AMC 8 2026 Problem 2
AMC8 2026 年第 2 题
数列与规律★★☆☆☆
In the array shown below, three 3s are surrounded by 2s, which are in turn surrounded by a border of 1s. What is the sum of the numbers in the array?
在下面的数组中，三个数字 3 被数字 2 包围，而这些 2 又被一圈 1 包围。该数组中所有数字之和是多少？
- A.49
- B.51
- C.53
- D.55
- E.57
正确答案 · Correct Answer
C
解析 · Solution
解法1
数出数组中各数字的数量：1 出现 20 次，2 出现 12 次，3 出现 3 次。
所有元素之和为
将五行标记为 ，由对称性知 ，。
各行之和：，，，总和为 2×7+2×14+21= 53。
由于每个选项的个位数都不同，只需看总和的个位即可确定答案，无需计算完整总和。
直观理解：第二、四行各有五个 2，2×5=10，个位贡献为 0。中间行 2+2+3+3=10，个位为 0，剩一个 3。边框 20 个 1，个位为 0。全部个位加起来等于 3，所以答案个位是 3，选 53。
严格来说，这利用了同余性质：
第二、四行贡献 20 ≡ 0 (mod 10)
中间行 2+2+3+3=10 ≡ 0 (mod 10)，剩余一个 3
边框 20 个 1，20 ≡ 0 (mod 10)
总和 ≡ 3 (mod 10)，因此答案是 53。
将 5×7 的数组视为三层同心矩形环，由外向内逐层计算：
外环：边框一周共 (6+4)×2 = 20 个格子，全是 1 → 20×1 = 20。
中环：去掉外框后剩下 3×5 的矩形，其边框一周共 (4+2)×2 = 12 个格子，全是 2 → 12×2 = 24。
内芯：最后剩下 1×3 的细条，共 3 个格子，全是 3 → 3×3 = 9。
合计 20 + 24 + 9 = 53。此方法避免了逐格去数 35 个格子。
换一个视角：将数组想象成三层值全为 1 的矩形叠加在一起。
第一层：完整的 5×7 矩形，每格为 1。贡献：5×7×1 = 35。
第二层：内部 3×5 矩形，每格再叠加 1。贡献：3×5×1 = 15。
第三层：最内 1×3 矩形，每格再叠加 1。贡献：1×3×1 = 3。
三层重叠处值相加：边框只有第一层（值=1），中间环有第一+第二层（值=2），中心有三层叠加（值=3）。总和 = 35+15+3 = 53。
Solution 1
Count how many of each digit appears: 20 ones, 12 twos, and 3 threes. The total sum is 20×1 + 12×2 + 3×3 = 20+24+9 = 53.
Label the rows A₁ through A₅ top to bottom. By symmetry, A₁ = A₅ and A₂ = A₄. Their sums: A₁ = 7 (all 1s), A₂ = 2+2+2+2+2+1+1 = 12, A₃ = 2+2+3+3+3+1+1 = 15. Total = 2×7 + 2×12 + 15 = 14+24+15 = 53.
Since each answer choice has a different units digit, we only need the sum modulo 10.
Rows 2 and 4 (five 2s each): contribution = 20 ≡ 0 (mod 10).
Middle row: 2+2+3+3 = 10 ≡ 0 (mod 10), leaving one 3.
Outer border: 20 ones = 20 ≡ 0 (mod 10).
Total sum ≡ 3 (mod 10), so the answer is 53.
View the 5×7 array as three concentric rectangular rings:
Outer ring (border): perimeter = (6+4)×2 = 20 cells, all 1s → 20×1 = 20.
Middle ring: after removing the border, the inner rectangle is 3×5. Its border has (4+2)×2 = 12 cells, all 2s → 12×2 = 24.
Center: the remaining 1×3 strip has 3 cells, all 3s → 3×3 = 9.
Total = 20 + 24 + 9 = 53. This method avoids counting all 35 cells individually.
Think of the array as three layers of 1s stacked together. The full 5×7 rectangle is one layer of 1s: 5×7×1 = 35. The inner 3×5 rectangle is a second layer of 1s stacked on top: 3×5×1 = 15. The innermost 1×3 rectangle is a third layer: 1×3×1 = 3. Where the layers overlap, the values add up: the border has 1 layer (value 1), the middle ring has 2 layers (value 2), and the center has 3 layers (value 3). Total = 35 + 15 + 3 = 53.
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