AMC12 2020 A

AMC12 2020 A · Q5

AMC12 2020 A · Q5. It mainly tests Arithmetic misc, Patterns & sequences (misc).

The 25 integers from −10 to 14, inclusive, can be arranged to form a 5-by-5 square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?

从 −10 到 14 的 25 个整数（包括两端）可以排列成一个 5×5 方阵，使得每行之和、每列之和以及两条主对角线之和都相同。这个公共和的值是多少？

(A) 2 2

(B) 5 5

(D) 25 25

(E) 50 50

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The sum of all 25 integers from $-10$ to $14$, inclusive, equals $11+12+13+14=50$ because the integers in that sum from $1$ to $10$ can be paired with their additive inverses to contribute $0$ to the sum. Because each of the $5$ rows in the $5$-by-$5$ square has the same sum and there are $5$ rows, that common sum is $50 \div 5 = 10$. Here is one such magic square, in which each row, each column, and each main diagonal sums to $10$. \[ \begin{array}{|c|c|c|c|c|} \hline 6 & 13 & -10 & -3 & 4\\ \hline 12 & -6 & -4 & 3 & 5\\ \hline -7 & -5 & 2 & 9 & 11\\ \hline -1 & 1 & 8 & 10 & -8\\ \hline 0 & 7 & 14 & -9 & -2\\ \hline \end{array} \]

答案（C）：从 $-10$ 到 $14$（含）这 $25$ 个整数的和等于 $11+12+13+14=50$，因为其中从 $1$ 到 $10$ 的整数可以与其加法逆元配对，从而对总和贡献 $0$。由于这个 $5\times 5$ 方阵的每一行和都相同，且共有 $5$ 行，所以公共行和为 $50 \div 5 = 10$。下面给出一个这样的幻方，其中每一行、每一列以及两条主对角线的和都为 $10$。 \[ \begin{array}{|c|c|c|c|c|} \hline 6 & 13 & -10 & -3 & 4\\ \hline 12 & -6 & -4 & 3 & 5\\ \hline -7 & -5 & 2 & 9 & 11\\ \hline -1 & 1 & 8 & 10 & -8\\ \hline 0 & 7 & 14 & -9 & -2\\ \hline \end{array} \]

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