2013COMC加拿大数学公开赛真题答案免费下载

历年 Canadian Open Mathematics Challenge加拿大数学公开赛

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2013 COMC真题答案免费下载

共计2.5小时考试时间

此套试卷由三部分题目组成

4题简答题，每题4分

4题挑战题，每题6分

4题解答题，每题10分

共计12题，满分80分

不可使用任何计算器

完整版下载链接见文末

部分真题预览：

Part A Introductory Questions' Solutions:

A3)A point is inside an odd number of circles if it is in the outermost ring, the third ring, or the middle circle. The area of the middle circle is : The third ring is the area contained in the circle of radius 3 but not contained in the circle of radius 2. The area of the third ring is 3²π - 2²π = 5π. The outer ring is the area contained in the circle of radius 5 but not contained in the circle of radius 4. The area of the fifth ring is 5²π - 4²π = 9π. Thus, the total area is π + 5π + 9π = 15π; so m = 15.

Part B Challenging Questions' Solutions:

B3) Two solutions:

Solution 1: When the sum of all the digits on the scoreboard is 10, the sum of the scores must be 1 more than a multiple of 9. The highest possible sum of the scores is 29 + 28 = 57: The numbers less than 57 that are 1 more than a multiple of 9 are 1, 10, 19, 28, 37, 46, and 55. If the sum of the scores is 1, then the sum of the digits is 1, not 10. If the sum of the scores is 55, then the scores are 26 and 29 or 27 and 28, both of which have a digit sum of 19. Thus, we cannot have this happen more than 5 times.
We see that the scores (5, 5); (5, 14); (14, 14); (23, 14); (23, 23) each have a digit sum of 10, and can all be acheived in the same game. Thus, the maximum number of times is 5.

Part C Long-form Proof Problems' Solutions:

C3)

1. Notice that for (a) m and n have greatest common divisor equal to 1, therefore on each turn a player can always make a move of replacing the number k with its divisor l strictly less than k, as long as l > 1, or as long as l = 1 and 1 has not yet appeared on the board. Instead of dealing with the actual numbers we will deal with the number of prime factors they have. Then, the game becomes equivalent to the following. Two numbers m and n are written on the board. On each turn a player can select a number k greater than 0 and replace it with any positive integer less than k, or replace it with 0, as long as 0 is not already written on the board. A player who cannot make a move loses.
   It immediately follows that m = 0; n = 1 is a losing position. Therefore, m = 0; n ≥ 2 is a winning position (since a player replaces n with 1 and wins). Furthermore, m = 1; n ≥ 1 is a winning position (since a player replaces n with 0 and wins). Hence m = 2; n = 2 is a losing position; m = 2; n ≥ 3 is a winning position; m = 3; n = 3 is a losing position, m = 3; n ≥4 is a winning position. By induction it follows that for k ≥ 2, m = k; n = k is a losing position, while m = k; n ≥ k + 1 is a winning position.
   We are in the case of m = 40; n = 51 ≥ 41 in the \transformed" game, thus this is a winning position and Alphonse wins.
   (b) This case is different, since now m and n have more than one divisor in common. We will deal with the original game and not make any transformations. Note that m and n are both powers of 2, so throughout the whole game only powers of 2 can appear on the board.
   We first note that the player who first writes down a number less than or equal to 2 loses. This is because if they write down 1, then 2 has not yet been written; the opponent on the next turn replaces the other number with 2 wins. (Note that this move is legal since at the start m > 2; n > 2 so at the time that 1 is written, the other number on the board must be greater than 2). If they write down 2, then 1 has not yet been written; the opponent on the next turn replaces the other number with 1 and wins.
   Similarly, the player who first writes down a number less than or equal to 8 loses. This is because if they write down 4, the other player writes 8 - thus forcing the original player to write down a number less than or equal to 2 (note they cannot replace 8 with 4 since 4 has already appeared on the board). Similarly, if they write down 8, the other player writes down 4 and wins.
   By induction it follows that if m, n > 2^2k-1 then the player who first writes down a number less than or equal to 2^2k-1 loses for every positive integer k. Thus for the case m = 2⁴⁰; n = 2⁵³, the player to first write down a number less than or equal to 2³⁹ loses. Therefore on his first turn, Alphonse replaces 253 with 241 and wins - because on her turn, Beryl is faced with 240 and 2⁴¹ on the board and has to write down a number less than or equal to 2³⁹.

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