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P2 2024 Oct Q11

A Level / Edexcel / P2

IAL 2024 Oct Paper · Question 11

题目

Problem

(i) Prove by counter example that the statement

“If nn is a prime number then 3n+23^n+2 is also a prime number.”

is false.

(2)

(ii) Use proof by exhaustion to prove that if mm is an integer that is not divisible by 33, then

m2−1m^2-1

is divisible by 33

(4)
题目中文翻译

(i) 用反例证明以下命题是错误的:(2)

“如果 nn 是质数,那么 3n+23^n+2 也是质数。”

(ii) 使用穷尽法证明:如果 mm 是不能被 33 整除的整数,那么 m2−1m^2-1 能被 33 整除。(4)

解答

(i)

解法一

思路

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反例只需要找一个 prime nn,使 3n+23^n+2 不是 prime。选 n=5n=5 很方便。

答题过程

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Take

n= 5.\begin{align*} n=&\,5. \end{align*}

This is prime. But

35+2= 243+2= 245= 5×49.\begin{align*} 3^5+2=&\,243+2\\ =&\,245\\ =&\,5\times49. \end{align*}

So 35+23^5+2 is not prime. Therefore, the statement is false.

(ii)

解法一

思路

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一个整数不被 33 整除,只可能形如 3k+13k+1 或 3k+23k+2。proof by exhaustion 就是把这两种情况都检查完。

答题过程

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If mm is not divisible by 33, then either

m= 3k+1\begin{align*} m=&\,3k+1 \end{align*}

or

m= 3k+2,\begin{align*} m=&\,3k+2, \end{align*}

where kk is an integer.

Case 1:

m= 3k+1m2−1= (3k+1)2−1= 9k2+6k+1−1= 9k2+6k= 3(3k2+2k).\begin{align*} m=&\,3k+1\\ m^2-1=&\,(3k+1)^2-1\\ =&\,9k^2+6k+1-1\\ =&\,9k^2+6k\\ =&\,3(3k^2+2k). \end{align*}

So m2−1m^2-1 is divisible by 33.

Case 2:

m= 3k+2m2−1= (3k+2)2−1= 9k2+12k+4−1= 9k2+12k+3= 3(3k2+4k+1).\begin{align*} m=&\,3k+2\\ m^2-1=&\,(3k+2)^2-1\\ =&\,9k^2+12k+4-1\\ =&\,9k^2+12k+3\\ =&\,3(3k^2+4k+1). \end{align*}

So m2−1m^2-1 is divisible by 33.

In all possible cases where mm is not divisible by 33, m2−1m^2-1 is divisible by 33.

解法二

思路

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先将 m2−1m^2-1 因式分解。仍穷尽两个余数类别,分别找出可被 33 整除的那个因子。

答题过程

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If mm is not divisible by 33, then m=3k+1m=3k+1 or m=3k+2m=3k+2, where k∈Zk\in\mathbb Z.

Factorise:

m2−1=(m−1)(m+1).m^2-1=(m-1)(m+1).

If m=3k+1m=3k+1, then

m2−1= (3k)(3k+2)= 3[k(3k+2)].\begin{align*} m^2-1=&\,(3k)(3k+2)\\ =&\,3\big[k(3k+2)\big]. \end{align*}

If m=3k+2m=3k+2, then

m2−1= (3k+1)(3k+3)= 3[(3k+1)(k+1)].\begin{align*} m^2-1=&\,(3k+1)(3k+3)\\ =&\,3\big[(3k+1)(k+1)\big]. \end{align*}

In each case the bracket is an integer, so m2−1m^2-1 is divisible by 33. These exhaust all integers not divisible by 33, hence the result is proved.