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Difference between revisions of "2020 AMC 10B Problems/Problem 24"

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(Quick solution if you’re in a hurry)
 
(Quick solution if you’re in a hurry)
  
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First notice that the graphs of <math>(x+1000)/70</math> and <math>\sqrt[]{n}</math> intersect at 2 points. Then, notice that <math>(n+1000)/70</math> must be an integer. This means that n is congruent to 50 (mod 70).  
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First notice that the graphs of <math>(x+1000)/70</math> and <math>\sqrt[]{n}</math> intersect at 2 points. Then, notice that <math>(n+1000)/70</math> must be an integer. This means that n is congruent to <math>50 (mod 70)</math>.  
  
−
For the first intersection, testing the first few values of <math>n</math> (adding <math>70</math> to n each time and noticing the left side increases by <math>1</math> each time) yields <math>n=20</math> and <math>n=21</math>.
+
For the first intersection, testing the first few values of <math>n</math> (adding <math>70</math> to <math>n</math> each time and noticing the left side increases by <math>1</math> each time) yields <math>n=20</math> and <math>n=21</math>.
  
 
For the second intersection, using binary search can narrow down the other cases, being <math>n=47</math>, <math>n=48</math>, <math>n=49</math>, and <math>n=50</math>. This results in a total of 6 cases, for an answer of <math>\boxed{\textbf{(C) }6}</math>.
 
For the second intersection, using binary search can narrow down the other cases, being <math>n=47</math>, <math>n=48</math>, <math>n=49</math>, and <math>n=50</math>. This results in a total of 6 cases, for an answer of <math>\boxed{\textbf{(C) }6}</math>.

Revision as of 12:17, 8 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

(Quick solution if you’re in a hurry)

First notice that the graphs of $(x+1000)/70$ and $\sqrt[]{n}$ intersect at 2 points. Then, notice that $(n+1000)/70$ must be an integer. This means that n is congruent to $50 (mod 70)$.

For the first intersection, testing the first few values of $n$ (adding $70$ to $n$ each time and noticing the left side increases by $1$ each time) yields $n=20$ and $n=21$.

For the second intersection, using binary search can narrow down the other cases, being $n=47$, $n=48$, $n=49$, and $n=50$. This results in a total of 6 cases, for an answer of $\boxed{\textbf{(C) }6}$.

~DrJoyo

See Also

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 10 Problems and Solutions

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