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

First notice that the graphs of $(n+1000)/70$ and $\sqrt[]{n}$ intersect at 2 points. Then, notice that $(n+1000)/70$ must be an integer, since it is equal to the floor of $n$ . This means that n is congruent to $50 \pmod{70}$ .

For the first intersection, testing the first few values of $n$ (adding $70$ to $n$ each time and noticing the left side increases by $1$ each time) yields $\lfloor \sqrt{n} \rfloor=20$ and $\lfloor \sqrt{n} \rfloor=21$ , so $n=400, 470$ respectively. Estimating from the graph can narrow down the other cases, being $\lfloor \sqrt{n} \rfloor=47$ , $\lfloor \sqrt{n} \rfloor=48$ , $\lfloor \sqrt{n} \rfloor=49$ , $\lfloor \sqrt{n} \rfloor=50$ , yielding $n=2290,2360,2430,2500$ respectively. This results in a total of 6 cases, for an answer of $\boxed{\textbf{(C) }6}$ .

~DrJoyo (edited by eagleye and vaporwave)

Solution 2 (Graphing)

One intuitive approach to the question is graphing. Obviously, you should know what the graph of the square root function is, and if any function is floored (meaning it is taken to the greatest integer less than a value), a stair-like figure should appear. The other function is simply a line with a slope of $1/70$ . If you precisely draw out the two regions of the graph where the derivative of the square function nears the derivative of the linear function, you can now deduce that $3$ values of intersection lay closer to the left side of the stair, and $3$ values lay closer to the right side of the stair.

With meticulous graphing, you can realize that the answer is $\boxed{\textbf{(C) }6}$ .

A in-depth graph with intersection points is linked below. https://www.desmos.com/calculator/e5wk9adbuk

Solution 3

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:

For $n = 330$ , left hand side $=19$ but $18^2 < 330 < 19^2$ so right hand side $=18$

For $n = 400$ , left hand side $=20$ and right hand side $=20$

For $n = 470$ , left hand side $=21$ and right hand side $=21$

For $n = 540$ , left hand side $=22$ but $540 > 23^2$ so right hand side $=23$

Now we move to $n = 2500$

For $n = 2430$ , left hand side $=49$ and $49^2 < 2430 < 50^2$ so right hand side $=49$

For $n = 2360$ , left hand side $=48$ and $48^2 < 2360 < 49^2$ so right hand side $=48$

For $n = 2290$ , left hand side $=47$ and $47^2 < 2360 < 48^2$ so right hand side $=47$

For $n = 2220$ , left hand side $=46$ but $47^2 < 2220$ so right hand side $=47$

For $n = 2500$ , left hand side $=50$ and right hand side $=50$

For $n = 2570$ , left hand side $=51$ but $2570 < 51^2$ so right hand side $=50$

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

Solution 4

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 5

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

~Rekt4

Solution 6

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 7 (Sequences - Rigorous)

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 8 (General Equation)

To solve an equation of the type $f(x, \lfloor x \rfloor) = 0$ , we proceed as follows.

. .
. .
. .

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

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2020 AMC 10B Problems/Problem 24

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