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2009COMC加拿大数学公开赛真题答案免费下载

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历年 Canadian Open Mathematics Challenge加拿大数学公开赛

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COMC加拿大数学公开赛

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2009 COMC真题答案免费下载

共计2.5小时考试时间

此套试卷由两部分题目组成

Part A共8题，每题5分

Part B共4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 Solutions:

A2)We write 5073 in place value notation as 5 × 1000 + 7 × 10 + 3 or 5 × 10³ + 7 × 10¹ + 3 × 10⁰.
Thus, if a = 0, b = 3 and c = 1, then the left side (3 × 10^a + 5 × 10^b + 7 × 10^c) equals 5073.
Any other combination of values for a, b and c will not give 5073.
Therefore, a + b + c = 0 + 3 + 1 = 4.
(We can show that the only possibility is a = 0, b = 3 and c = 1. We start by noting that the remainder when the right side is divided by 10 is 3, so the remainder when the left side is divided by 10 must also be 3. If a = 0 and each of b and c is larger than 0, then the remainder on the left side will be 3. If more than one of a, b and c equals 0, then we can see by trying the possibilities that the remainder on the left side cannot be 3.
Therefore, a = 0 and b and c are positive.
We can then subtract 3 from both sides and divide by 10 to obtain the new equation 5 × 10^b-1 + 7 × 10^c-1 = 507 and repeat the argument to show that c = 1 and then b = 3.)
Answer: 4

Part B Solutions:

B1)

We want to find all positive integers a for which the smallest positive s with the property that a divides into 1 + 2 + 3 + ... + s is s = 8.
Note that 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36. (We use the term triangular number to mean a positive integer of the form 1 + 2 + 3 + ... + s.) The previous triangular numbers are 1, 3, 6, 10, 15, 21, 28. Therefore, we want to nd all positive integers a that are divisors of 36 but not of any of 1, 3, 6, 10, 15; 21; 28.
The divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.
The divisors 1, 2, 3, 4, 6 each divide into at least one of 6 and 28, so f(1) ≠ 8, f(2) ≠ 8,f(3) ≠ 8, f(4) ≠ 8, and f(6) ≠ 8.
The divisors 9, 12, 18, 36 do not divide into any triangular number smaller than 36.
Therefore, the complete solution to f(a) = 8 is a = 9,12, 18, 36.
For m a positive integer, we define T(m) = 1 + 2 + ... + m = 1/2m(m + 1). (T(m) is the mth triangular number.)
First, we show that if T(z) is a multiple of w, then f(w) z:
We know that f(w) is the smallest integer m for which 1+2+3+...+m = T(m) is a multiple of w.
Suppose that T(z) is a multiple of w.
If z is the smallest positive integer with this property, then f(w) = z; otherwise, z is not the smallest such positive integer, so f(w) < z.
In either case, f(w) < z.
Next, we show that if y is an odd positive integer with y > 1, then f(y) ≤y - 1:
Suppose that y = 2Y + 1 for some positive integer Y .
Then T(y - 1) = 1/2 (y - 1)y = 1/2(2Y )(2Y + 1) = Y (2Y + 1) = Y y.
Thus, T(y - 1) is a multiple of y, and so by the first fact above, f(y) ≤y - 1.
Next, we show that f(2^a) = 2^a+1 - 1 for every positive integer a:
Suppose that f(2^a) = m.
Then 1/2m(m+1) is a multiple of 2^a, so 1/2m(m+1) = q2^a for some positive integer q or 2^a+1q = m(m + 1).
Since one of m and m+1 is even and the other is odd, then the even one of these must contain at least a + 1 factors of 2 and so must be at least 2a+1.
The smallest m for which this is possible is m = 2^a+1-1 which makes m+1 = 2^a+1.
This tells us that f(2^a) ≥2^a+1 - 1.
But T(2^a+1 - 1) = 1/2(2^a+1 - 1)2^a+1 = 2^a(2^a+1 - 1), which is divisible by 2^a, so f(2^a) ≤ 2^a+1 - 1.
Therefore, f(2^a) = 2^a+1 - 1.
We can now look at f(b + 1) - f(b) when b = 2^a - 1 for some positive integer a.
Note that b is odd.
In this case,
f(b + 1) - f(b) = f(2^a) - f(2^a - 1) = 2^a+1 - 1 - f(2^a - 1) ≥ 2^a+1 - 1 - (2^a - 2) = 2^a + 1
If a ≥ 11, then 2^a + 1 ≥ 2049 > 2009.
Therefore, if b = 2^a-1 and a is a positive integer with a ≥ 11, then f(b+1)-f(b) > 2009, so there are infinitely many odd positive integers b for which f(b + 1) - f(b) > 2009.