2014COMC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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2014 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)Two solutions:

1. Solution 1: For a number to be divisible by 3, the sum of the digits must be a multiple of 3, so 3|(A + 4), which means A ∈{2, 5, 8}.
   For a number to be divisible by 4, the number formed by the last two digits must be divisible by 4, so 4|(10A + 2): This gives A∈ {1, 3, 5, 7, 9}.
   The only number in common is A = 5; so n = 2052.
2. Solution 2: The number n - 12 is divisible by 5, so n is of the form 60k + 12 for some k.60 × 34 + 12 = 2052; so n = 2052. It is easy to verify that no other value of k works.

Part B Challenging Questions' Solutions:

B3) If we have a horizontal (or vertical) line of all Os, then since there are 5Xs for the other two lines, there must be a horizontal (or vertical) line of all Xs. Thus, our line of 3 Os must be a diagonal.
When one of the diagonal lines is all Os, then no other line can be all Xs, since each diagonal line intersects all other lines. Thus, each configuration with one diagonal line of Os is a
desired solution.
When we have the diagonal line 1 5 9; there are 6 places that the last O could be: 2, 3, 4, 6, 7, 8. Each of these will give a valid solution. Similarly, we have 6 solutions when we have the diagonal line 7 - 5 - 3:
It is not possible for both diagonal lines to have only Os, since there are only 4 Os, thus we have not counted the same configuration twice. Thus 12 of the 126 ways contain a line of 3 Os and no line of 3 Xs.

Part C Long-form Proof Problems' Solutions:

C3)

1. There are 11 students in the club other than Jeffrey and each field trip that Jeffrey is on has 5 other students. In order for Jeffrey to go on a field trip with each other student, he must go on at least 11/5=2.2 = 3 field trips .
   To see that 3 trips is sufficient, we let the other students be {s₁, s₂, ..., s₁₁}.On the first trip, Jeffrey can go with {s_1,... _; s₅}, on the second trip with{s₆, ... , s₁₀} and on the third trip with s₁₁ and any other 4 students. .
2. From part (a), we know that each student must go on at least 3 trips. Since there are 12 students in total, if we count the number of trips that each student went on, we would get a minimum of 12×3 = 36: Since 6 students attend each field trip, that means there must be at least 36/6 = 6 trips.
   We now show that it is possible to do this with exactly 6 trips. Divide the students into 4 groups of 3 students each (groups A, B, C, D). There are 6 different pairs of groups (AB, AC, AD, BC, BD, CD). Let these pairs of groups be our 6 field trips. We see that since each group goes on a trip with each other group, that each pair of students goes on a trip together. Hence, this can be done with 6 trips.

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