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

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==Solution==
 
==Solution==
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*Not a reliable or in-depth solution (for the guess and check students)
 +
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We can first consider the equation without a floor function:
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 +
<cmath>\dfrac{n+1000}{70} = \sqrt{n} </cmath>
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 +
Multiplying both sides by 70 and then squaring:
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 +
<cmath>n^2 + 2000n + 1000000 = 4900n</cmath>
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 +
Moving all terms to the left:
 +
 +
<cmath>n^2 - 2900n + 1000000 = 0</cmath>
 +
 +
Now we can use wishful thinking to determine the factors:
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 +
<cmath>(n-400)(n-2500) = 0</cmath>
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 +
This means that for <math>n = 400</math> and <math>n = 2500</math>, 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 <math>n = 330</math>, left hand side = <math>19</math> but <math>18^2 < 330 < 19^2</math> so right hand side = <math>18</math>
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 +
For <math>n = 400</math>, left hand side = <math>20</math> and right hand side = <math>20</math>
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 +
For <math>n = 470</math>, left hand side = <math>21</math> and right hand side = <math>21</math>
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 +
For <math>n = 540</math>, left hand side = <math>22</math> but <math>540 > 23^2</math> so right hand side = <math>23</math>
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 +
Now we move to <math>n = 2500</math>
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For <math>n = 2430</math>, left hand side = <math>49</math> and <math>49^2 < 2430 < 50^2</math> so right hand side = <math>49</math>
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For <math>n = 2360</math>, left hand side = <math>48</math> and <math>48^2 < 2360 < 49^2</math> so right hand side = <math>48</math>
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For <math>n = 2290</math>, left hand side = <math>47</math> and <math>47^2 < 2360 < 48^2</math> so right hand side = <math>47</math>
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For <math>n = 2220</math>, left hand side = <math>46</math> but <math>47^2 < 2220</math> so right hand side = <math>47</math>
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 +
For <math>n = 2500</math>, left hand side = <math>50</math> and right hand side = <math>50</math>
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For <math>n = 2570</math>, left hand side = <math>51</math> but <math>2570 < 51^2</math> so right hand side = <math>50</math>
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Therefore we have 6 total solutions, <math>n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 23:54, 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, $n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}$

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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