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

(Solution)
(Solution 1)
(45 intermediate revisions by 21 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #24]] and [[2020 AMC 12B Problems|2020 AMC 12B #21]]}}
 +
 
==Problem==
 
==Problem==
  
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>.)
+
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>
 
<math>\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32</math>
  
==Solution==
+
==Solution 1==
(Quick solution if you’re in a hurry)
+
First notice that the graphs of <math>(n+1000)/70</math> and <math>\sqrt[]{n}</math> intersect at 2 points. Then, notice that <math>(n+1000)/70</math> must be an integer, since it is equal to the floor of <math>n</math>. This means that n is congruent to <math>50 \pmod{70}</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>\lfloor \sqrt{n} \rfloor=20</math> and <math>\lfloor \sqrt{n} \rfloor=21</math>, so <math>n=400, 470</math> respectively. Estimating from the graph can narrow down the other cases, being <math>\lfloor \sqrt{n} \rfloor=47</math>, <math>\lfloor \sqrt{n} \rfloor=48</math>, <math>\lfloor \sqrt{n} \rfloor=49</math>, <math>\lfloor \sqrt{n} \rfloor=50</math>, yielding <math>n=2290,2360,2430,2500</math> respectively. This results in a total of 6 cases, for an answer of <math>\boxed{\textbf{(C) }6}</math>.
 +
 
 +
~DrJoyo (edited by eagleye and vaporwave)
 +
 
 +
==Solution 2 (Graphing)==
 +
 
 +
One intuitive approach to the question is graphing. Obviously, you should know what the graph of the square root function is, and if any function is floored (meaning it is taken to the greatest integer less than a value), a stair-like figure should appear. The other function is simply a line with a slope of <math>1/70</math>. If you precisely draw out the two regions of the graph where the derivative of the square function nears the derivative of the linear function, you can now deduce that <math>3</math> values of intersection lay closer to the left side of the stair, and <math>3</math> values lay closer to the right side of the stair.
 +
 
 +
With meticulous graphing, you can realize that the answer is <math>\boxed{\textbf{(C) }6}</math>.
 +
 
 +
A in-depth graph with intersection points is linked below.
 +
https://www.desmos.com/calculator/e5wk9adbuk
 +
 
 +
==Solution 3==
 +
 
 +
*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>
 +
 
 +
==Solution 4==
 +
 
 +
This is my first solution here, so please forgive me for any errors.
 +
 
 +
We are given that <cmath>\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor</cmath>
 +
 
 +
<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>
 +
 
 +
Also, we can say that <math>\sqrt{70k+50}-1 < k+15</math> and <math>k+15\leq\sqrt{70k+50}</math>
 +
 
 +
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 < 19-\sqrt{155}</math> or <math>k > 19+\sqrt{155}</math>
 +
 
 +
<math>\sqrt{155}</math> is larger than <math>12</math> and smaller than <math>13</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{\textbf{(C) 6}}</math>.
 +
-Solution By Qqqwerw
 +
 
 +
==Solution 5==
 +
 
 +
We start with the given equation<cmath>\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor</cmath>From there, we can start with the general inequality that <math>\lfloor \sqrt{n} \rfloor \leq \sqrt{n} < \lfloor \sqrt{n} \rfloor + 1</math>. This means that<cmath>\dfrac{n+1000}{70} \leq \sqrt{n} < \dfrac{n+1070}{70}</cmath>Solving each inequality separately gives us two inequalities:<cmath>n - 70\sqrt{n} +1000 \leq 0 \rightarrow (\sqrt{n}-50)(\sqrt{n}-20)\leq 0 \rightarrow 20\leq \sqrt{n} \leq 50</cmath><cmath>n-70\sqrt{n}+1070 > 0 \rightarrow \sqrt{n} < 35-\sqrt{155} , \sqrt{n} > 35+\sqrt{155}</cmath>Simplifying and approximating decimals yields 2 solutions for one inequality and 4 for the other. Hence <math>2+4 = \boxed{\textbf{(C)} 6}</math>.
 +
 
 +
~Rekt4
 +
 
 +
==Solution 6==
 +
 
 +
Let <math>n</math> be uniquely of the form <math>n=k^2+r</math> where <math>0 \le r \le 2k \; \bigstar</math>. Then, <cmath> \frac{k^2+r+1000}{70} = k</cmath> Rearranging and completeing the square gives <cmath>(k-35)^2 + r = 225</cmath> <cmath>\Rightarrow r = (k-20)(50-k)\; \smiley</cmath> This gives us <cmath>(k-35)^2 \le (k-35)^2+r=225 \le (k-35)^2 + 2k</cmath> Solving the left inequality shows that <math>20 \le k \le 50</math>. Combing this with the right inequality gives that <cmath>(k-35)^2+r=225 \le (k-35)^2 + 2k \le (k-35)^2+100 </cmath> which implies either <math>k \ge 47</math> or <math>k \le 23</math>. By directly computing the cases for <math>k = 20, 21, 22, 23, 47, 48, 49, 50</math> using <math>\smiley</math>, it follows that only <math> k = 22, 23</math> yield and invalid <math>r</math> from <math>\bigstar</math>. Since each <math>k</math> corresponds to one <math>r</math> and thus to one <math>n</math> (from <math>\smiley</math> and the original form), there must be 6 such <math>n</math>. 
  
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).
 
  
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>.
+
~the_jake314
  
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>.
+
==Video Solutions==
 +
===Video Solution 1===
 +
On The Spot STEM:
 +
https://youtu.be/BEJybl9TLMA
  
~DrJoyo
+
===Video Solution 2===
 +
https://www.youtube.com/watch?v=VWeioXzQxVA&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9 ~ MathEx
 +
 
 +
===Video Solution 3 by the Beauty of Math===
 +
https://youtu.be/4RVYoeiyC4w?t=62
  
 
==See Also==  
 
==See Also==  
  
 
{{AMC10 box|year=2020|ab=B|num-b=23|num-a=25}}
 
{{AMC10 box|year=2020|ab=B|num-b=23|num-a=25}}
 +
{{AMC12 box|year=2020|ab=B|num-b=20|num-a=22}}
 +
 +
[[Category:Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:57, 24 October 2021

The following problem is from both the 2020 AMC 10B #24 and 2020 AMC 12B #21, so both problems redirect to this page.

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 1

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

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 $\lfloor \sqrt{n} \rfloor=20$ and $\lfloor \sqrt{n} \rfloor=21$, so $n=400, 470$ respectively. Estimating from the graph can narrow down the other cases, being $\lfloor \sqrt{n} \rfloor=47$, $\lfloor \sqrt{n} \rfloor=48$, $\lfloor \sqrt{n} \rfloor=49$, $\lfloor \sqrt{n} \rfloor=50$, yielding $n=2290,2360,2430,2500$ respectively. This results in a total of 6 cases, for an answer of $\boxed{\textbf{(C) }6}$.

~DrJoyo (edited by eagleye and vaporwave)

Solution 2 (Graphing)

One intuitive approach to the question is graphing. Obviously, you should know what the graph of the square root function is, and if any function is floored (meaning it is taken to the greatest integer less than a value), a stair-like figure should appear. The other function is simply a line with a slope of $1/70$. If you precisely draw out the two regions of the graph where the derivative of the square function nears the derivative of the linear function, you can now deduce that $3$ values of intersection lay closer to the left side of the stair, and $3$ values lay closer to the right side of the stair.

With meticulous graphing, you can realize that the answer is $\boxed{\textbf{(C) }6}$.

A in-depth graph with intersection points is linked below. https://www.desmos.com/calculator/e5wk9adbuk

Solution 3

  • 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 4

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 $\sqrt{70k+50}-1 < k+15$ 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 $k < 19-\sqrt{155}$ or $k > 19+\sqrt{155}$

$\sqrt{155}$ is larger than $12$ and smaller than $13$, 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

Solution 5

We start with the given equation\[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor\]From there, we can start with the general inequality that $\lfloor \sqrt{n} \rfloor \leq \sqrt{n} < \lfloor \sqrt{n} \rfloor + 1$. This means that\[\dfrac{n+1000}{70} \leq \sqrt{n} < \dfrac{n+1070}{70}\]Solving each inequality separately gives us two inequalities:\[n - 70\sqrt{n} +1000 \leq 0 \rightarrow (\sqrt{n}-50)(\sqrt{n}-20)\leq 0 \rightarrow 20\leq \sqrt{n} \leq 50\]\[n-70\sqrt{n}+1070 > 0 \rightarrow \sqrt{n} < 35-\sqrt{155} , \sqrt{n} > 35+\sqrt{155}\]Simplifying and approximating decimals yields 2 solutions for one inequality and 4 for the other. Hence $2+4 = \boxed{\textbf{(C)} 6}$.

~Rekt4

Solution 6

Let $n$ be uniquely of the form $n=k^2+r$ where $0 \le r \le 2k \; \bigstar$. Then, \[\frac{k^2+r+1000}{70} = k\] Rearranging and completeing the square gives \[(k-35)^2 + r = 225\] \[\Rightarrow r = (k-20)(50-k)\; \smiley\] This gives us \[(k-35)^2 \le (k-35)^2+r=225 \le (k-35)^2 + 2k\] Solving the left inequality shows that $20 \le k \le 50$. Combing this with the right inequality gives that \[(k-35)^2+r=225 \le (k-35)^2 + 2k \le (k-35)^2+100\] which implies either $k \ge 47$ or $k \le 23$. By directly computing the cases for $k = 20, 21, 22, 23, 47, 48, 49, 50$ using $\smiley$, it follows that only $k = 22, 23$ yield and invalid $r$ from $\bigstar$. Since each $k$ corresponds to one $r$ and thus to one $n$ (from $\smiley$ and the original form), there must be 6 such $n$.


~the_jake314

Video Solutions

Video Solution 1

On The Spot STEM: https://youtu.be/BEJybl9TLMA

Video Solution 2

https://www.youtube.com/watch?v=VWeioXzQxVA&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9 ~ MathEx

Video Solution 3 by the Beauty of Math

https://youtu.be/4RVYoeiyC4w?t=62

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
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
2020 AMC 12B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png