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IAL 2023 Oct Q1

A Level / Edexcel / P2

IAL 2023 Oct Paper · Question 1

题目

Problem

Given that aa, bb and cc are integers greater than 0 such that

  • c=3a+1c = 3a + 1
  • a+b+c=15a + b + c = 15

prove, by exhaustion, that the product abcabc is always a multiple of 4.

You may use the table below to illustrate your answer.

aabbccabcabc
(3)
题目中文翻译

已知 aabbcc 是正整数,且满足:

  • c=3a+1c = 3a + 1
  • a+b+c=15a + b + c = 15

用穷举法证明 abcabc 一定是4的倍数。

可以使用下表来说明你的答案。

(3分)

解答

解法一

思路

穷举证明题。由 c=3a+1c = 3a + 1 代入 a+b+c=15a + b + c = 15,得到 b=144ab = 14 - 4a。因为 a,b,ca, b, c 都是正整数,aa 只能取 1,2,31, 2, 3,逐一验证即可。

答题过程

Given c=3a+1c = 3a + 1, substitute into a+b+c=15a + b + c = 15:

a + b + (3a + 1) =&\, 15 \\ 4a + b + 1 =&\, 15 \\ b =&\, 14 - 4a \end{align*}$$ Since $a, b, c$ are positive integers: - $a \geqslant 1$ - $b = 14 - 4a \geqslant 1 \Rightarrow a \leqslant 3$ Therefore $a$ can only take the values $1, 2, 3$. Check each case: | $a$ | $b = 14 - 4a$ | $c = 3a + 1$ | $abc$ | |-----|---------------|--------------|-------| | 1 | 10 | 4 | 40 | | 2 | 6 | 7 | 84 | | 3 | 2 | 10 | 60 | $$\begin{align*} 40 =&\, 4 \times 10 \\ 84 =&\, 4 \times 21 \\ 60 =&\, 4 \times 15 \end{align*}$$ All values of $abc$ are multiples of 4. $\quad \blacksquare$