New Crowdin updates

crowdin/cards/am-ET/abundant-numbers.json

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+{
+  "Abundant Numbers": "Abundant Numbers",
+  "The integer 12 is the first <strong>abundant number</strong>. Its proper divisors (factors not including 12 itself)  are 1, 2, 3, 4, and 6 which add up to 16 which is more than 12.": "The integer 12 is the first <strong>abundant number</strong>. Its proper divisors (factors not including 12 itself)  are 1, 2, 3, 4, and 6 which add up to 16 which is more than 12.",
+  "You could make a 10 by 10 grid of the numbers 1 to 100 and colour in the abundant numbers.": "You could make a 10 by 10 grid of the numbers 1 to 100 and colour in the abundant numbers.",
+  "The next abundant number is 20, because the proper divisors of 20 are 1,2,4,5, and 10 which add up to 22. Can you explain why prime numbers can never be abundant?": "The next abundant number is 20, because the proper divisors of 20 are 1,2,4,5, and 10 which add up to 22. Can you explain why prime numbers can never be abundant?",
+  "The first known classification of numbers as deficient, perfect or abundant was by Nicomachus in his Introductio Arithmetica (circa 100)": "The first known classification of numbers as deficient, perfect or abundant was by Nicomachus in his Introductio Arithmetica (circa 100)",
+  "Numbers whose proper divisors add up to less than the number are called deficient numbers.": "Numbers whose proper divisors add up to less than the number are called deficient numbers.",
+  "You could make a 10 by 10 grid of the numbers 1 to 100 and colour in the deficient numbers.": "You could make a 10 by 10 grid of the numbers 1 to 100 and colour in the deficient numbers.",
+  "An example of a deficient number is 15, because 1,3, and 5 add up to 9 which is less than 15.": "An example of a deficient number is 15, because 1,3, and 5 add up to 9 which is less than 15.",
+  "Take the proper divisors of 6, they are 1,2, and 3 which adds up perfectly to 6.": "Take the proper divisors of 6, they are 1,2, and 3 which adds up perfectly to 6.",
+  "6 is the smallest perfect number. Can you find the only other perfect number below 100?": "6 is the smallest perfect number. Can you find the only other perfect number below 100?"
+}

crowdin/cards/am-ET/alphanumerics.json

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+{
+  "Alphanumerics": "Alphanumerics",
+  "In the sum CAR + CAT = RARE each letter represents a different digit from 0 to 9. What is the value of the word RARE?": "In the sum CAR + CAT = RARE each letter represents a different digit from 0 to 9. What is the value of the word RARE?",
+  "What digit must the letter R represent? If you add two 3-digit numbers together, what is the largest sum you could make?": "What digit must the letter R represent? If you add two 3-digit numbers together, what is the largest sum you could make?",
+  "R must be equal to 1 since only the digit 1 can carry over from the hundreds column into the thousands column. So we now have CA1 + CAT = 1A1E.": "R must be equal to 1 since only the digit 1 can carry over from the hundreds column into the thousands column. So we now have CA1 + CAT = 1A1E.",
+  "If we look at the tens column we have A + A = 1 (or ends in 1). Since A + A is even, the only way for it to end in the digit 1 is for there to be a carry over from the ones column.": "If we look at the tens column we have A + A = 1 (or ends in 1). Since A + A is even, the only way for it to end in the digit 1 is for there to be a carry over from the ones column.",
+  "This gives us A = 5 or A = 0.": "This gives us A = 5 or A = 0.",
+  "Since we know there was a carry over from the ones column, T = 9 and hence E = 0. This means A = 5 since each letter represents a different digit. Therefore the word RARE = 1510. (Though not needed, it can be verified that C = 7). Finally we have 751 + 759 = 1510.": "Since we know there was a carry over from the ones column, T = 9 and hence E = 0. This means A = 5 since each letter represents a different digit. Therefore the word RARE = 1510. (Though not needed, it can be verified that C = 7). Finally we have 751 + 759 = 1510.",
+  "These types of puzzles are called 'Cryptarithms' or 'Verbal Arithmetic'. The most classic example was published in the July 1924 issue of Strand Magazine by English mathematician Henry Dudeney.": "These types of puzzles are called 'Cryptarithms' or 'Verbal Arithmetic'. The most classic example was published in the July 1924 issue of Strand Magazine by English mathematician Henry Dudeney.",
+  "SEND + MORE = MONEY": "SEND + MORE = MONEY",
+  "Given the potentially large number of possibilities, computer programs are often used to generate and solve these types of puzzles.": "Given the potentially large number of possibilities, computer programs are often used to generate and solve these types of puzzles.",
+  "In the sum TWO + TWO = FOUR each letter represents a different digit from 0 to 9. There are 7 different solutions to this sum. For these 7 solutions, what is the largest value of the word FOUR?": "In the sum TWO + TWO = FOUR each letter represents a different digit from 0 to 9. There are 7 different solutions to this sum. For these 7 solutions, what is the largest value of the word FOUR?",
+  "Once you figured out what digit F represents, you then want to consider what the largest possible value of O could be.": "Once you figured out what digit F represents, you then want to consider what the largest possible value of O could be.",
+  "F must be equal to 1 since only the digit 1 can carry over from the hundreds column.": "F must be equal to 1 since only the digit 1 can carry over from the hundreds column.",
+  "We now have TWO + TWO = 1OUR.": "We now have TWO + TWO = 1OUR.",
+  "As we are trying to find the largest possible value of FOUR, we next try O = 9. Which would give TW9 + TW9 = 19UR. Then the only possibility is T = 9 but this is not allowed since O = 9, making O = T.": "As we are trying to find the largest possible value of FOUR, we next try O = 9. Which would give TW9 + TW9 = 19UR. Then the only possibility is T = 9 but this is not allowed since O = 9, making O = T.",
+  "We next try O = 8. So TW8 + TW8 = 18UR. Which leads to R = 6 and T = 9.  And we then have 9W8 + 9W8 = 18U6.": "We next try O = 8. So TW8 + TW8 = 18UR. Which leads to R = 6 and T = 9.  And we then have 9W8 + 9W8 = 18U6.",
+  "We still have a choice of answers for W and U.": "We still have a choice of answers for W and U.",
+  "We require W to be less than 5 to avoid carrying over into the hundreds column. If we tried W = 4, this would lead to U = 9 but this is not possible as T = 9. So we try W = 3.": "We require W to be less than 5 to avoid carrying over into the hundreds column. If we tried W = 4, this would lead to U = 9 but this is not possible as T = 9. So we try W = 3.",
+  "We end up with  938 + 938 = 1876, and so the largest possible value of the word FOUR is 1876.": "We end up with  938 + 938 = 1876, and so the largest possible value of the word FOUR is 1876.",
+  "In the sum MATHS + GAMES = DECODE each letter represents a different digit from 0 to 9. What is the value of the word DECODE?": "In the sum MATHS + GAMES = DECODE each letter represents a different digit from 0 to 9. What is the value of the word DECODE?",
+  "There are many ways to solve this puzzle. First D = 1. To continue, we must use all the given information to eliminate possibilities. For example, we observe that E must be an even digit. S cannot be equal to 0 or 1 or 5. A lot of brute force is required to work through this problem, but with a table to organise all of the information, you should be able to whittle down the possible solutions until you have reached the correct answer. T = 0, D = 1, A = 2, S = 3, C = 4, H = 5, E = 6, M = 7, O = 8, G = 9.": "There are many ways to solve this puzzle. First D = 1. To continue, we must use all the given information to eliminate possibilities. For example, we observe that E must be an even digit. S cannot be equal to 0 or 1 or 5. A lot of brute force is required to work through this problem, but with a table to organise all of the information, you should be able to whittle down the possible solutions until you have reached the correct answer. T = 0, D = 1, A = 2, S = 3, C = 4, H = 5, E = 6, M = 7, O = 8, G = 9."
+}

crowdin/cards/am-ET/avoid-the-river.json

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+{
+  "Avoid the River": "Avoid the River",
+  "Start from home (H) and go to school (S). You can walk along the 4 by 4 grid by making steps only towards east or north. You cannot cross the river (blue line)! In how many ways can you reach your school?": "Start from home (H) and go to school (S). You can walk along the 4 by 4 grid by making steps only towards east or north. You cannot cross the river (blue line)! In how many ways can you reach your school?",
+  "In how many ways you can go from home to school in a 2 by 2 grid? What about in a 3 by 3 grid?": "In how many ways you can go from home to school in a 2 by 2 grid? What about in a 3 by 3 grid?",
+  "This is not an easy combinatorial problem. It is better to start with a simpler version and familiarise with the problem. It is indeed intuitive to find the result for 2 by 2 grid: there are indeed only 2 possible ways. As far as the 3 by 3 case is concerned, students might easily draw all 5 ways explicitly. However, when it comes to the 4 by 4 case, it is easy to lose the count. Therefore the problem aims at making students think how to tackle this counting in a more systematic way. In particular, students should think whether they can decompose their counting in a way they can employ results known for a lower number of people. They may notice for example that the result for a 3 by 3 grid is: 2 + 1 + 2 = 5, 2 is the result for a 2 by 2 grid. Moreover, the result for a 4 by 4 grid is 5 + 2 + 2 + 5 = 14, where 2 is the result for a 2 by 2 and 5 is the result for a 3 by 3 grid.": "This is not an easy combinatorial problem. It is better to start with a simpler version and familiarise with the problem. It is indeed intuitive to find the result for 2 by 2 grid: there are indeed only 2 possible ways. As far as the 3 by 3 case is concerned, students might easily draw all 5 ways explicitly. However, when it comes to the 4 by 4 case, it is easy to lose the count. Therefore the problem aims at making students think how to tackle this counting in a more systematic way. In particular, students should think whether they can decompose their counting in a way they can employ results known for a lower number of people. They may notice for example that the result for a 3 by 3 grid is: 2 + 1 + 2 = 5, 2 is the result for a 2 by 2 grid. Moreover, the result for a 4 by 4 grid is 5 + 2 + 2 + 5 = 14, where 2 is the result for a 2 by 2 and 5 is the result for a 3 by 3 grid.",
+  "Cₙ are called <em>Catalan numbers</em> and they are crucial in the world of combinatorics. There is an explicit formula for them, but finding it goes much beyond our scopes: the recursive definition we find out in this puzzle is more than enough for now!": "Cₙ are called <em>Catalan numbers</em> and they are crucial in the world of combinatorics. There is an explicit formula for them, but finding it goes much beyond our scopes: the recursive definition we find out in this puzzle is more than enough for now!",
+  "In how many ways you can go from home to school in a 5 by 5 grid? Try using previous results to compute this!": "In how many ways you can go from home to school in a 5 by 5 grid? Try using previous results to compute this!",
+  "Counting explicitly might be too long. Can you come up with a way to reduce the problem in simpler cases you have already counted? Focus on when a path last touches the river before reaching the school. Do you know anything about all possible paths that can reach that point? What about all possible paths after that point?": "Counting explicitly might be too long. Can you come up with a way to reduce the problem in simpler cases you have already counted? Focus on when a path last touches the river before reaching the school. Do you know anything about all possible paths that can reach that point? What about all possible paths after that point?",
+  "This part is aimed at finding a recursive way to solve the problem. It might not be easy. \nLet us number the points in which the paths might intersect the river from 1 to 6 starting from home to school. Let us focus our attention on the paths which last touch the river in point 6: these are all paths (so not useful); let us consider all paths that last touch the river in point 5. The bottom part of the problem is equivalent to counting all paths from home up to point 5 (such that they do not cross the river), these are exactly the one counted in a 4 by 4 grid previously: there are 14. For the top part, there is only one way to reach school from point 5. Therefore we have 14×1 = 14 paths that touch the river in point 5 for the last time before reaching school. Now let us consider the number of paths that last touches the river in point 4. The bottom part of the problem is equivalent to counting all paths from home up to point 4 (such that they do not cross the river), these are exactly the one counted in a 3 by 3 grid previously: there are 5. Starting from point 4 there is now only one way to reach school without touching the river again. Therefore we have a total of 5×1 = 5 paths.\nNow let us consider the number of paths that last touches the river in point 3. The bottom part of the problem is equivalent to counting all paths from home up to point 3 (such that they do not cross the river), these are exactly the one counted in a 2 by 2 grid previously: there are 2. Starting from point 3, you need to move one step east. From here, you can not go north (otherwise you touch the river), but can only go East again. From here you can go one step East or one step North, but not two steps North (otherwise you would touch the river). And so on. You might realise you are solving the problem for a 2 by 2 grid whose bottom left corner is the point one step East to point 3 and the top right corner one step south the point 6. Therefore you have 2 paths. \nIn total you have 2×2 = 4 paths which last touch the river in the point 3. You then continue by considering the number of paths that last touch the river in point 2. You realise that you split the problem in the one for a 1 by 1 grid and for a 3 by 3 grid (whose bottom left corner is the point at one step East to point 2 and the top right corner one step south the point 6). Therefore we have in total 1×5 paths. Finally the number of paths which last touch the river in point 1 (i.e. only touch the river in point 1) are the same as the paths for the problem with 4 by 4 grid (whose bottom left corner is the point at one step East to point 1 and the top right corner one step south the point 6). Therefore we have 14 paths.\nWe covered all possible paths which do not cross the river. In total they add up as: 14 + 1×5 + 2×2 + 5×1 + 14 = 42.": "This part is aimed at finding a recursive way to solve the problem. It might not be easy. \nLet us number the points in which the paths might intersect the river from 1 to 6 starting from home to school. Let us focus our attention on the paths which last touch the river in point 6: these are all paths (so not useful); let us consider all paths that last touch the river in point 5. The bottom part of the problem is equivalent to counting all paths from home up to point 5 (such that they do not cross the river), these are exactly the one counted in a 4 by 4 grid previously: there are 14. For the top part, there is only one way to reach school from point 5. Therefore we have 14×1 = 14 paths that touch the river in point 5 for the last time before reaching school. Now let us consider the number of paths that last touches the river in point 4. The bottom part of the problem is equivalent to counting all paths from home up to point 4 (such that they do not cross the river), these are exactly the one counted in a 3 by 3 grid previously: there are 5. Starting from point 4 there is now only one way to reach school without touching the river again. Therefore we have a total of 5×1 = 5 paths.\nNow let us consider the number of paths that last touches the river in point 3. The bottom part of the problem is equivalent to counting all paths from home up to point 3 (such that they do not cross the river), these are exactly the one counted in a 2 by 2 grid previously: there are 2. Starting from point 3, you need to move one step east. From here, you can not go north (otherwise you touch the river), but can only go East again. From here you can go one step East or one step North, but not two steps North (otherwise you would touch the river). And so on. You might realise you are solving the problem for a 2 by 2 grid whose bottom left corner is the point one step East to point 3 and the top right corner one step south the point 6. Therefore you have 2 paths. \nIn total you have 2×2 = 4 paths which last touch the river in the point 3. You then continue by considering the number of paths that last touch the river in point 2. You realise that you split the problem in the one for a 1 by 1 grid and for a 3 by 3 grid (whose bottom left corner is the point at one step East to point 2 and the top right corner one step south the point 6). Therefore we have in total 1×5 paths. Finally the number of paths which last touch the river in point 1 (i.e. only touch the river in point 1) are the same as the paths for the problem with 4 by 4 grid (whose bottom left corner is the point at one step East to point 1 and the top right corner one step south the point 6). Therefore we have 14 paths.\nWe covered all possible paths which do not cross the river. In total they add up as: 14 + 1×5 + 2×2 + 5×1 + 14 = 42.",
+  "You might be able now to find a recursion relation for the number Cₙ of east and north paths in an n by n grid which do not cross the diagonal. Generalising what you did before we have that:\nCₙ = Cₙ₋₁ + C₁×Cₙ₋₂ + C₂×Cₙ₋₃ + … + Cₙ₋₂×C₁ + Cₙ₋₁": "You might be able now to find a recursion relation for the number Cₙ of east and north paths in an n by n grid which do not cross the diagonal. Generalising what you did before we have that:\nCₙ = Cₙ₋₁ + C₁×Cₙ₋₂ + C₂×Cₙ₋₃ + … + Cₙ₋₂×C₁ + Cₙ₋₁",
+  "The iᵗʰ summand corresponds to the number of paths which last touch the diagonal at the point i, (where i is a number from 1 to n).": "The iᵗʰ summand corresponds to the number of paths which last touch the diagonal at the point i, (where i is a number from 1 to n).",
+  "What if the grid has the size of 7 by 7? Try to use previous results to calculate the number of ways you can go to school in this grid.": "What if the grid has the size of 7 by 7? Try to use previous results to calculate the number of ways you can go to school in this grid.",
+  "We already calculated C₄ = C₃ + C₁×C₂ + C₂×C₁ + C₃ = 5 + 1×2 +2×1 + 5 = 14": "We already calculated C₄ = C₃ + C₁×C₂ + C₂×C₁ + C₃ = 5 + 1×2 +2×1 + 5 = 14",
+  "and C₅ = 42": "and C₅ = 42",
+  "We can carry on this pattern to give": "We can carry on this pattern to give",
+  "C₆ = C₅ +C₁×C₄ + C₂×C₃ +…+ C₅= 132": "C₆ = C₅ +C₁×C₄ + C₂×C₃ +…+ C₅= 132",
+  "Likewise, we calculate C₇ = 429": "Likewise, we calculate C₇ = 429"
+}