Difference between revisions of "2020 AMC 12B Problems/Problem 21"

Revision as of 23:25, 7 February 2020

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, $n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}$

Solution 2

This is my first solution here, so please forgive me for any errors.

We are given that $\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor$

$\lfloor\sqrt{n}\rfloor$ must be an integer, which means that $n+1000$ is divisible by $70$ As $1000\equiv 20\pmod{70}$ , this means that $n\equiv 50\pmod{70}$ , so we can write $n=70k+50$ for an integer $k$ .

Therefore, $\frac{n+1000}{70}=\frac{70k+1050}{70}=k+15=\lfloor\sqrt{70k+50}\rfloor$

Also, we can say that $\sqrt{70k+50}-1\leq k+15$ and $k+15\leq\sqrt{70k+50}$

Squaring the second inequality, we get $k^{2}+30k+225\leq70k+50\implies k^{2}-40k+175\leq 0\implies (k-5)(k-35)\leq0\implies 5\leq k\leq 35$ .

Similarly solving the first inequality gives us $k\leq 19-\sqrt{155}$ or $k\geq 19+\sqrt{155}$

$\sqrt{155}$ is slightly greater than $12$ , so instead, we can say $k\leq 6$ or $k\geq 32$ .

Combining this with $5\leq k\leq 35$ , we get $k=5,6,32,33,34,35$ are all solutions for $k$ that give a valid solution for $n$ , meaning that our answer is $\boxed{6}$ .

@@ Line 65: / Line 65: @@
 Squaring the second inequality, we get <math>k^{2}+30k+225\leq70k+50\implies k^{2}-40k+175\leq 0\implies (k-5)(k-35)\leq0\implies 5\leq k\leq 35</math>.
+Similarly solving the first inequality gives us <math>k\leq 19-\sqrt{155}</math> or <math>k\geq 19+\sqrt{155}</math>
+<math>\sqrt{155}</math> is slightly greater than <math>12</math>, so instead, we can say <math>k\leq 6</math> or <math>k\geq 32</math>.
+Combining this with <math>5\leq k\leq 35</math>, we get <math>k=5,6,32,33,34,35</math> are all solutions for <math>k</math> that give a valid solution for <math>n</math>, meaning that our answer is <math>\boxed{6}</math>.
 ==See Also==

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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Difference between revisions of "2020 AMC 12B Problems/Problem 21"

Revision as of 23:25, 7 February 2020

Contents

Problem

Solution

Solution 2

See Also