2020 AMC 10B Problem 24

Question

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.

)

2

4

6

30

32

Let

k=\lfloor\sqrt n\rfloor,

so

n=70k-1000

Use

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

to bound the possible integers

k

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Let

k=\left\lfloor\sqrt n\right\rfloor.

The equation gives

n=70k-1000.

Also, by the definition of the floor function,

k^2\le n<(k+1)^2.

Substituting

n=70k-1000,

we get

k^2\le 70k-1000<(k+1)^2.

The left inequality is

\begin{aligned} &k^2-70k+1000\le0 \\ &\quad \Longrightarrow (k-20)(k-50)\le0, \end{aligned}

so

20\le k\le50.

The right inequality is

\begin{aligned} &70k-1000<k^2+2k+1 \\ &\quad \Longrightarrow k^2-68k+1001>0. \end{aligned}

The roots of

k^2-68k+1001

are

34\pm\sqrt{155},

which are approximately

21.55

and

46.45.

Thus, together with

20\le k\le50,

the possible integer values are

k=20,21,47,48,49,50.

There are

6

such values.

Thus, C is the correct answer.

Problem 24 in other years