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

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We are given that <cmath>\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor</cmath>
 
We are given that <cmath>\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor</cmath>
  
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<math>\lfloor\sqrt{n}\rfloor</math> must be an integer, which means that <math>n+1000</math> is divisible by <math>70</math>. As <math>1000\equiv 20\pmod{70}</math>, this means that <math>n\equiv 50\pmod{70}</math>, so we can write <math>n=70k+50</math> for an integer <math>k</math>.
+
<math>\lfloor\sqrt{n}\rfloor</math> must be an integer, which means that <math>n+1000</math> is divisible by <math>70</math>. As <math>1000\equiv 20\pmod{70}</math>, this means that <math>n\equiv 50\pmod{70}</math>, so we can write <math>n=70k+50</math> for <math>k\in\mathbb{Z}</math>.
  
 
Therefore, <cmath>\frac{n+1000}{70}=\frac{70k+1050}{70}=k+15=\lfloor\sqrt{70k+50}\rfloor</cmath>
 
Therefore, <cmath>\frac{n+1000}{70}=\frac{70k+1050}{70}=k+15=\lfloor\sqrt{70k+50}\rfloor</cmath>

Revision as of 23:27, 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 $k\in\mathbb{Z}$.

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 $\boxed{\textbf{(C)} 6}$.

-Solution By Qqqwerw

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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