题目
Problem
In this question you must show detailed reasoning.
Use algebra to prove by exhaustion that,
for all positive integers m that are not multiples of 3, the value of
m2+3m+2
is always a multiple of 3
(4)
题目中文翻译
本题必须展示详细的推理过程。
用代数方法,通过穷举证明:对于所有不是 3 的倍数的正整数 m,
m2+3m+2
的值总是 3 的倍数。(4分)
解答
解法一
思路
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m 不是 3 的倍数,所以除以 3 的余数只能是 1 或 2。因此只需检查 m=3p+1 和 m=3p+2 两类。
答题过程
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Since m is a positive integer that is not a multiple of 3, there are two cases:
m=3p+1orm=3p+2,
where p is a non-negative integer.
Case 1: m=3p+1.
m2+3m+2====(3p+1)2+3(3p+1)+29p2+6p+1+9p+3+29p2+15p+63(3p2+5p+2).
So m2+3m+2 is a multiple of 3 in this case.
Case 2: m=3p+2.
m2+3m+2====(3p+2)2+3(3p+2)+29p2+12p+4+9p+6+29p2+21p+123(3p2+7p+4).
So m2+3m+2 is also a multiple of 3 in this case.
Therefore, for all positive integers m that are not multiples of 3, m2+3m+2 is always a multiple of 3.
解法二
思路
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先因式分解:m2+3m+2=(m+1)(m+2)。如果 m 不是 3 的倍数,那么接下来的两个整数 m+1,m+2 之中必有一个是 3 的倍数。
答题过程
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Factorise:
m2+3m+2=(m+1)(m+2).
Since m is a positive integer not divisible by 3, m=3p+1 or m=3p+2, where p is a non-negative integer.
If m=3p+1, then
m+2=3p+3=3(p+1),
so m+2 is a multiple of 3.
If m=3p+2, then
m+1=3p+3=3(p+1),
so m+1 is a multiple of 3.
In both cases, one factor of
(m+1)(m+2)
is a multiple of 3.
Therefore,
m2+3m+2
is always a multiple of 3.