2020 AMC 12B Problems/Problem 21

Problem

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 $x$ .)

$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$

Solution

Not a reliable or in-depth solution (for the guess and check students)

We can first consider the equation without a floor function:

$\dfrac{n+1000}{70} = \sqrt{n}$

Multiplying both sides by 70 and then squaring:

$n^2 + 2000n + 1000000 = 4900n$

Moving all terms to the left:

$n^2 - 2900n + 1000000 = 0$

Now we can use wishful thinking to determine the factors:

$(n-400)(n-2500) = 0$

This means that for $n = 400$ and $n = 2500$ , the equation will hold without the floor function.

Now we can simply check the multiples of 70 around 400 and 2500 in the original equation:

For $n = 330$ , left hand side = $19$ but $18^2 < 330 < 19^2$ so right hand side = $18$

For $n = 400$ , left hand side = $20$ and right hand side = $20$

For $n = 470$ , left hand side = $21$ and right hand side = $21$

For $n = 540$ , left hand side = $22$ but $540 > 23^2$ so right hand side = $23$

Now we move to $n = 2500$

For $n = 2430$ , left hand side = $49$ and $49^2 < 2430 < 50^2$ so right hand side = $49$

For $n = 2360$ , left hand side = $48$ and $48^2 < 2360 < 49^2$ so right hand side = $48$

For $n = 2290$ , left hand side = $47$ and $47^2 < 2360 < 48^2$ so right hand side = $47$

For $n = 2220$ , left hand side = $46$ but $47^2 < 2220$ so right hand side = $47$

For $n = 2500$ , left hand side = $50$ and right hand side = $50$

For $n = 2570$ , left hand side = $51$ but $2570 < 51^2$ so right hand side = $50$

Therefore we have 6 total solutions, $n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}$

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2020 AMC 12B Problems/Problem 21

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