AMC8 2007

AMC8 2007 · Q9

AMC8 2007 · Q9. It mainly tests Basic counting (rules of product/sum), Logic puzzles.

To complete the grid below, each of the digits 1 through 4 must occur once in each row and once in each column. What number will occupy the lower right-hand square?

要完成下面的网格，每个数字1到4必须在每行和每列中各出现一次。右下角的方格应该是多少？

(A) 1 1

(B) 2 2

(D) 4 4

(E) cannot be determined 无法确定

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) The number in the last column of the second row must be 1 because there are already a 2 and a 3 in the second row and a 4 in the last column. By similar reasoning, the number above the 1 must be 3. So the number in the lower right-hand square must be 2. This is not the only way to find the solution. \[ \begin{array}{|c|c|c|c|} \hline 1 & & 2 & 3 \\ \hline 2 & 3 & & 1 \\ \hline & & & 4 \\ \hline & & & 2 \\ \hline \end{array} \] The completed square is \[ \begin{array}{|c|c|c|c|} \hline 1 & 4 & 2 & 3 \\ \hline 2 & 3 & 4 & 1 \\ \hline 3 & 2 & 1 & 4 \\ \hline 4 & 1 & 3 & 2 \\ \hline \end{array} \]

（B）第二行最后一列的数字必须是 1，因为第二行已经有了 2 和 3，而最后一列已经有了 4。用类似的推理，1 上面的数字必须是 3。因此右下角的方格必须是 2。这并不是找到解的唯一方法。 \[ \begin{array}{|c|c|c|c|} \hline 1 & & 2 & 3 \\ \hline 2 & 3 & & 1 \\ \hline & & & 4 \\ \hline & & & 2 \\ \hline \end{array} \] 完成后的方阵为 \[ \begin{array}{|c|c|c|c|} \hline 1 & 4 & 2 & 3 \\ \hline 2 & 3 & 4 & 1 \\ \hline 3 & 2 & 1 & 4 \\ \hline 4 & 1 & 3 & 2 \\ \hline \end{array} \]

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