# 2002COMC加拿大数学公开赛真题答案免费下载

### 2002 COMC真题答案免费下载

Part A共8题，每题5分

Part B共4题，每题10分

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Part A Solutions:

A4) Three solutions:

1. Solution 2: We determine the first 5 terms in the sequence and then add up the 3rd, 4th and 5th terms.
From the formula given, the sum of the first 1 terms is 11.
This tells us that the first term is 11.
From the formula given, the sum of the first 2 terms is 32. Since the first term is 11, then the second term is 21.
Next, the sum of the first 3 terms is 63, and so the third term is 31, since the first two terms are 11 and 21. (We could use the fact that the sum of the first two terms is 32, instead.)
Next, the sum of the first 4 terms is 104, and so the fourth term is 41.
Lastly, the sum of the first 5 terms is 155, and so the fifth term is 51.
Therefore, the sum of the 3rd, 4th and 5th terms is 31+ 41+ 51= 123.

Part B Solutions:

B2)

1. We consider the possible cases. On his first turn, Alphonse can take either 1 marble or 2 marbles.
If Alphonse takes 1 marble, Beryl can take 2 marbles and then Colleen 1 marble, to leave Alphonse with 1 marble left in the bowl. Therefore, Alphonse loses. (Note that Beryl and Colleen can agree on their strategy before the game starts.)
If Alphonse takes 2 marbles, Beryl can take 1 marble and then Colleen 1 marble, to leave Alphonse again with 1 marble left in the bowl. Therefore, Alphonse loses.
In either case, Beryl and Colleen can work together and force Alphonse to lose.
2. Two Solutions:
1. Solution 1: On their two consecutive turns, Beryl and Colleen remove in total 2, 3 or 4 marbles. On his turn, Alphonse removes either 1 marble or 2 marbles. Therefore, by working together, Beryl and Colleen can ensure that the total number of marbles removed on any three consecutive turns beginning with Alphonse’s turn is 4 or 5. (Totals of 3 and 6 cannot be guaranteed because of Alphonse’s choice.)
Therefore, if N is a number of marbles in which Alphonse can be forced to lose, then so are N + 4 and N + 5, because Beryl and Colleen can force Alphonse to choose from N
marbles on his second turn.
From (a), we know that 5 is a losing position for Alphonse. Also, 1 is a losing position for Alphonse. (Since 1 is a losing position, then 5 and 6 are both losing positions,
based on our earlier comment.)
Since 5 and 6 are losing positions, then we can determine that 9, 10 and 11 are also losing positions, as are 13, 14, 15, and 16. If we add 4 to each of these repeatedly, we see that N is a losing position for every N ≥ 13.
What about the remaining possibilities, i.e. 2, 3, 4, 7, 8, and 12?
For N = 2 or N = 3, if Alphonse chooses 1 marble, then either Beryl or Colleen is forced to take the last marble, so these are not losing positions for Alphonse, i.e. they
are winning positions.
For N = 4, if Alphonse chooses 2 marbles, then either Beryl or Colleen is forced to take the last marble, so this is also not a losing position for Alphonse.
Next, we notice that if Alphonse chooses 1 marble, then the total number of marbles chosen by the three players will be 3, 4 or 5, and if Alphonse chooses 2 marbles, then
the total number chosen will be 4, 5 or 6.
So if N = 7, then Alphonse can choose 1 marble and ensure that he receives 2, 3 or 4 marbles on his next turn. So 7 is a winning position for Alphonse.
If N = 8, then Alphonse can choose 2 marbles and ensure that he receives 2, 3 or 4 marbles on his next turn. So 8 is also a winning position for Alphonse.
Lastly, we consider N = 12.
If Alphonse chooses 1 marble, Beryl and Colleen can choose 1 each and return 9 marbles to Alphonse. As we have shown, this is a losing position for Alphonse.
If Alphonse chooses 2 marbles, Beryl and Colleen can choose 2 each and return 6 marbles to Alphonse. This is a losing position for Alphonse.
Therefore, the values of N for which Beryl and Colleen can force Alphonse to lose are 1, 5, 6, and all N for which N ≥ 9.