AMC8 2020

AMC8 2020 · Q16

AMC8 2020 · Q16. It mainly tests Basic counting (rules of product/sum), Logic puzzles.

Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $B?$

图中有点 $A,B,C,D,E,$ 和 $F$，每个点代表从 $1$ 到 $6$ 的不同数字。图中显示的五条直线每条都经过其中一些点。每条直线上的数字相加得到五个和，这五个和的总和是 $47$。$B$ 代表的数字是多少？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (E)

正确答案：（E）

Solution

We can form the following expressions for the sum along each line: \[ \begin{cases} A+B+C\\ A+E+F\\ C+D+E\\ B+D\\ B+F \end{cases} \] Adding these together, we must have $2A+3B+2C+2D+2E+2F=47$, i.e. $2(A+B+C+D+E+F)+B=47$. Since $A,B,C,D,E,F$ are unique integers between $1$ and $6$, we obtain $A+B+C+D+E+F=1+2+3+4+5+6=21$ (where the order doesn't matter as addition is commutative), so our equation simplifies to $42 + B = 47$. This means $B = \boxed{\textbf{(E) }5}$.

我们可以为每条直线上的和建立以下表达式： \[ \begin{cases} A+B+C\\ A+E+F\\ C+D+E\\ B+D\\ B+F \end{cases} \] 将这些相加，我们得到 $2A+3B+2C+2D+2E+2F=47$，即 $2(A+B+C+D+E+F)+B=47$。由于 $A,B,C,D,E,F$ 是 $1$ 到 $6$ 的唯一整数，所以 $A+B+C+D+E+F=1+2+3+4+5+6=21$（加法满足交换律，顺序无关），因此方程简化为 $42 + B = 47$。这意味着 $B = \boxed{\textbf{(E) }5}$。

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