How many positive integers n satisfy (n + 1000)/(70) = ⌊ √n ⌋? (Recall that ⌊ x ⌋ is the greatest integer not exceeding x. )

2020 AMC 12B Problem 21

Attempt Problem 21 of the 2020 AMC 12B below, then check your answer against the professionally curated solution 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 problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

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

Let

k = \lfloor \sqrt{n} \rfloor,

n = 70k - 1000

must satisfy

k^2 \le n \lt (k + 1)^2

This gives

k^2 - 70k + 1000 \le 0

and

k^2 - 68k + 1001 \gt 0;

count integer

k

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

k \ge 47.

Intersecting,

k \in \{20, 21, 47, 48, 49, 50\},

giving

6

values of

n.

Thus, the correct answer is C.