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SMC2012_web_solutions

So for three copies of a given tile to form an

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So, for three copies of a given tile to form an equilateral triangle, the number of triangles which comprise the tile must be one third of a square number. n n 2 Only the tiles made up of three equilateral triangles and twelve equilateral triangles satisfy this condition. However, it is still necessary to show that three copies of these tiles can indeed make equilateral triangles. The diagrams above show how they can do this. 9. C If Pierre is telling the truth then Qadr is not telling the truth. However, this means that Ratna is telling the truth, so this leads to a contradiction as Pierre stated that just one person is telling the truth. So Pierre is not telling the truth, which means that Qadr is telling the truth, but Ratna is not telling the truth. This in turn means that Sven is telling the truth, but Tanya is not. So only Qadr and Sven are telling the truth. 10. E It can be deduced that must consist of at least 224 digits since the largest 223-digit positive integer consists of 223 nines and has a digit sum of 2007. It is possible to find 224-digit positive integers which have a digit sum of 2012. N The largest of these is 99 999 ...999 995 and the smallest is 59 999 ...999 999. So and (223 zeros). N = 59 999 999 999 N + 1 = 60 000 000 000 11. D Let the radius of the circular piece of cardboard be . The diagram shows a sector of the circle which would make one hat, with the minor arc shown becoming the circumference of the base of the hat. The circumference of the circle is . Now . This shows that we can cut out 6 hats in this fashion and also shows that the area of cardboard unused in cutting out any 6 hats is less than the area of a single hat. Hence there is no possibility that more than 6 hats could be cut out. r 2 π r 6 r < 2 π r < 7 r r r 12. E Two different ways of expressing 5 are 1 + 4 and 4 + 1. In the following list these are denoted as {1, 4: two ways}. The list of all possible ways is {5: one way}, {2, 3: two ways}, {1, 4: two ways}, {1, 2, 2: three ways}, {1, 1, 3: three ways}, {1, 1, 1, 2: four ways}, {1, 1, 1, 1, 1: one way}. So in total there are 16 ways. { Different expressions of a positive integer in the above form are known as ‘partitions’. It may be shown that the number of distinct compositions of a positive integer is .} n 2 n - 1 13. B The table below shows the position of the face marked with paint when the base of the cube is on the 25 squares. Code: T - top, B - base; F - front; H - hidden (rear); L - left; R - right. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 T H B F T R B L T F B H T L B R R R R B L L L B F So the required sum is 3 + 7 + 11 + 15 + 20 + 24 = 80. 14. D Note that each student has a language in common with exactly four of the other five students. For instance, Jean-Pierre has a language in common with each of Ina, Karim, Lionel and Mary. Only Helga does not have a language in common with Jean-Pierre. So whichever two students are chosen, the probability that they have a language in common is 4/5.
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