2015COMC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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2015 COMC真題答案免費(fèi)下載

共計(jì)2.5小時(shí)考試時(shí)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:

A1)Since 2015 has 4 digits, any number larger than 2015 will also have 4 digits. Given two starting digits, in order, there is a unique 4-digit palindrome that begins with those two starting digits. For starting digits 20, the palindrome is 2002, which is less than 2015. The next smallest will start with 21 and is 2112. Since 2112 > 2015 this is the smallest palindrome greater than 2015.

Part B Challenging Questions' Solutions:

B1) For a positive integer n; let n' be the smallest prime that divides n. Then we can see that f(n) = n/n' and also that n = f(n) ?n' . We are given that f(n) = 35, so to maximize n we must maximize n': Since n' is the smallest prime factor of n; it cannot be larger than any factor of f(n): Thus, the largest possible value of n' is 5, so the largest possible value of n is 5 ? 35 = 175.

Part C Long-form Proof Problems' Solutions:

C4)

1. Suppose there are 5 students and the starting amounts are: (0, 0, 0, 1, 2). As the game progresses, the amounts of money become: (1, 1, 1, 1,-1) then (0, 0, 0, 0,3) and finally (1, 1, 1, 1,-1) at which point the game ends.
2. Consider a game with n players that starts with the initial distribution of one player having n - 2 dollars and all the rest having n - 3. After the first turn, the first player will have -1 dollars, thus n - 2 + (n - 3)(n - 1) = n² - 3n + 1 is not suficient.
   We observe that the only way for a person to end up with negative money is if they were the giver and gave away more money than they had. Suppose instead that the sum of all the money is n² -3n+2. By the pigeonhole principle, at every stage of the game, we must either have at least one student with at least n-1 dollars or at least two students with exactly n-2 dollars. In the first case, since the giver is giving one dollar to at most n - 1 people, they will not end the turn with a negative amount of money. Similarly, if there are at least two students with n - 2 dollars, then they will each give one dollar to at most n - 2 students and will not end the turn with a negative amount of money. Thus kn = n² - 3n + 2:

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