AMC8 1998

AMC8 1998 · Q24

AMC8 1998 · Q24. It mainly tests Casework, Remainders & modular arithmetic.

A rectangular board of 8 columns has squares numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and hades square 3, skip two squares and shades square 6, ships 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that ¯rst achieves this result?

一个有 8 列的矩形棋盘，方格从左上角开始编号，从左到右，行一是 1 到 8，行二是 9 到 16，依此类推。学生涂黑方格 1，然后跳过一个涂黑 3，跳过两个涂黑 6，跳过三个涂黑 10，并以此类推，直到每列至少有一个涂黑方格。第一个实现此结果的涂黑方格编号是多少？

(A) 36 36

(B) 64 64

(D) 91 91

(E) 120 120

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): The numbers in the first column all have remainders of 1 when divided by 8, those of the second column have remainders of 2 when divided by 8, and so on. We need to find numbered squares so that each remainder 0 through 7 appears at least once. The squares that are shaded are numbered 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, and the remainders upon dividing by 8 are 1, 3, 6, 2, 7, 5, 4, 4, 5, 7, 2, 6, 3, 1, 0. Thus, we must shaded square 120 to obtain the first shaded square in the last column.

答案（E）：第一列中的数除以$8$都余$1$，第二列中的数除以$8$都余$2$，依此类推。我们需要找到一些编号的方格，使得余数$0$到$7$每个至少出现一次。被涂色的方格编号为$1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120$，它们分别除以$8$的余数为$1, 3, 6, 2, 7, 5, 4, 4, 5, 7, 2, 6, 3, 1, 0$。因此，我们必须将方格$120$涂色，才能得到最后一列中的第一个被涂色方格。

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