Question

How many positive integers $n$ satisfy

$\frac{n + 1000}{70} = \lfloor \sqrt{n} \rfloor?$

(Recall that $\lfloor x \rfloor$ is the greatest integer not exceeding $x.$ )

$2$

$4$

$6$

$30$

$32$

Answer: C

Concepts:floor and ceiling functions substitution inequality

Difficulty rating: 1800

Solution:

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

The lower bound $k^2 \le 70k - 1000$ gives $k^2 - 70k + 1000 \le 0,$ i.e. $20 \le k \le 50.$ The upper bound $70k - 1000 \lt (k + 1)^2$ gives $k^2 - 68k + 1001 \gt 0,$ i.e. $k \le 21$ or $k \ge 47.$

Intersecting, $k \in \{20, 21, 47, 48, 49, 50\},$ giving $6$ values of $n.$

Thus, the correct answer is C.

2020 AMC 12B Problem 21

Answer: C

Difficulty rating: 1800

Solution:

Problem 21 in Other Years