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

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==Problem==
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#REDIRECT [[2020 AMC 10B Problems/Problem 24]]
How many positive integers <math>n</math> satisfy<cmath>\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?</cmath>(Recall that <math>\lfloor x\rfloor</math> is the greatest integer not exceeding <math>x</math>.)
 
 
 
<math>\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32</math>
 
 
 
==Solution==
 
 
 
*Not a reliable or in-depth solution (for the guess and check students)
 
 
 
We can first consider the equation without a floor function:
 
 
 
<cmath>\dfrac{n+1000}{70} = \sqrt{n} </cmath>
 
 
 
Multiplying both sides by 70 and then squaring:
 
 
 
<cmath>n^2 + 2000n + 1000000 = 4900n</cmath>
 
 
 
Moving all terms to the left:
 
 
 
<cmath>n^2 - 2900n + 1000000 = 0</cmath>
 
 
 
Now we can use wishful thinking to determine the factors:
 
 
 
<cmath>(n-400)(n-2500) = 0</cmath>
 
 
 
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>
 
 
 
For <math>n = 400</math>, left hand side = <math>20</math> and right hand side = <math>20</math>
 
 
 
For <math>n = 470</math>, left hand side = <math>21</math> and right hand side = <math>21</math>
 
 
 
For <math>n = 540</math>, left hand side = <math>22</math> but <math>540 > 23^2</math> so right hand side = <math>23</math>
 
 
 
Now we move to <math>n = 2500</math>
 
 
 
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>
 
 
 
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>
 
 
 
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>
 
 
 
For <math>n = 2220</math>, left hand side = <math>46</math> but <math>47^2 < 2220</math> so right hand side = <math>47</math>
 
 
 
For <math>n = 2500</math>, left hand side = <math>50</math> and right hand side = <math>50</math>
 
 
 
For <math>n = 2570</math>, left hand side = <math>51</math> but <math>2570 < 51^2</math> so right hand side = <math>50</math>
 
 
 
Therefore we have 6 total solutions, <math>n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}</math>
 
 
 
==See Also==
 
 
 
{{AMC12 box|year=2020|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 

Revision as of 16:34, 8 February 2020