2020 AMC 10B Problems/Problem 24

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

(Quick solution if you’re in a hurry)

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

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 $n=20$ and $n=21$ .

For the second intersection, using binary search can narrow down the other cases, being $n=47$ , $n=48$ , $n=49$ , and $n=50$ . This results in a total of 6 cases, for an answer of $\boxed{\textbf{(C) }6}$ .

~DrJoyo

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2020 AMC 10B Problems/Problem 24

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