AMC10 2021 B

AMC10 2021 B · Q8

AMC10 2021 B · Q8. It mainly tests Basic counting (rules of product/sum), Coordinate geometry.

Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?

周先生将整数从$1$到$225$放入一个$15$乘$15$的网格中。他将$1$放在中间的方格（第八行第八列），然后顺时针逐一放置其他数字，如下图部分所示。从顶部第二行中出现的最小数和最大数的和是多少？

(A) 367 367

(B) 368 368

(D) 379 379

(E) 380 380

Answer

Correct choice: (A)

正确答案：（A）

Solution

In the diagram below, the red arrows indicate the progression of numbers. In the second row from the top, the greatest number and the least number are $D$ and $E,$ respectively. Note that the numbers in the yellow cells are consecutive odd perfect squares, as we can prove by induction. By observations, we proceed as follows: \begin{alignat*}{6} A=15^2=225, \ B=13^2=169 \quad &\implies \quad &C &= \hspace{1mm}&&A-14\hspace{1mm} &= 211& \\ \quad &\implies \quad &D &= &&C-1 &= 210& \\ \quad &\implies \quad &E &= &&B-12 &= 157&. \end{alignat*} Therefore, the answer is $D+E=\boxed{\textbf{(A)} ~367}.$

下图中，红色箭头表示数字的递增方向。从顶部第二行中，最大数和最小数分别是$D$和$E$。注意黄色单元格中的数字是连续的奇完全平方数，可通过归纳法证明。通过观察，我们按以下方式计算： \begin{alignat*}{6} A=15^2=225, \ B=13^2=169 \quad &\implies \quad &C &= \hspace{1mm}&&A-14\hspace{1mm} &= 211& \\ \quad &\implies \quad &D &= &&C-1 &= 210& \\ \quad &\implies \quad &E &= &&B-12 &= 157&. \end{alignat*} 因此，答案是$D+E=\boxed{\textbf{(A)} ~367}$。

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