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JMO mentoring scheme answers November 2012 paper 1 Ans : 24 He also received an allowance of wood for heating, beer (and possibly wine) from the Prince's cellars and other perks. In his last job in Leipzig, Bach was always very insistent that he got paid the proper rate for the job! 2 Ans : February 9th 1649 This can be a great source of confusion to historians comparing continental and British documents in this period. 3 Ans : Any value of x with y = 0 or x = 0 with any value of y. The equation multiplies out to x2 + 2xy + y2 = x2 + y2 so 2xy = 0 4 Ans : 8(1 + Ö 3) cm2 If we join the centres of the coin, we obtain a right-angled triangle with hypotenuse 2 cm and another side of 1 cm. The third side is Ö3 cm using Pythagoras’s rule. The length of the box is 2(1 + Ö3) cm and the width is 4 cm. 5 The key to this answer is that the property of divisibility of a number by 3 (or by 9) implies and is implied by the property that the sum of the digits of the number is divisible by 3 (or by 9). [We hope you know this!] Since 3N is divisible by 3, it follows that the sum of the digits of 3N is divible by 3. Since it was given that the sum of the digits did not alter when N was multiplied by 3, it follows that N must be divisible by 3. We repeat this argument because we now know that 3N is divisible by 9. Again we can conclude that the sum of the digits of 3N is divisible by 9 and so the sum of the digits of N is divisible by 9. 6 This proof depends on the fact that if you join mid-points of two of the sides of a triangle, the resulting line is parallel to the third side. This can be proved by similar triangles. The converse is also true, namely that a line through a mid-point of a side of a triangle and parallel to one of the other sides passes through the mid-point of the third side. Let E, F, G, H be midpoints of AB, BD, AC and CD. Since DAEG and DABC are similar, EG is parallel to BC. For similar reasons EF is parallel to AD. Since EFG is one line, BC is parallel to AD. By using the converse, we can see that GH is parallel to AD and hence parallel to BC. Thus EFGH is one line. 7 Ans : 62 There are … pairs of numbers which add to 125, namely (25, 100), (26, 99), …, (61, 64), (62, 63), that is 38 pairs. For Jack to achieve his objective, he must not write down both numbers from any of these pairs, i.e. he must exclude at least 38 numbers. 8 Ans : 198 100 cards are removed on the first step leaving 9900. The largest square not exceeding 9900 is 9801 = 992. Hence 99 cards can be removed on the second step leaving exactly 9801. Since this is a perfect square we can continue the process 2 steps at a time where on every second step N is a perfect square. We need 99 pairs of steps to get from N from 1002 down to 12. Proof : Suppose we start with n2 cards. After the first step we have n2 − n cards. This is greater than (n − 1)2 = n2 − 2n + 1 so we can remove n − 1 cards on the next step. After two steps we have n2 − n − (n − 1) which is exactly (n − 1)2 cards.