# Difference between revisions of "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

## 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 $\boxed{\textbf{(C) }6}$.

 2020 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions