题目
Problem
(i) Relative to a fixed origin O, the points A, B and C have position vectors a, b and c respectively.
Points A, B and C lie in a straight line, with B lying between A and C.
Given that AB:AC=1:3 show that
c=3b−2a
(3)
(ii) Given that n∈N, prove by contradiction that if n2 is a multiple of 3 then n is a multiple of 3
(5)
题目中文翻译
(i) 相对于固定原点 O,点 A、B 和 C 的位置向量分别为 a,b,c。
点 A、B 和 C 共线,且 B 在线段 AC 之间。
已知 AB:AC=1:3,证明
c=3b−2a
(ii) 已知 n∈N,用反证法证明:如果 n2 是 3 的倍数,那么 n 也是 3 的倍数。
解答
(i)
解法一
思路
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因为 AB:AC=1:3 且 B 位于 A,C 之间,所以 BC=2AB。分别用位置向量写出 AB 与 BC,再整理即可得到目标式。
答题过程
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Since AB:AC=1:3, we have
BC=2AB.
In vector form,
BC=c−b
and
AB=b−a.
Therefore,
c−b==2(b−a)2b−2a.
Hence,
c=3b−2a,
as required.
(ii)
解法一
思路
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反设 n2 是 3 的倍数,但 n 不是。任何不是 3 的倍数的自然数都可写成 3p+1 或 3p+2;分别平方后,结果除以 3 都余 1,与假设矛盾。
答题过程
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Assume, for a contradiction, that n2 is a multiple of 3 but n is not a multiple of 3.
Then, for some integer p, either
n=3p+1
or
n=3p+2.
In the first case,
n2===(3p+1)29p2+6p+13(3p2+2p)+1.
Thus n2 leaves remainder 1 when divided by 3.
In the second case,
n2===(3p+2)29p2+12p+43(3p2+4p+1)+1.
Again, n2 leaves remainder 1 when divided by 3.
In neither case is n2 a multiple of 3, contradicting the assumption. Therefore,
n2 is a multiple of 3⟹n is a multiple of 3.
解法二
思路
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官方资料也接受用模 3 算术表达同一个反证。若 n 不是 3 的倍数,则它模 3 只能为 1 或 2;两种余数平方后都模 3 余 1。
答题过程
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Assume, for a contradiction, that
n2≡0(mod3)
but
n≡0(mod3).
Then n≡1(mod3) or n≡2(mod3). However,
12≡1(mod3)
and
22=4≡1(mod3).
Thus n2≡1(mod3), contradicting n2≡0(mod3). Hence,
3∣n2 ⟹ 3∣n.