AMC12 2023 A

AMC12 2023 A · Q20

AMC12 2023 A · Q20. It mainly tests Sequences & recursion (algebra), Digit properties (sum of digits, divisibility tests).

Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below. Each row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digits of the sum of the 2023 numbers in the 2023rd row?

一个整数三角形的第 1、2、3、4 和 5 行如下所示。每行之后的行在行首尾放置 1，内部每个条目比前一行对角线上方的两个数的和大 1。求第 2023 行 2023 个数的和的个位数是多少？

(A) 1 1

(B) 3 3

(D) 7 7

(E) 9 9

Answer

Correct choice: (C)

正确答案：（C）

Solution

First, let $R(n)$ be the sum of the $n$th row. Now, with some observation and math instinct, we can guess that $R(n) = 2^n - n$. Now we try to prove it by induction, $R(1) = 2^n - n = 2^1 - 1 = 1$ (works for base case) $R(k) = 2^k - k$ $R(k+1) = 2^{k+1} - (k + 1) = 2(2^k) - k - 1$ By definition from the question, the next row is always$:$ Double the sum of last row (Imagine each number from last row branches off toward left and right to the next row), plus # of new row, minus 2 (minus leftmost and rightmost's 1) Which gives us $:$ $2(2^k - k) + (k + 1) - 2 = 2(2^k) - k - 1$ Hence, proven Last, simply substitute $n = 2023$, we get $R(2023) = 2^{2023} - 2023$ Last digit of $2^{2023}$ is $8$, $8-3 = \boxed{\textbf{(C) } 5}$

首先，令 $R(n)$ 为第 $n$ 行的和。现在，通过一些观察和数学直觉，我们可以猜测 $R(n) = 2^n - n$。现在我们用归纳法证明它， $R(1) = 2^n - n = 2^1 - 1 = 1$（基例成立） $R(k) = 2^k - k$ $R(k+1) = 2^{k+1} - (k + 1) = 2(2^k) - k - 1$ 根据题目定义，下一行总是：前一行和的两倍（想象前一行每个数向左右分支到下一行），加上新行的数量，减去 2（减去最左和最右的 1）这给出： $2(2^k - k) + (k + 1) - 2 = 2(2^k) - k - 1$ 故证明。最后，简单代入 $n = 2023$，得到 $R(2023) = 2^{2023} - 2023$ $2^{2023}$ 的个位数是 8，$8-3 = \boxed{\textbf{(C) } 5}$

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