2011COMC加拿大數(shù)學(xué)公開賽真題答案免費(fèi)下載

歷年 Canadian Open Mathematics Challenge加拿大數(shù)學(xué)公開賽

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2011 COMC真題答案免費(fèi)下載

共計(jì)2.5小時考試時間

此套試卷由三部分題目組成

4題簡答題，每題4分

4題挑戰(zhàn)題，每題6分

4題解答題，每題10分

共計(jì)12題，滿分80分

不可使用任何計(jì)算器

完整版下載鏈接見文末

部分真題預(yù)覽：

Part A Introductory Questions' Solutions:

A3)The answer is 343. Two solutions:

1. Solution 1: The left picture shows the corners and the faces touching the side labeled 1 and the right picture shows the opposite side of the die, whose label is 6, which is incident to the other four corners.
   We compute the eight numbers individually and sum the eight numbers. The eight triplets of integers at the eight corners are (1, 2, 3), (1, 2, 4), (1, 3, 5), (1, 4, 5), (6, 2, 3), (6, 2, 4), (6, 3, 5), (6, 4, 5). These eight values are
   1 × 2 × 3 = 6
   1 × 2 × 4 = 8
   1 × 3 × 5 = 15
   1 × 4 × 5 = 20
   6 × 2 × 3 = 36
   6 × 2 × 4 = 48
   6 × 3 × 5 = 90
   6 × 4 × 5 = 120 .
   The sum of these eight positive integers is 343.
2. Solution 2: Since no corner contains two numbers that sum to 7, the sum in solution 1 can be computed as (1 + 6)(2 + 5)(3 + 4) = 73 = 343.

Part B Challenging Questions' Solutions:

B4) The answer is 10. 3 solutions:

1. Solution 1: The sum of all of the numbers written is the sum of all of the number of pairs of problems of the form (S_i, F_j) each student solved. The contribution of each student to this
   sum is the product of the number of problems this student solved in the short-answer part and the number of problems this student solved in the full-solution part. Since each student solved 11 problems, each student solved either 8 short-answer problems and 3 full-solution problems, or 7 short-answer problems and 4 full-solution problems. Let x be the number of students who solved 8 short-answer problems and 3 full-solution problems and y the number of students who solved 7 short-answer problems and 4 full-solution problems. Then the sum of the numbers written down is 8×3×x+7×4×y = 256. Hence, 24x+28y = 256. Dividing both sides by 4 yields 6x + 7y = 64. Note that 0  x  10. Substituting each such value of x, we get the following values of y:? We note that only (x, y) = (6, 4) yields a non-negative integer solution for x and y. Hence, the number of students is x + y = 6 + 4 = 10.

 

Part C Long-form Proof Problems' Solutions:

C1)

1. The answer is 60.
   Split the 100 problems into groups of 5, namely 1?5, 6?10, 11?15, . . . , 91?95, 96?100. Since there are 100 problems and five problems per group and every set of five consecutive
   problems contain exactly three problems whose answer is TRUE, each group contains three problems whose answers are TRUE. Since there are 20 groups, there are 20×3 = 60 problems whose answers are TRUE on the test.
2. Consider the problems 1, 2, 3, 4, 5, 6. Among problems 1?5, there are exactly three problems whose answer is TRUE. Since the answer to the first problem is FALSE, among problems 2?5, exactly three of these problems have answer TRUE. Now consider problem 6. Since problems 2 ? 6 contains exactly three problems whose answers are TRUE and problems 2?5 already contain 3 such problems, the answer to problem 6 is FALSE.
3. Solution 1: We claim that the answer to problem n has the same answer as problem n + 5. Consider the problems n, n + 1, n + 2, n + 3, n + 4, n + 5. Note that problems n, n+1, n+2, n+3, n+4 contain three problems whose answers are TRUE and problems n + 1, n + 2, n + 3, n + 4, n + 5 contain three problems whose answers are TRUE. Note that problems n + 1, n + 2, n + 3, n + 4 contain either 2 or 3 problems whose answers are TRUE. In the former case, the answers to both problem n and problem n + 5 are TRUE. In the latter case, the answers to both problem n and problem n+5 are FALSE. In either case, problems n and n + 5 have the same answer.
   By this claim, problems {1, 6, 11, 16, . . . , 91, 96} have the same answers. So do {2, 7, 12, 17, . . . , 92, 97}, {3, 8, 13, 18, . . . , 93, 98}, {4, 9, 14, 19, . . . , 94, 99} and {5, 10, 15, 20, . . . , 95, 100}. For each of these five groups of problems, if we can determine the answer to one problem in the group, we can determine the answers to every problem in the group. Since the answer to problem 1 is FALSE, the answers to problems {1, 6, 11, 16, . . . , 91, 96} are all FALSE. Since problem 100 is FALSE, then the answers to problems {5, 10, 15, 20, . . . , 95, 100} are also FALSE. Since problems 1 and 5 have answers FALSE, and exactly three of problems 1, 2, 3, 4, 5 have answer TRUE, problems 2, 3, 4 have answer TRUE. Therefore, the answers to the remaining problems {2, 7, 12, 17, . . . , 92, 97}, {3, 8, 13, 18, . . . , 93, 98}, {4, 9, 14, 19, . . . , 94, 99} are all TRUE. Having determined the correct answer to each question, Charlotte achieves a perfect score by answering FALSE, TRUE, TRUE, TRUE, FALSE to the first five questions, and repeating this pattern for each block of five consecutive questions.

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