longest a sequence can be using numbers which are integer multiples of $\frac{1}{12}$ and less than 10, above is such an example) + +## Extension + +How many sequences of different length are possible from the same four starting numbers? + +If we take a general starting square, using the numbers `a, b, c, d`. + +Say that if we start with line `(a, b, c, d)`, this means that `a` is in the top-left corner, `b` is in the top-right, `c` the bottom-right and `d` the bottom-left. + +By rotating the square, we will not change the length of the sequence generated, and we can see therefore that the starting lines `(b, c, d, a)`, `(c, d, a, b)` and `(d, a, b, c)` will give sequences of the same length as that given by `(a, b, c, d)`. + + + +We could also reflect the square in any of its four lines of symmetry without changing the length of the resulting square. So the starting lines `(a, d, c, b)`, `(d, c, b, a)`, `(c, b, a, d)` and `(b, a, d, c)` will all give sequences of the same length as `(a, b, c, d)`. + + + +So for each starting line `(a, b, c, d)` there are 8 lines in total using the numbers `a`, `b`, `c` and `d` which give sequences of the same length. + +But how many starting lines are possible in total from the numbers `a`, `b`, `c` and `d`? + +If we choose the number in the top-left corner first, we have four choices `a`, `b`, `c` or `d`. Next, if we choose the number in the top-right we have a choice of the three remaining number. Likewise, for the bottom-right, we have a choice from two and the bottom-left must be the one remaining number. + +So, the total number of possible lines is `4 x 3 x 2 x 1 = 24` + +But since we know that these lines come in groups of 8, all of which have the same length. The largest number of different lengths possible must be $\frac{24}{3}$ which is 3. + +Another way to see this would be to fix the position of `a` to the top-left, which we can do since rotation does not change the length of the sequence as we have seen. We can then ask, of the remaining numbers `b`, `c` and `d`, how many different ways can we choose the two numbers which `a` will share an edge with? + +The answer is 3. Either `b` and `c`, `b` and `d` or `c` and `d`. diff --git a/translations/en/facilitator/kaprekars-number.md b/translations/en/facilitator/kaprekars-number.md new file mode 100644 index 00000000..fe1f99eb --- /dev/null +++ b/translations/en/facilitator/kaprekars-number.md @@ -0,0 +1,54 @@ +# Kaprekar's Number + +## Introduction + +This problem is a good way to practice subtraction, it also demonstrates an interesting result. + +After trying a few examples, it becomes quite apparent that all four-digit numbers seem to lead to the same number under repeated application of this opperation, namely 6174 or *Kaprekar's Number*. + +Students will notice a similar pattern for three-digit numbers, with the resulting number being 495. They may also be able to demonstrate that this will happen for any three-digit number (see Extension section), although this is a difficult task for the four-digit problem. + +## Solution + +Starting with 4252, as in the question, and applying the operation defined in the question (known as the *Kaprekar Mapping*) the process would be as follows: + +$5422 - 2245 = 3177$ +$7731 - 1377 = 6354$ +$6543 - 3456 = 3087$ +$8730 - 0378 = 8352$ +$8532 - 2358 = 6174$ +$7641 - 1467 = 6174$ +... + +So we can see that the sequence eventually reaches 6174 at which point, if we continued, it would repeat 6174 indefinately. + +## Extension + +For three-digit numbers, we can demonstrate that repeated application of the Kaprekar mapping will always lead to 495. + +Consider a general three-digit number $*abc* = 100a + 10b + c$ +where $9 \geq a \geq b \geq c \geq 0$ + +Applying the Kaprekar mapping gives the second number in our sequence: + +$(100a + 10b + c) - (100c + 10b + a) = 100(a - c) + (c - a)$ + +We can simplify this to: $99(a - c)$ + +So, after one application of the Kaprekar mapping, the only possible results are the multiples of 99: + +099, 198, 297, 396, 495, 594, 693, 792, 891, 990 + +In fact, each of these numbers uses the same digits as exactly one other, and will give the same result under the Kaprekar mapping. So, we only need to consider how half of them will be changed by a second application. + +$990 - 099 = 891$ +$981 - 189 = 792$ +$972 - 279 = 693$ +$963 - 369 = 594$ +$954 - 459 = 495$ + +We could express this information as a flow diagram: + + + +After one application of the Kaprekar mapping, the sequence will be one of the numbers on the flow diagram and then will follow it to reach 495. In total, any number will reach 495 in a maximum of 5 applications of the Kaprekar mapping. For example, 545 will require 5 applications. diff --git a/translations/en/facilitator/konigsberg-bridges.md b/translations/en/facilitator/konigsberg-bridges.md new file mode 100644 index 00000000..2fe823ad --- /dev/null +++ b/translations/en/facilitator/konigsberg-bridges.md @@ -0,0 +1,41 @@ +# Bridges of Konigsberg + +## Introduction + +This is a classic mathematics problem. Its resolution in 1736 by Leonhard Euler is considered a foundation of the mathematical field of graph theory. + +Without knowing the problem, it is initially quite enjoyable to try any find a solution. However, there is in fact no way to cross every bridge once, as Euler showed (see Solution section). + +You could say that the ‘solution’ to this problem is to wait for the second world war, where two of the bridges were destroyed, making it possible to find a route crossing each remaining bridge once. + +## Solution + +As mentioned, there is no solution to the problem as it is outlined in the question. To demonstrate this, it is helpful to reduce the situation to a mathematical structure called a *graph* (a different type of graph to what you are probably used to). + + + +We have replaced the sections of land with circles and the bridges with connecting lines. We usually refer to these circles as *vertices*, and the lines as *edges*. + +Say we start our walk on the middle vertex and walk to the top one, and then to the vertex on the right. Our path might look like this: + + + +Notice how, since we have walked both to and from the top vertex, we have used two of the edges connected to it. Similarly, once we leave the rightmost vertex, we will use another edge and so will have removed two of its connected edges: + + + +We can see that, as we walk, we always use pairs of edges connected to a particular vertex. The only time that this is not true is for the vertices that we begin and end on. + +Because of this, after walking we will always have used an even number of edges connected to a given vertex (unless they are one that we start or end on). So, if we want to remove all the edges, the vertices in the graph must have an even number of edges connected to them. The only two vertices that can have an odd number of edges are the starting and ending ones. + +This was Euler's discovery, that to be able to pass through all the edges in a graph exactly once, there can be no more than two vertices which have an odd number of edges connected to them. + +In the case of Konigsberg, three of the vertices have 3 edges and one (the central island) has 5. Since all four of the vertices have an odd number of edges, it is impossible to find a path which passes over each edge exactly once. + +## Extension + +Test this result by creating other graphs and trying to find a route passing over each edge once, so-called *Eulerian paths*. See if you can predict whether a given graph will have a Eulerian path. + +Where a Eulerian path is possible, do you notice anything else about the number of vertices which can have an odd number of edges? Try finding a path for a graph which only has one vertex with an odd number of edges, if you cannot, can you see why? + +*Hint: we said that only the starting and ending vertices can have an odd number of connected edges, what happens in the two possible cases where these are different vertices and where these are the same vertex?* diff --git a/translations/en/facilitator/tetris-shapes.md b/translations/en/facilitator/tetris-shapes.md new file mode 100644 index 00000000..ac187110 --- /dev/null +++ b/translations/en/facilitator/tetris-shapes.md @@ -0,0 +1,50 @@ +# Tetris Shapes + +## Introduction + +This creative puzzle is a good exercise in spatial visualtisation and deduction. For most of the +shapes, there is a clear starting point where a certain space can only be filled by one of the +blocks, and a solution can then be logically worked through. + +It should be noted that the choice of blocks is arbitrary and only done in reference to the +game ‘Tetris’. The essence of this puzzle can be maintained using blocks of smaller or larger +size, depending on the age and ability of students. + +## Solution + +The shape which cannot be made by fitting together the blocks given is the Star. It may be +interesting for students to try and reason this logically (see Extension). + +The other four solutions, for the Tree, Boat, Spaceship, and Flower, are as follows: + + + +Note that there may be other solutions not given here, any which uses each block once is equally valid. + +## Extension + +In the case of the star, we can reason as to why is cannot be created from the blocks given: + + + +Firstly, looking at the bottom two points, there is no way to fill this space other than to use +the **S** and **Z** blocks. + +If we fill these bottom points, we can consider the remaining three, which have a similar +three-square line structure. We cannot fit the **O** or **T** blocks in these spaces, so we need to +use the **I**, **J** and **L** blocks to fill these three points. + +We could attempt to do this in a few ways: + +Firstly, we could fill the top point with the **I** block and use the **J** and **L** blocks in the two side points (left on the diagram). At this point, we will not be able to fit the **O** block in the remaining space, so this way will not work. + +If we instead fill the top point with either the **J** or **L** block and use the remaining one to fill the side points, along with the **I** block, we will get a situation like on the right of the +diagram. Notice that, had we used the **L** block at the top, rather that the **J**, this would have +given a situation which is a reflection of the case shown in the diagram. In either case, there +is only one space where we can fit the **O** block and after this the remaining space will not fit +the **T** block, so this way will also not work. + +Therefore, we have exhausted all possibilities and can conclude that the star shape is +impossible to make from the blocks given. + + diff --git a/translations/en/student/diffy-squares.md b/translations/en/student/diffy-squares.md new file mode 100644 index 00000000..db4be8cb --- /dev/null +++ b/translations/en/student/diffy-squares.md @@ -0,0 +1,30 @@ +# Diffy Squares + +Create a Diffy Square by following this process: + +1. Draw a square and choose four numbers to place on each vertex. + +2. Along each edge, write the positive difference of the numbers at the vertices that the edge connects. Join these four numbers up to form a new square. + +3. Now repeat this process with the new square. Find the positive difference between each pair of numbers which share an edge, write these differences on the edge and connect up the new numbers. + +4. At some point, you will reach a square with four zeros on the edges. At this point, you should stop. + +5. That's it. You should now count the number of squares which you drew, not including the largest square on the outside. This is the *length* of the sequence of squares. + +Here is an example, starting with the numbers 7, 15, 9 and 13: + + + +We drew two smaller squares, so this sequence has length 2. + +Try following this process with different starting numbers. + +What is the longest sequence of squares that you can find? + + +Tip: For longer sequences, it may be easier to write your sequences as a column of numbers, where each row represents a square. + +For our example of length 2, we could have written: + + diff --git a/translations/en/student/kaprekars-number.md b/translations/en/student/kaprekars-number.md new file mode 100644 index 00000000..0b7c675b --- /dev/null +++ b/translations/en/student/kaprekars-number.md @@ -0,0 +1,19 @@ +# Kaprekar's Number + +Choose a four digit number where the digits are not all the same (so not 1111, 2222, ...). + +Rearrange the digits to get the largest and smallest numbers that these digits can make. + +Subtract the smallest number from the largest to get a new number, and carry on repeating this operation for each new number. + +For example, starting with 4252: + +5422 - 2245 = 3177 +7731 - 1377 = 6354 +6543 - 3456 = 3087 +... + +What do you notice as you continue this process? +What happens when you use different starting numbers? + +Does something similar happen for three digit numbers? diff --git a/translations/en/student/konigsberg-bridges.md b/translations/en/student/konigsberg-bridges.md new file mode 100644 index 00000000..bb2b89f9 --- /dev/null +++ b/translations/en/student/konigsberg-bridges.md @@ -0,0 +1,13 @@ +# Bridges of Konigsberg + +The Prussian city of Konigsberg (now Kaliningrad, Russia) is divided into four landmasses by the river Pregel. + + + +You can see that there are seven bridges connecting these sections of the city. + +Can you find a way to walk through the city crossing each bridge exactly once? + +Do you think that it is possible? + +Assume that there are no other ways to cross the river than by walking over one of these bridges. You may begin and end anywhere in the city. diff --git a/translations/en/student/tetris-shapes.md b/translations/en/student/tetris-shapes.md new file mode 100644 index 00000000..ea68d932 --- /dev/null +++ b/translations/en/student/tetris-shapes.md @@ -0,0 +1,16 @@ +# Tetris Shapes + +Using these seven blocks (from the arcade game *Tetris*), can you create the shapes at the bottom of this page? + + + +You are allowed to rotate the blocks but not to reflect them, so you cannot change the 'L-block' into a 'J-block' by flipping it over. + +Four of these five shapes can be created using exactly one of each block. +(left to right: Tree, Boat, Spaceship, Star, Flower) + + + +Can you spot which one is not possible? + +What other shapes can you make using these blocks?