Merge pull request #23 from IDEMSInternational/v2-release

V3 release

src/app/components/cards.component.scss

@@ -189,6 +189,9 @@ body {
 .House_image_1 {
   width: 30%;
 }
+.House_image_3 {
+  width: 30%
+}
 .Colouring_image_0 {
   width: 100%;
 }

src/assets/card-content/en/cards/10C.json

@@ -7,7 +7,7 @@
     "statement": "<p>How many ways are there to draw four dots on a piece of paper such that whichever two dots you choose, the distance between these two points is one of only two possible values?</p>\n<p>Here is one correct example:</p>\n<p><img alt=\"\" src=\"images/Two_Distances_image_0.png\" class=\"Two_Distances_image_0\" /></p>",
     "correct_answer": "6",
     "hint": "<p>Try drawing an equilateral triangle. That will give you three points that are all the same distance from each other. Where could you put the fourth point?</p>",
-    "explanation": "<p>If we call the two distances \"short\" and \"long\", then any such drawing of points must have a triangle with one long and two short edges (it's not quite obvious, but if you think about it a bit you can eliminate all other possibilities). Now where does the last point go? To each of the other 3 points, it's connected either via a long or a short edge, giving 8 possibilities. Try out each of them, you will notice that 2 of them do not work out, leaving 6 solutions.</p>\n<p><img alt=\"\" src=\"assets/images/Two_Distances_image_1.png\" class=\"Two_Distances_image_1\" /></p>\n<p>Short edges are represented by <strong>blue</strong> while long edges are represented by <strong>green</strong>. One of these solutions can be extended to 5 points that only use two distances, see bottom right picture.</p>"
+    "explanation": "<p>If we call the two distances \"short\" and \"long\", then any such drawing of points must have a triangle with one long and two short edges (it's not quite obvious, but if you think about it a bit you can eliminate all other possibilities). Now where does the last point go? To each of the other 3 points, it's connected either via a long or a short edge, giving 8 possibilities. Try out each of them, you will notice that 2 of them do not work out, leaving 6 solutions.</p>\n<p><img alt=\"\" src=\"assets/images/Two_Distances_image_1.png\" class=\"Two_Distances_image_1\" /></p>\n<p>Short edges are represented by <strong>blue</strong> while long edges are represented by <strong>green</strong>. One of these solutions can be extended to 5 points that only use two distances, see the pictures above.</p>"
   },
   "extension_1": {
     "statement": "<p>In how many different ways can you draw 4 points in the plane whose pairwise distances are \u221a2, \u221a2, 2, \u221a10, \u221a10 and 4?</p>",

src/assets/card-content/en/cards/4C.json

@@ -16,7 +16,7 @@
     "explanation": "<p>If you start with the point at the top, you will notice that after following all the lines, you end up at the start again. But this means you went in a loop! So you can start anywhere along this loop and simply follow the lines the same way you did before, and you will end up where you started. This means you can start from any of the vertices and visit all lines. </p>"
   },
   "extension_2": {
-    "statement": "<p>Remember that an Euler Path visits every edge once, and an Euler circuit does the same but starts and ends in the same point. How many of these graphs are Euler circuits?</p>",
+    "statement": "<p>Remember that an Euler Path visits every edge once, and an Euler circuit does the same but starts and ends in the same point. How many of these graphs are Euler circuits?</p>\n<p><img alt=\"\" src=\"assets/images/House_image_3.png\" class=\"House_image_3\" /></p>",
     "hint": "<p>Look to see how many nodes have an odd number of edges.</p>",
     "correct_answer": "4",
     "explanation":"<p>If there are no odd nodes or if there are two odd nodes, that means that the network is traversable. Graphs with only two odd nodes are in an Euler path and graphs with no odd nodes are in an Euler circuit. 2, 3, 7 and 9 have no odd nodes so they are the four graphs with Euler circuits.</p>"

src/assets/card-content/en/cards/5C.json

@@ -22,7 +22,7 @@
     "explanation": "<p>We can see this with an example with the number of hexagons in the hive. If there are 120 hexagons, then the worker bee makes 2 per minute and the queen bee makes 3 per minute. So together they make 5 per minute. 120 \u2215 5 = 24 . </p>\n<p>The same argument works for any number of hexagons in a hive. Take the total number of hexagons as h. The worker bee will be working at a rate of (h  \u2215 60) hexagons per minute, and the queen bee at a rate of (h  \u2215 40) per minute. By simply adding the rates together, </p>\n<p>(h  \u2215 60) + (h  \u2215 40) = 100\u00d7h  \u2215 2400 = h \u2215 24, we can find the answer, 24 minutes. </p>"
   },
   "additional_information": {
-    "about": "<p>Problems such as \"If A people can dig a trench in B<strong> </strong>days, then how long will it take C<strong> </strong>people to dig the trench?\" can be solved by calculating (A\u00d7B  \u2215 C). This ancient technique is called \"the rule of three\" and was written about in a <u>nursery rhyme</u> in 1570: <strong> \"</strong>Multiplication is vexation;</p>\n<p>Division is as bad; The Rule of Three doth puzzle me, And Practice drives me mad.\"</p>",
+    "about": "<p>Problems such as \"If A people can dig a trench in B<strong> </strong>days, then how long will it take C<strong> </strong>people to dig the trench?\" can be solved by calculating (A\u00d7B  \u2215 C). This ancient technique is called \"the rule of three\" and was written about in a <u>nursery rhyme</u> in 1570: <strong> \"</strong>Multiplication is vexation; Division is as bad; The Rule of Three doth puzzle me, And Practice drives me mad.\"</p>",
     "references": [
       "<p><a href=\"https://nrich.maths.org/5668\" target=”_blank”>https://nrich.maths.org/5668</a> </p>",
       "<p><a href=\"https://en.wikipedia.org/wiki/Cross-multiplication#Rule_of_Three\" target=”_blank”>https://en.wikipedia.org/wiki/Cross-multiplication#Rule_of_Three</a> </p>"

src/assets/card-content/en/cards/7D.json

@@ -6,8 +6,8 @@
   "main_version": {
     "statement": "<p>What is the biggest number of pieces of a circular cake you can obtain with 5 straight cuts? Cuts must be made on the top of the cake.</p>",
     "correct_answer": "16",
-    "hint": "<p>Start with a small number of cuts, e.g. 1, 2, 3. How can you add another cut to get the biggest number of pieces? Remember it doesn't say that the pieces have to be the same size! What happens if many cuts pass through the same point?</p>",
-    "explanation": "<p>With one cut you always get 2 pieces. With two cuts you can get 4 pieces. With three cuts, if all cuts pass through the same point you get 6 pieces. However, you can get 7 pieces if the third cut doesn't pass through the same point as the previous two. Use this idea to get the maximum number with four cuts. So to get the most pieces, make sure that no more than two cuts pass through the same point, otherwise you get less pieces. With four cuts you can get 11 pieces. Do the same again for five cuts, start with the four cuts that give 7 pieces and add a cut without passing through any meeting points (intersections) and you can get 16 pieces.</p>\n<p><img alt=\"\" src=\"assets/images/Pieces_of_Cake_image_0.png\" class=\"Pieces_of_Cake_image_0\" /></p>"
+    "hint": "<p>Start with a small number of cuts, e.g. 1, 2, 3. How can you add another cut to get the biggest number of pieces? Remember it doesn't say that the pieces have to be of the same size! What happens if many cuts pass through the same point?</p>",
+    "explanation": "<p>It is best to start with a simpler question and look for patterns. Start by looking at a square. There are two different ways we could slice the cake. </p>\n<p><img alt=\"\" src=\"images/Triangular_Slices_image_1.png\" class=\"Triangular_Slices_image_1\" /></p>\n<p>For a pentagon there are five ways:</p>\n<p><img alt=\"\" src=\"images/Triangular_Slices_image_2.png\" class=\"Triangular_Slices_image_2\" /></p>\n<p>For a hexagon there are 14 ways as you can see below:</p>\n<p><img alt=\"\" src=\"images/Triangular_Slices_image_3.png\" class=\"Triangular_Slices_image_3\" /></p>\n<p>In this diagram you can see the solution is found by thinking of 6 + 3 + 2 + 3 = 14, but you can arrive at the solution by using the solutions for smaller polygons: If we consider the side of the hexagon connecting points 1 and 2, it can form a triangle with any of the other 4 points. If we pick point 3 or 6, we are left with a pentagon, for which we know there are 5 ways of triangulating it. If we pick point 4 or 5, we are left with a triangle and a square, for which there are 1\u00d72 ways. Adding up the number of choices gives us 5 + 2 + 2 + 5 = 14. The equivalent combination for the pentagon is 2 + 1 + 2 = 5, where 1 is the result of the triangle (which has of course only one option) and 2 is the result for a square.</p>\n<p>This is a more useful technique for moving on to thinking of polygons with more vertices.</p>"
   },
   "extension_1": {
     "statement": "<p>What is the biggest number of pieces of cake you can obtain with 7 cuts?</p>",

src/assets/card-content/en/cards/8C.json

@@ -7,7 +7,7 @@
     "statement": "<p>Pegs numbered 1 to 52 are placed in a circle. Starting with number 2, alternate pegs are knocked down until only one is left.</p>\n<p>What is the number of the last peg to be knocked down?</p>",
     "correct_answer": "41",
     "hint": "<p>What happens if you only had 2 pegs? What about 3, 4, 5, 6, 7, or 8 pegs? Can you spot the pattern after considering these cases?</p>",
-    "explanation": "<p>Starting with only 2 pegs, we know that the last peg to be knocked down would be number 1. We can work out which would be the last peg if we increase the number of pegs. To start, we record our results and look for a pattern.</p>\n<p>2 - 1</p>\n<p>3 - 3</p>\n<p>4 - 1</p>\n<p>5 - 3</p>\n<p>6 - 5</p>\n<p>7 - 7</p>\n<p>8 - 1</p>\n<p>9 - 3</p>\n<p>And so on. Each power of two (1, 2, 4, 8, \u2026) \"resets\" the last peg standing to be number 1. You can see that the number of the last peg standing goes up in odd numbers after this. To work out the answer for 52, we know the last reset was the largest power of two. The largest power of two below 52 is 32, and 52 - 32 = 20. So it is the twentieth odd number that is calculated as 2\u00d720 + 1 = 41.  </p>"
+    "explanation": "<p>Starting with 2 pegs, we know that the last peg to be knocked down would be number 1. We can work out which would be the last peg if we increase the number of pegs. To start, we record our results and look for a pattern.</p>\n<p>2 - 1,</p>\n<p>3 - 3,</p>\n<p>4 - 1,</p>\n<p>5 - 3,</p>\n<p>6 - 5,</p>\n<p>7 - 7,</p>\n<p>8 - 1,</p>\n<p>9 - 3</p>\n<p>and so on. Each power of two (1, 2, 4, 8, \u2026) \"resets\" the last peg standing to be number 1. You can see that the number of the last peg standing goes up in odd numbers after this. To work out the answer for 52, we know the last reset was the largest power of two. The largest power of two below 52 is 32, and 52 - 32 = 20. So it is the twentieth odd number that is calculated as 2\u00d720 + 1 = 41.  </p>"
   },
   "extension_1": {
     "statement": "<p>Pegs numbered 1 to 2000 are placed in a circle. Starting with number 2, alternate pegs are knocked down until only one is left.</p>\n<p>What is the number of the last peg to be knocked down?</p>",

src/assets/card-content/en/cards/AC.json

@@ -7,7 +7,7 @@
     "statement": "<p>Pegs numbered 1 to 50 are placed in order in a line, with number 1 on the left.</p>\n<p>They are then knocked over, one at a time, following these two rules:</p>\n<p>(1.) Starting with the first standing peg on the left, alternate pegs are knocked down, until the end of the row is reached.</p>\n<p>(2.) Each time the end of the row is reached, repeat the previous rule.</p>\n<p>What is the number of the last peg to be knocked down?</p>",
     "correct_answer": "32",
     "hint": "<p>What happens if you only had 5 pegs? What about 10 or 20? Can you spot the pattern?</p>",
-    "explanation": "<p>Start: 1 2 3 4 5 6 7 8 9 ...</p>\n<p>First round:  \u2715 2 \u2715 4 \u2715 6 \u2715 8 \u2715  ... only even numbers left</p>\n<p>Second round: \u2715 4 \u2715 8 \u2715 12 ...  only multiples of 4 left</p>\n<p>Third round: \u2715 8 \u2715 16 \u2715 24  ...  only multiples of 8 left</p>\n<p>Fourth round: Only multiples of 16 will be left</p>\n<p>Fifth round: Only multiples of 32 will be left.</p>\n<p>32 is the only multiple of 32 under 50, so 32 is the last peg left.</p>"
+    "explanation": "<p>Start: 1 2 3 4 5 6 7 8 9 ...</p>\n<p>First round:  \u2715 2 \u2715 4 \u2715 6 \u2715 8 \u2715  ... only even numbers left</p>\n<p>Second round: \u2715 4 \u2715 8 \u2715 12 ...  only multiples of 4 left</p>\n<p>Third round: \u2715 8 \u2715 16 \u2715 24  ...  only multiples of 8 left</p>\n<p>Fourth round: \u2715 16 \u2715 32 ... only multiples of 16 will be left</p>\n<p>Fifth round: \u2715 32 ... only multiples of 32 will be left.</p>\n<p>32 is the only multiple of 32 under 50, so 32 is the last peg left.</p>"
   },
   "extension_1": {
     "statement": "<p>Pegs numbered 1 to 1050 are placed in order in a line, with number 1 on the left.</p>\n<p>They are then knocked over, one at a time, following these two rules:</p>\n<p>(1.) Starting with the first standing peg on the left, alternate pegs are knocked down, until the end of the row is reached.</p>\n<p>(2.) Each time the end of the row is reached, repeat the previous rule.</p>\n<p>What is the number of the last peg to be knocked down?</p>",

src/assets/card-content/en/cards/AH.json

@@ -6,7 +6,7 @@
   "main_version": {
     "statement": "<p>In a room with 23 people the probability that two people have the same birthday is more than 50%.</p>",
     "hint": "<p>The probability that two people have the same birthday is just 1 minus the probability of everyone's birthday being different.</p>\n<p>You should ask yourself the following questions:</p>\n<p>If we only have 2 people what is the probability their birthday is different? </p>\n<p>If we had 5 people, in how many ways can we select two to compare their birthdays?</p>\n<p>What do we get if we use the same approach for 23 people?</p>",
-    "explanation": "<p>The probability of 2 people not having the same birthday is 364 \u2215 365 because the second person must have a different birthday so there are only 364 days to choose from. The probability of 3 people not having the same birthday is 364*363 \u2215 (365)^2 because the third person can't share a birthday with either of them, so there are only 363 days to choose from. Therefore the probability of 23 people not having the same birthday would be 364*363*362*...*343\u2215(365)^22=0.4927. Hence, the probability of 2 or more people sharing the same birthday must be 1-0.4927=0.5073"
+    "explanation": "<p>The probability of any 2 people not having the same birthday is 364 \u2215 365: Given the birthday of the first person, there are 364 possible different days for the second person. The probability of 3 people not having the same birthday is 364\u00d7363 / 365\u00b2 because the third person can't share a birthday with either of the first two, so there are only 363 days to choose from. Therefore the probability of 23 people not having the same birthday would be 364\u00d7363\u00d7362\u00d7...\u00d7343 / 365\u00b2\u00b2 = 0.4927. Hence, the probability of 2 or more people sharing the same birthday must be 1\u22120.4927 = 0.5073, or 50.73%. (The fun fact is still true even if we consider leap years in our computation).</p>"
   },
   "extension_1": {
     "statement": "<p>If there are 50 people in a room, what is the chance that two people share a birthday?</p>",

src/assets/card-content/en/cards/QC.json

@@ -6,7 +6,7 @@
   "main_version": {
     "statement": "<p>A small number of cards has been lost from a complete pack of 52. I deal cards to players until all have the same number of cards and some cards are left. If I deal to four players, three cards remain. If I deal to three players, two cards remain and if I deal to five players, two cards remain. How many cards are there?</p>",
     "correct_answer": "47",
-    "hint": "<p>Think about remainders when dividing. If there are two left when divided amongst five people then the number must be two more than a multiple of 5, e.g. 7 or 32</p>",
+    "hint": "<p>Think about remainders when dividing. If there are two left when divided amongst five people, then the number must be two more than a multiple of 5, e.g. 7 or 32.</p>",
     "explanation": "<p>As only a small number of cards have been lost, let us start by listing the numbers that work that are close to 52.</p>\n<p>If there were no cards remaining when dealt between four people we would know that the number of cards was in the four times table, i.e \u2026, 36, 40, 44, 48. But there are three left over, so the number of cards must be in the sequence \u2026, 39, 43, 47, 51.</p>\n<p>For dealing among three people with two left over we get the sequence \u2026, 41,44,47,50</p>\n<p>For dealing among five people with two left over we get the sequence \u2026,32, 37,42,47</p>\n<p>The number that appears on all three lists is 47.</p>\n<p>We could also solve the puzzle starting with the last two clues. There is the same number of cards leftover when dealing to three or five people, so the answer must be two more than a multiple of 15, so 17, 32 or 47. The only one of these numbers that fits with the first clue is 47.</p>"
   },
   "extension_1": {