AMC10 2010 A

AMC10 2010 A · Q25

AMC10 2010 A · Q25. It mainly tests Exponents & radicals, Constructive proofs / constructions.

Jim starts with a positive integer n and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with n = 55, then his sequence contains 5 numbers: 55, 55−7²=6, 6−2²=2, 2−1²=1, 1−1²=0. Let N be the smallest number for which Jim’s sequence has 8 numbers. What is the units digit of N?

Jim 从正整数 $n$ 开始，生成数列。每个后续数由减去当前数小于或等于的最大完全平方整数得到，直到达到零。例如，若 Jim 从 $n = 55$ 开始，则数列含 5 个数：55, 55−7²=6, 6−2²=2, 2−1²=1, 1−1²=0。设 $N$ 为 Jim 的数列有 8 个数的数的最小值。$N$ 的个位数是多少？

(A) 1 1

(B) 3 3

(D) 7 7

(E) 9 9

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let the sequence be $(a_1, a_2, \ldots, a_8)$. For $j>1$, $a_{j-1}=a_j+m^2$ for some $m$ such that $a_j<(m+1)^2-m^2=2m+1$. To minimize the value of $a_1$, construct the sequence in reverse order and choose the smallest possible value of $m$ for each $j$, $2\le j\le 8$. The terms in reverse order are $a_8=0$, $a_7=1$, $a_6=1+1^2=2$, $a_5=2+1^2=3$, $a_4=3+2^2=7$, $a_3=7+4^2=23$, $a_2=23+12^2=167$, and $N=a_1=167+84^2=7223$, which has the unit digit 3.

答案（B）：设数列为 $(a_1, a_2, \ldots, a_8)$。对 $j>1$，有 $a_{j-1}=a_j+m^2$，其中 $m$ 满足 $a_j<(m+1)^2-m^2=2m+1$。为了使 $a_1$ 取到最小值，从后往前构造该数列，并对每个 $j$（$2\le j\le 8$）选取尽可能小的 $m$。按逆序得到各项为：$a_8=0$，$a_7=1$，$a_6=1+1^2=2$，$a_5=2+1^2=3$，$a_4=3+2^2=7$，$a_3=7+4^2=23$，$a_2=23+12^2=167$，以及 $N=a_1=167+84^2=7223$，其个位数字为 3。

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