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

Not a reliable or in-depth solution (for the guess and check students)

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 use wishful thinking to 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)

Right away, we realize that since the floor function returns only integers, $\frac{1000 + n}{70}$ is also an integer.

$1000\equiv 20\pmod{70}$ , so $n\equiv 50\pmod{70}$ to make the numerator divisible by 70.

Let $S$ be the sequence of all positive integers $n$ such that $n\equiv 50\pmod{70}$ , and let $S_j$ be the $j^{th}$ element of $S$ , where $j \in \mathbb{N}$

As $S$ is an arithmetic sequence with initial term 50 and common difference 70, the explicit formula of the sequence is $S_j = 50 + 70(j-1) = 70j - 20$ .

$70j-20$ calculates $n$ for the $j^{th}$ element in the sequence, so we can plug in this formula for $n$ ;

$\frac{1000 + 70j - 20}{70} = \lfloor \sqrt{70j-20} \rfloor$

Now we can simplify the left-hand-side:

$\frac{1000+70j-20}{70} = \frac{1000}{70} + \frac{70j-20}{70} = 14\frac{2}{7} + \frac{70j}{70} - \frac{20}{70} = 14 + \frac{2}{7} + j - \frac{2}{7} = 14 + j$

We know that for any x, $\lfloor x \rfloor \leq x < \lfloor x \rfloor +1$ . Therefore:

$14+j \leq \sqrt{70j-20} < 15+j$

We can square the inequality to get $196+28j+j^{2} \leq 70j-20 < 225 + 30j + j^{2}$

We can split this three-part inequality into two inequalities:

$j^{2} + 28j + 196 \leq 70j-20$ $j^{2} + 30j + 225 > 70j-20$

Now we can solve for each inequality and bound $j$ ;

$j^{2} + 28j + 196 \leq 70j - 20 \implies j^{2}-42j+216 \leq 0$ $\implies (j-36)(j-6) \leq 0 \implies 6 \leq j \leq 36$ (by considering signs of $j-36$ and $j-6$ ).

The next inequality cannot be factored so we will have to use the quadratic equation to find j: $j^{2} +30j+225 > 70j - 20 \implies j^{2}-40j+245 > 0$ To find j, we treat the inequality as an equation: this becomes $j^{2} - 40j + 245 = 0 \implies j = \frac{40 \pm \sqrt{40^{2} - 4\cdot245}}{2}$ $\implies j = \frac{40 \pm \sqrt{1600-980}}{2} \implies j = \frac{40 \pm \sqrt{620}}{2} \implies j=20 \pm \sqrt{155}$

Now we can turn it back into an inequality. These roots are a bit harder to work with than the first inequality, but we recognize that the factored form becomes $(j-(20+\sqrt{155}))(j-(20-\sqrt{155})) > 0$ , and we can use signs as before to determine that the solutions for this inequality are

$j < 20-\sqrt{155}$ and $j > 20+\sqrt{155}$ .

Because $j$ can only have positive integer values, we can approximate $20-\sqrt{155}$ as between 7 and 8, and we can approximate $20 + \sqrt{155}$ as between 32 and 33.

We can then write the stricter inequalities $j \leq 7$ and $j \geq 33$ .

Overall, $6 \leq j \leq 36$ , $j \leq 7$ , and $j \geq 33$ , so the only possible solutions for $j$ are 6, 7, 33, 34, 35, and 36. Because each valid solution for $j$ is a valid solution for $n$ , we have 6 solutions for $n$ that satisfy the given equation, or $\boxed{\textbf{(C)} 6}$ .

~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 $k < 34 - \sqrt{155} \approx 22 \quad \text{or} \quad k > 34 + \sqrt{155} \approx 46$ 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 • Answer Key • Resources)
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2020 AMC 10B Problems/Problem 24

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