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
p A – 4 marks each
A1. Suppose x is a real number such that x(x + 3) = 154. Determine the value of (x + 1)(x + 2).Solution 1: The answer is 156 .
Solution 1: The answer is 156 .
Expanding x(x + 3) = 154 yields x
2 + 3x = 154.
Hence, (x + 1)(x + 2) = x
2 + 3x + 2 = 154 + 2 = 156.
Solution 2: The answer is 156 .
154 = 11 × 14 = (−11) × (−14). In the former case x = 11 and 12 × 13 = 156. In the latter case x = −14
and (−13) × (−12) = 156.
A2. Let v, w, x, y, and z be five distinct integers such that 45 = v × w × x × y × z. What is the sum of the integers?
Solution: The answer is 5 .
Notice that 45 = 3 × 3 × 5. It stands to reason that, to write 45 as a product of five integer factors, each of its prime factors must appear, along with ±1 (we can’t use fractions). Further, to have exactly 5 distinct integers −3 and −1 must each appear once. We have 45 = (−1) × 1 × (−3) × 3 × 5. The sum of these five factors is 5.
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