AMC10 2020 B

AMC10 2020 B · Q24

AMC10 2020 B · Q24. It mainly tests Vieta / quadratic relationships (basic), Inequalities with floors/ceilings (basic).

How many positive integers $n$ satisfy $\frac{n + 1000}{70} = \lfloor \sqrt{n} \rfloor$? (Recall that $\lfloor x \rfloor$ is the greatest integer not exceeding $x$.)

有几个正整数 $n$ 满足 $\frac{n + 1000}{70} = \lfloor \sqrt{n} \rfloor$？（回想 $\lfloor x \rfloor$ 是不超过 $x$ 的最大整数。）

(A) 2 2

(B) 4 4

(D) 30 30

(E) 32 32

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Each positive integer $n$ can be uniquely expressed as $k^2+j$, where $j$ and $k$ are integers with $k\ge 1$ and $0\le j\le 2k$. When $n$ is so expressed, $\lfloor\sqrt{n}\rfloor=k$, and the given equation can be written as $$ k^2-(70k-j)+1000=0. $$ For a fixed value of $k$, there is a solution for $j$ if and only if $$ k^2-70k+1000\le 0 \quad\text{and}\quad k^2-68k+1000\ge 0. $$ Solving the first inequality gives $20\le k\le 50$, and solving the second gives $k\le 34-\sqrt{156}\approx 21\frac12$ or $k\ge 34+\sqrt{156}\approx 46\frac12$. Thus the suitable values of $k$ are $20, 21, 47, 48, 49,$ and $50$, and, for each of these, the unique value of $j$ is $(k-20)(50-k)$. Hence the given equation has 6 solutions for $n$, namely $400, 470, 2290, 2360, 2430,$ and $2500$.

答案（C）：每个正整数 $n$ 都可以唯一地表示为 $k^2+j$，其中 $j$ 和 $k$ 为整数，满足 $k\ge 1$ 且 $0\le j\le 2k$。当 $n$ 如此表示时，$\lfloor\sqrt{n}\rfloor=k$，所给方程可写为 $$ k^2-(70k-j)+1000=0. $$ 对固定的 $k$，存在关于 $j$ 的解当且仅当 $$ k^2-70k+1000\le 0 \quad\text{且}\quad k^2-68k+1000\ge 0. $$ 解第一个不等式得 $20\le k\le 50$，解第二个得 $k\le 34-\sqrt{156}\approx 21\frac12$ 或 $k\ge 34+\sqrt{156}\approx 46\frac12$。因此合适的 $k$ 值为 $20, 21, 47, 48, 49, 50$；并且对每个这样的 $k$，唯一的 $j$ 值为 $(k-20)(50-k)$。因此所给方程对 $n$ 有 6 个解，分别为 $400, 470, 2290, 2360, 2430, 2500$。

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