历年 Canadian Open Mathematics Challenge加拿大数学公开赛
Part A Solutions:
A2)This problem can be handled by guessing and verifying the guess. It is also nicely solved by using the fact that 7 divides 49a+7b, so the remainder on the left side is c, while on the right side division by 7 gives remainder 6;hence c is 6. Continuing in the same way, a=b=5, and the answer is 556.
A5) The solution to this problem depends on two observations. First, there are identical numbers of routes going above the park or below the park; second, going above the park required going through either X or Y and one cannot go through both.
There are 5 routes from A to X and for each there is only one way of proceeding to B. There are 10 routes from A to Y and for each there are 5 routes to B. Hence there are 110 routes in total.
A8) Any question of this type is handled by solving for one variable in terms of the other. This leads to a faction in which the denominator is an expression that must divide the numerator if integers are to result. Here, solving for y in terms of x forces x to have values 2 or 1, resulting in y being 9 or -2.
Part B Solutions:
B1) The given conditions lead to three numbers that must form a geometric sequence. Using these and the fact that the ratio of consecutive terms is constant leads to a quadratic equation which has two solutions and hence two possible arithmetic sequences 14,25,36, or -26, -15, -4.