AMC12 2011 B

AMC12 2011 B · Q13

AMC12 2011 B · Q13. It mainly tests Linear equations, Logic puzzles.

Brian writes down four integers $w > x > y > z$ whose sum is $44$. The pairwise positive differences of these numbers are $1, 3, 4, 5, 6$ and $9$. What is the sum of the possible values of $w$?

Brian 写下四个整数 $w > x > y > z$，它们的和为 $44$。这四个数两两之间的正差分别为 $1, 3, 4, 5, 6$ 和 $9$。所有可能的 $w$ 的取值之和是多少？

(A) 16 16

(B) 31 31

(D) 62 62

(E) 93 93

Answer

Correct choice: (B)

正确答案：（B）

Solution

Assume that $y-z=a, x-y=b, w-x=c.$ $w-z$ results in the greatest pairwise difference, and thus it is $9$. This means $a+b+c=9$. $a,b,c$ must be in the set ${1,3,4,5,6}$. The only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$. $a+b$, and $b+c$ then must be the remaining two numbers which are $4$ and $6$. The ordering of $(a,b,c)$ must be either $(3,1,5)$ or $(5,1,3)$. \begin{align*} z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ 4z + a + (a + b) + 9 &= 44\\ \text{if} \hspace{1cm} a &= 3 \\ a + b &= 4\\ 4z &= 44 - 9 - 3 - 4\\ z &= 7\\ w &= 16\\ \end{align*} \begin{align*} \text{if} \hspace{1cm} a &= 5\\ a + b &= 6\\ 4z &= 44 - 9 - 5 - 6\\ z &= 6\\ w &= 15\\ \end{align*} The sum of the two w's is $15+16=31$ $\boxed{B}$

设 $y-z=a, x-y=b, w-x=c.$ 最大的成对差是 $w-z$，因此它等于 $9$。这意味着 $a+b+c=9$。$a,b,c$ 必须取自集合 ${1,3,4,5,6}$。该集合中三个数相加等于 $9$ 的唯一方式是 $1,3,5$。于是 $a+b$ 与 $b+c$ 必须是剩下的两个数 $4$ 和 $6$。因此 $(a,b,c)$ 的顺序只能是 $(3,1,5)$ 或 $(5,1,3)$。 \begin{align*} z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ 4z + a + (a + b) + 9 &= 44\\ \text{if} \hspace{1cm} a &= 3 \\ a + b &= 4\\ 4z &= 44 - 9 - 3 - 4\\ z &= 7\\ w &= 16\\ \end{align*} \begin{align*} \text{if} \hspace{1cm} a &= 5\\ a + b &= 6\\ 4z &= 44 - 9 - 5 - 6\\ z &= 6\\ w &= 15\\ \end{align*} 两个 $w$ 的和为 $15+16=31$ $\boxed{B}$。

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