2020 AMC 12B Problem 21

Below is the professionally curated solution for Problem 21 of the 2020 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:floor and ceiling functionssubstitutioninequality

Difficulty rating: 1800

21.

How many positive integers nn satisfy

n+100070=n?\frac{n + 1000}{70} = \lfloor \sqrt{n} \rfloor?

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

22

44

66

3030

3232

Solution:

The right side is an integer, so let k=n.k = \lfloor \sqrt{n} \rfloor. Then n=70k1000,n = 70k - 1000, and k=nk = \lfloor \sqrt{n} \rfloor requires k2n<(k+1)2.k^2 \le n \lt (k + 1)^2.

The lower bound k270k1000k^2 \le 70k - 1000 gives k270k+10000,k^2 - 70k + 1000 \le 0, i.e. 20k50.20 \le k \le 50. The upper bound 70k1000<(k+1)270k - 1000 \lt (k + 1)^2 gives k268k+1001>0,k^2 - 68k + 1001 \gt 0, i.e. k21k \le 21 or k47.k \ge 47.

Intersecting, k{20,21,47,48,49,50},k \in \{20, 21, 47, 48, 49, 50\}, giving 66 values of n.n.

Thus, the correct answer is C.

Problem 21 in Other Years