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 $2 + 4 = \boxed{\textbf{(C) }6}$

~isabelchen

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)

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2020 AMC 10B Problems/Problem 24

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Video Solutions

Video Solution 1

Video Solution 2

Video Solution 3 by the Beauty of Math

See Also