2020 AMC 10B Problem 24

P21-25~8 minPrint

Number Properties Absolute Value & Inequalities Quadratics Bound the quantity above/below or estimate to pin it down Substitute to simplify (u = x+1/x, shifting, scaling)

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 $x$ .)

Show solution without answering

Key insight

Call the common value k. Then n = 70k - 1000, and n must be at least k squared but less than (k+1) squared. Solve both conditions.

Solution

1. Name the integer. $\lfloor \sqrt n \rfloor$ is a whole number, so call it $k$ . Then $\frac{n + 1000}{70} = k$ , so
$n = 70k - 1000.$
2. What " $\lfloor \sqrt n \rfloor = k$ " means. It means $\sqrt n$ is at least $k$ but less than $k + 1$ . Squaring, $n$ is at least $k^2$ but less than $(k+1)^2$ :
$k^2 \le 70k - 1000 < (k + 1)^2.$
These are two conditions, and we need both.
3. First condition: $k^2 \le 70k - 1000$ , that is, $k^2 - 70k + 1000 \le 0$ . This factors as $(k - 20)(k - 50) \le 0$ , which holds exactly for $20 \le k \le 50$ .
4. Second condition: $70k - 1000 < k^2 + 2k + 1$ , that is, $k^2 - 68k + 1001 > 0$ . Rather than solve it exactly, test values near the ends of $20$ to $50$ :

$k$	$n = 70k - 1000$	$(k+1)^2$	$n < (k+1)^2$ ?
$20$	$400$	$441$	yes
$21$	$470$	$484$	yes
$22$	$540$	$529$	no
$46$	$2220$	$2209$	no
$47$	$2290$	$2304$	yes
$48$	$2360$	$2401$	yes
$49$	$2430$	$2500$	yes
$50$	$2500$	$2601$	yes

Why the middle fails. $n$ goes up by $70$ each time $k$ goes up by $1$ , while $(k+1)^2$ goes up by $2k + 3$ . So the gap $(k+1)^2 - n$ keeps shrinking while $2k + 3 < 70$ and keeps growing after that. It is negative from $k = 22$ to $k = 46$ (both ends checked in the table) and positive outside. (Solving $k^2 - 68k + 1001 = 0$ gives $k \approx 21.6$ and $k \approx 46.4$ , which agrees.)
Collect. $k = 20, 21, 47, 48, 49, 50$ : six values, each giving a different positive $n$ : $400, 470, 2290, 2360, 2430, 2500$ .

The answer is $\boxed{\textbf{(C)}\ 6}$ .

Why this works

An equation with a floor is really two inequalities: at least $k^2$ , and less than $(k+1)^2$ . Naming the whole-number value $k$ makes $n$ a simple expression in $k$ . The straight line $70k - 1000$ climbs at a steady rate while the squares speed up, so the solutions sit in two short groups at the ends, not one long run. The choices $30$ and $32$ are there to catch anyone who checks only the first inequality.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The right side is an integer, so let $k = \lfloor \sqrt n \rfloor = \dfrac{n + 1000}{70}$ . Then $n = 70k - 1000$ , and the floor condition $k \le \sqrt n < k + 1$ becomes
$k^2 \le 70k - 1000 < (k+1)^2.$

Left inequality: $k^2 - 70k + 1000 \le 0$ . The roots of $k^2 - 70k + 1000$ are $k = \frac{70 \pm \sqrt{4900 - 4000}}{2} = \frac{70 \pm 30}{2} = 20, 50$ , so $20 \le k \le 50$ .

Right inequality: $70k - 1000 < k^2 + 2k + 1$ , i.e. $k^2 - 68k + 1001 > 0$ . The roots are $k = \frac{68 \pm \sqrt{4624 - 4004}}{2} = 34 \pm \frac{\sqrt{620}}{2}$ . Since $\sqrt{620} \approx 24.9$ , the roots are about $21.55$ and $46.45$ , so we need $k \le 21$ or $k \ge 47$ .

Combining: $k \in \{20, 21, 47, 48, 49, 50\}$ , six values. Each gives a positive $n = 70k - 1000$ (namely $400, 470, 2290, 2360, 2430, 2500$ ), and distinct $k$ give distinct $n$ .

Spot check: $k = 21$ gives $n = 470$ , $\sqrt{470} \approx 21.7$ , floor $21$ . Good. $k = 22$ gives $n = 540$ , $\sqrt{540} \approx 23.2$ , floor $23 \ne 22$ . Fails, as predicted.

The answer is $\boxed{\textbf{(C)}\ 6}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.2, §10.3, §10.4)
Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)