# 2020 AMC 10B Problems/Problem 24

The following problem is from both the 2020 AMC 10B #24 and 2020 AMC 12B #21, so both problems redirect to this page.

## Problem

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

$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$

## Solution 1

We can first consider the equation without a floor function:

$$\dfrac{n+1000}{70} = \sqrt{n}$$

Multiplying both sides by 70 and then squaring:

$$n^2 + 2000n + 1000000 = 4900n$$

Moving all terms to the left:

$$n^2 - 2900n + 1000000 = 0$$

Now we can determine the factors:

$$(n-400)(n-2500) = 0$$

This means that for $n = 400$ and $n = 2500$, the equation will hold without the floor function.

Now we can simply check the multiples of 70 around 400 and 2500 in the original equation, which we abbreviate as $L=R$.

For $n = 330$, $L=19$ but $18^2 < 330 < 19^2$ so $R=18$

For $n = 400$, $L=20$ and $R=20$

For $n = 470$, $L=21$, $R=21$

For $n = 540$, $L=22$ but $540 > 23^2$ so $R=23$

Now we move to $n = 2500$

For $n = 2430$, $L=49$ and $49^2 < 2430 < 50^2$ so $R=49$

For $n = 2360$, $L=48$ and $48^2 < 2360 < 49^2$ so $R=48$

For $n = 2290$, $L=47$ and $47^2 < 2360 < 48^2$ so $R=47$

For $n = 2220$, $L=46$ but $47^2 < 2220$ so $R=47$

For $n = 2500$, $L=50$ and $R=50$

For $n = 2570$, $L=51$ but $2570 < 51^2$ so $R=50$

Therefore we have 6 total solutions, $n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}$

## Solution 2

This is my first solution here, so please forgive me for any errors.

We are given that $$\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor$$

$\lfloor\sqrt{n}\rfloor$ must be an integer, which means that $n+1000$ is divisible by $70$. As $1000\equiv 20\pmod{70}$, this means that $n\equiv 50\pmod{70}$, so we can write $n=70k+50$ for $k\in\mathbb{Z}$.

Therefore, $$\frac{n+1000}{70}=\frac{70k+1050}{70}=k+15=\lfloor\sqrt{70k+50}\rfloor$$

Also, we can say that $\sqrt{70k+50}-1 < k+15$ and $k+15\leq\sqrt{70k+50}$

Squaring the second inequality, we get $k^{2}+30k+225\leq70k+50\implies k^{2}-40k+175\leq 0\implies (k-5)(k-35)\leq0\implies 5\leq k\leq 35$.

Similarly solving the first inequality gives us $k < 19-\sqrt{155}$ or $k > 19+\sqrt{155}$

$\sqrt{155}$ is larger than $12$ and smaller than $13$, so instead, we can say $k\leq 6$ or $k\geq 32$.

Combining this with $5\leq k\leq 35$, we get $k=5,6,32,33,34,35$ are all solutions for $k$ that give a valid solution for $n$, meaning that our answer is $\boxed{\textbf{(C) 6}}$. -Solution By Qqqwerw

## Solution 3

We start with the given equation$$\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor$$From there, we can start with the general inequality that $\lfloor \sqrt{n} \rfloor \leq \sqrt{n} < \lfloor \sqrt{n} \rfloor + 1$. This means that$$\dfrac{n+1000}{70} \leq \sqrt{n} < \dfrac{n+1070}{70}$$Solving each inequality separately gives us two inequalities:$$n - 70\sqrt{n} +1000 \leq 0 \rightarrow (\sqrt{n}-50)(\sqrt{n}-20)\leq 0 \rightarrow 20\leq \sqrt{n} \leq 50$$$$n-70\sqrt{n}+1070 > 0 \rightarrow \sqrt{n} < 35-\sqrt{155} , \sqrt{n} > 35+\sqrt{155}$$Simplifying and approximating decimals yields 2 solutions for one inequality and 4 for the other. Hence, the answer is $2+4 = \boxed{\textbf{(C) } 6}$.

~Rekt4

## Solution 4

Let $n$ be uniquely of the form $n=k^2+r$ where $0 \le r \le 2k \; \bigstar$. Then, $$\frac{k^2+r+1000}{70} = k$$ Rearranging and completeing the square gives $$(k-35)^2 + r = 225$$ $$\Rightarrow r = (k-20)(50-k)\; \smiley$$ This gives us $$(k-35)^2 \le (k-35)^2+r=225 \le (k-35)^2 + 2k$$ Solving the left inequality shows that $20 \le k \le 50$. Combing this with the right inequality gives that $$(k-35)^2+r=225 \le (k-35)^2 + 2k \le (k-35)^2+100$$ which implies either $k \ge 47$ or $k \le 23$. By directly computing the cases for $k = 20, 21, 22, 23, 47, 48, 49, 50$ using $\smiley$, it follows that only $k = 22, 23$ yield and invalid $r$ from $\bigstar$. Since each $k$ corresponds to one $r$ and thus to one $n$ (from $\smiley$ and the original form), there must be 6 such $n$.

~the_jake314

## Solution 5

Since the right-hand-side is an integer, so must be the left-hand-side. Therefore, we must have $n\equiv -20\pmod{70}$; let $n=70j-20$. The given equation becomes$$j+14 = \lfloor \sqrt{70j-20} \rfloor$$

Since $\lfloor x \rfloor \leq x < \lfloor x \rfloor +1$ for all real $x$, we can take $x=\sqrt{70j-20}$ with $\lfloor x \rfloor =j+14$ to get $$j+14 \leq \sqrt{70j-20} < j+15$$ We can square the inequality to get$$196+28j+j^{2} \leq 70j-20 < 225 + 30j + j^{2}$$ The left inequality simplifies to $(j-36)(j-6) \leq 0$, which yields $$6 \le j \le 36.$$ The right inequality simplifies to $(j-20)^2 - 155 > 0$, which yields $$j < 20 - \sqrt{155} < 8 \quad \text{or} \quad j > 20 + \sqrt{155} > 32$$

Solving $j < 8$, and $6 \le j \le 36$, we get $6 \le j < 8$, for $2$ values $j\in \{6, 7\}$.

Solving $j >32$, and $6 \le j \le 36$, we get $32 < j \le 36$, for $4$ values $k\in \{33, \ldots , 36\}$.

Thus, our answer is $2 + 4 = \boxed{\textbf{(C) }6}$

~KingRavi

## Solution 6

Set $x=\sqrt{n}$ in the given equation and solve for $x$ to get $x^2 = 70 \cdot \lfloor x \rfloor - 1000$. Set $k = \lfloor x \rfloor \ge 0$; since $\lfloor x \rfloor^2 \le x^2 < (\lfloor x \rfloor + 1)^2$, we get $$k^2 \le 70k - 1000 < k^2 + 2k + 1.$$ The left inequality simplifies to $(k-20)(k-50) \le 0$, which yields $$20 \le k \le 50.$$ The right inequality simplifies to $(k-34)^2 > 155$, which yields $$k < 34 - \sqrt{155} < 22 \quad \text{or} \quad k > 34 + \sqrt{155} > 46$$ Solving $k < 22$, and $20 \le k \le 50$, we get $20 \le k < 22$, for $2$ values $k\in \{20, 21\}$.

Solving $k >46$, and $20 \le k \le 50$, we get $46 < k \le 50$, for $4$ values $k\in \{47, \ldots , 50\}$.

Thus, our answer is $2 + 4 = \boxed{\textbf{(C) }6}$

## Solution 7

If $n$ is a perfect square, we can write $n = k^2$ for a positive integer $k$, so $\lfloor \sqrt{n} \rfloor = \sqrt{n} = k.$ The given equation turns into

\begin{align*} \frac{k^2 + 1000}{70} &= k \\ k^2 - 70k + 1000 &= 0 \\ (k-20)(k-50) &= 0, \end{align*}

so $k = 20$ or $k= 50$, so $n = 400, 2500.$

If $n$ is not square, then we can say that, for a positive integer $k$, we have \begin{align*} k^2 < &n < (k+1)^2 \\ k^2 + 1000 < &n + 1000 = 70\lfloor \sqrt{n} \rfloor = 70k< (k+1)^2 + 1000 \\ k^2 + 1000 < &70k < (k+1)^2 + 1000. \end{align*}

To solve this inequality, we take the intersection of the two solution sets to each of the two inequalities $k^2 + 1000 < 70k$ and $70k < (k+1)^2 + 1000$. To solve the first one, we have

\begin{align*} k^2 - 70k + 1000 &< 0 \\ (k-20)(k-50) &< 0\\ \end{align*} $k\in (20, 50),$ because the portion of the parabola between its two roots will be negative.

The second inequality yields

\begin{align*} 70k &< k^2 + 2k + 1 + 1000 \\ 0 &< k^2 -68k + 1001. \end{align*} This time, the inequality will hold for all portions of the parabola that are not on or between the its two roots, which are $34 + \sqrt{155}>46$ and $34-\sqrt{155}<22$ (they are roughly equal, but this is to ensure that we do not miss any solutions).

Notation wise, we need all integers $k$ such that

$$k \in \left(20, 50\right) \cap \left(-\infty,34 - \sqrt{155} \right)$$ or $$k \in \left(20, 50\right) \cap \left(34 + \sqrt{155}, \infty \right).$$

For the first one, since our uppoer bound is a little less than $22$, the $k$ that works is $21$. For the second, our lower bound is a little more than $46$, so the $k$ that work are $47, 48,$ and $49$.

$\boxed{\textbf{(C) }6}$ total solutions for $n$, since each value of $k$ corresponds to exactly one value of $n$.

-Benedict T (countmath1)

## Video Solutions

### Video Solution 1

On The Spot STEM: https://youtu.be/BEJybl9TLMA