AMC10 2025 A

AMC10 2025 A · Q5

AMC10 2025 A · Q5. It mainly tests Sequences & recursion (algebra), Patterns & sequences (misc).

Consider the sequence of positive integers $1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2 \dots$ What is the $2025$th term in the sequence?

考虑正整数序列 $1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2 \dots$ 该序列的第 2025 项是多少？

(A) 5 5

(B) 15 15

(D) 44 44

(E) 45 45

Answer

Correct choice: (E)

正确答案：（E）

Solution

Group the numbers by their hill pattern: $(12)(1232)(123432)(12345432)...$ The maximums of each hill occur at terms $n = 2, 5, 10, 17...$ These terms correspond to maximums of $2, 3, 4,...$ Let $a$ be the maximum at term $N$. Since the sum of the first $x$ odd numbers is $x^2$ we have $1 + (a-1)^2 = N.$ So for example, if a = $4,$ then $N = 10,$ telling us that the peak of the hill with maximum $4$ occurs at the $10$th term. Now, we know $2025 = 45^2,$ so let $a = 46.$ Then $N = 2026,$ so the $2026$th term is $46.$ Then the $2025$th term must be $\boxed{\text{(E) }45}.$

按“山峰”模式分组： $(12)(1232)(123432)(12345432)...$ 每个山峰的最大值出现在第 $n = 2, 5, 10, 17...$ 项，对应最大值 $2, 3, 4,...$。设第 $N$ 项的最大值为 $a$。由于前 $x$ 个奇数之和为 $x^2$，有 $1 + (a-1)^2 = N$。例如，若 $a = 4$，则 $N = 10$，表示最大值为 4 的山峰峰值在第 10 项。现在，$2025 = 45^2$，设 $a = 46$。则 $N = 2026$，所以第 2026 项是 46。第 2025 项必为 $\boxed{\text{(E) }45}$。

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