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

Latest revision as of 09:26, 9 March 2024

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

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 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, which we abbreviate as $L=R$ .

For $n = 330$ , $L=19$ but $18^2 < 330 < 19^2$ so $R=18$

For $n = 400$ , $L=20$ and $R=20$

For $n = 470$ , $L=21$ , $R=21$

For $n = 540$ , $L=22$ but $540 > 23^2$ so $R=23$

Now we move to $n = 2500$

For $n = 2430$ , $L=49$ and $49^2 < 2430 < 50^2$ so $R=49$

For $n = 2360$ , $L=48$ and $48^2 < 2360 < 49^2$ so $R=48$

For $n = 2290$ , $L=47$ and $47^2 < 2360 < 48^2$ so $R=47$

For $n = 2220$ , $L=46$ but $47^2 < 2220$ so $R=47$

For $n = 2500$ , $L=50$ and $R=50$

For $n = 2570$ , $L=51$ but $2570 < 51^2$ so $R=50$

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 $\sqrt{70k+50}-1 < k+15$ 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 3

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, the answer is $2+4 = \boxed{\textbf{(C) } 6}$ .

~Rekt4

Solution 4

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

Solution 5

Since the right-hand-side is an integer, so must be the left-hand-side. Therefore, we must have $n\equiv -20\pmod{70}$ ; let $n=70j-20$ . The given equation becomes $j+14 = \lfloor \sqrt{70j-20} \rfloor$

Since $\lfloor x \rfloor \leq x < \lfloor x \rfloor +1$ for all real $x$ , we can take $x=\sqrt{70j-20}$ with $\lfloor x \rfloor =j+14$ to get $j+14 \leq \sqrt{70j-20} < j+15$ We can square the inequality to get $196+28j+j^{2} \leq 70j-20 < 225 + 30j + j^{2}$ The left inequality simplifies to $(j-36)(j-6) \leq 0$ , which yields $6 \le j \le 36.$ The right inequality simplifies to $(j-20)^2 - 155 > 0$ , which yields $j < 20 - \sqrt{155} < 8 \quad \text{or} \quad j > 20 + \sqrt{155} > 32$

Solving $j < 8$ , and $6 \le j \le 36$ , we get $6 \le j < 8$ , for $2$ values $j\in \{6, 7\}$ .

Solving $j >32$ , and $6 \le j \le 36$ , we get $32 < j \le 36$ , for $4$ values $k\in \{33, \ldots , 36\}$ .

Thus, our answer is $2 + 4 = \boxed{\textbf{(C) }6}$

~KingRavi

Solution 6

Set $x=\sqrt{n}$ in the given equation and solve for $x$ to get $x^2 = 70 \cdot \lfloor x \rfloor - 1000$ . Set $k = \lfloor x \rfloor \ge 0$ ; since $\lfloor x \rfloor^2 \le x^2 < (\lfloor x \rfloor + 1)^2$ , we get $k^2 \le 70k - 1000 < k^2 + 2k + 1.$ The left inequality simplifies to $(k-20)(k-50) \le 0$ , which yields $20 \le k \le 50.$ The right inequality simplifies to $(k-34)^2 > 155$ , which yields $k < 34 - \sqrt{155} < 22 \quad \text{or} \quad k > 34 + \sqrt{155} > 46$ Solving $k < 22$ , and $20 \le k \le 50$ , we get $20 \le k < 22$ , for $2$ values $k\in \{20, 21\}$ .

Solving $k >46$ , and $20 \le k \le 50$ , we get $46 < k \le 50$ , for $4$ values $k\in \{47, \ldots , 50\}$ .

Thus, our answer is $2 + 4 = \boxed{\textbf{(C) }6}$

~isabelchen

Solution 7

If $n$ is a perfect square, we can write $n = k^2$ for a positive integer $k$ , so $\lfloor \sqrt{n} \rfloor = \sqrt{n} = k.$ The given equation turns into

\begin{align*} \frac{k^2 + 1000}{70} &= k \\ k^2 - 70k + 1000 &= 0 \\ (k-20)(k-50) &= 0, \end{align*}

so $k = 20$ or $k= 50$ , so $n = 400, 2500.$

If $n$ is not square, then we can say that, for a positive integer $k$ , we have \begin{align*} k^2 < &n < (k+1)^2 \\ k^2 + 1000 < &n + 1000 = 70\lfloor \sqrt{n} \rfloor = 70k< (k+1)^2 + 1000 \\ k^2 + 1000 < &70k < (k+1)^2 + 1000. \end{align*}

To solve this inequality, we take the intersection of the two solution sets to each of the two inequalities $k^2 + 1000 < 70k$ and $70k < (k+1)^2 + 1000$ . To solve the first one, we have

\begin{align*} k^2 - 70k + 1000 &< 0 \\ (k-20)(k-50) &< 0\\ \end{align*} $k\in (20, 50),$ because the portion of the parabola between its two roots will be negative.

The second inequality yields

\begin{align*} 70k &< k^2 + 2k + 1 + 1000 \\ 0 &< k^2 -68k + 1001. \end{align*} This time, the inequality will hold for all portions of the parabola that are not on or between the its two roots, which are $34 + \sqrt{155}>46$ and $34-\sqrt{155}<22$ (they are roughly equal, but this is to ensure that we do not miss any solutions).

Notation wise, we need all integers $k$ such that

$k \in \left(20, 50\right) \cap \left(-\infty,34 - \sqrt{155} \right)$ or $k \in \left(20, 50\right) \cap \left(34 + \sqrt{155}, \infty \right).$

For the first one, since our uppoer bound is a little less than $22$ , the $k$ that works is $21$ . For the second, our lower bound is a little more than $46$ , so the $k$ that work are $47, 48,$ and $49$ .

$\boxed{\textbf{(C) }6}$ total solutions for $n$ , since each value of $k$ corresponds to exactly one value of $n$ .

-Benedict T (countmath1)

Video Solutions

@@ Line 1: / Line 1: @@
+{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #24]] and [[2020 AMC 12B Problems|2020 AMC 12B #21]]}}
 ==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>
-==Solution==
+==Solution 1==
-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 <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>.
+We can first consider the equation without a floor function:
-For the second intersection, using binary search and/or estimating from the graph 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>.
+<cmath>\dfrac{n+1000}{70} = \sqrt{n} </cmath>
-~DrJoyo
+Multiplying both sides by 70 and then squaring:
-==Solution 2 (Graphing)==
+<cmath>n^2 + 2000n + 1000000 = 4900n</cmath>
-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.
+Moving all terms to the left:
-With meticulous graphing, you can realize that the answer is <math>\boxed{\textbf{(C) }6}</math>.
+<cmath>n^2 - 2900n + 1000000 = 0</cmath>
+Now we can 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, which we abbreviate as <math>L=R</math>.
+For <math>n = 330</math>, <math>L=19</math> but <math>18^2 < 330 < 19^2</math> so <math>R=18</math>
+For <math>n = 400</math>,  <math>L=20</math> and  <math>R=20</math>
+For <math>n = 470</math>,  <math>L=21</math>,  <math>R=21</math>
+For <math>n = 540</math>, <math>L=22</math> but <math>540 > 23^2</math> so   <math>R=23</math>
+Now we move to <math>n = 2500</math>
+For <math>n = 2430</math>,  <math>L=49</math> and <math>49^2 < 2430 < 50^2</math> so <math>R=49</math>
+For <math>n = 2360</math>,  <math>L=48</math> and <math>48^2 < 2360 < 49^2</math> so <math>R=48</math>
+For <math>n = 2290</math>,  <math>L=47</math> and <math>47^2 < 2360 < 48^2</math> so  <math>R=47</math>
+For <math>n = 2220</math>, <math>L=46</math> but <math>47^2 < 2220</math> so  <math>R=47</math>
+For <math>n = 2500</math>,  <math>L=50</math> and <math>R=50</math>
+For <math>n = 2570</math>, <math>L=51</math> but <math>2570 < 51^2</math> so <math>R=50</math>
+Therefore we have 6 total solutions, <math>n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}</math>
+==Solution 2==
+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 3==
+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, the answer is <math>2+4 = \boxed{\textbf{(C) } 6}</math>.
+~Rekt4
+==Solution 4==
+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>.
+~the_jake314
+==Solution 5==
+Since the right-hand-side is an integer, so must be the left-hand-side. Therefore, we must have <math>n\equiv -20\pmod{70}</math>; let <math>n=70j-20</math>. The given equation becomes<cmath>j+14 = \lfloor \sqrt{70j-20} \rfloor</cmath>
+Since <math>\lfloor x \rfloor \leq x < \lfloor x \rfloor +1</math> for all real <math>x</math>, we can take <math>x=\sqrt{70j-20}</math> with <math>\lfloor x \rfloor =j+14</math> to get
+<cmath>j+14 \leq \sqrt{70j-20} < j+15</cmath>
+We can square the inequality to get<cmath>196+28j+j^{2} \leq 70j-20 < 225 + 30j + j^{2}</cmath>
+The left inequality simplifies to <math>(j-36)(j-6) \leq 0</math>, which yields <cmath>6 \le j \le 36.</cmath>
+The right inequality simplifies to <math>(j-20)^2 - 155 > 0</math>, which yields <cmath>j < 20 - \sqrt{155} < 8 \quad \text{or} \quad j > 20 + \sqrt{155} > 32</cmath>
+Solving <math>j < 8</math>, and <math>6 \le j \le 36</math>, we get <math>6 \le j < 8</math>, for <math>2</math> values <math>j\in \{6, 7\}</math>.
+Solving <math>j >32</math>, and <math>6 \le j \le 36</math>, we get <math>32 < j \le 36</math>, for <math>4</math> values <math>k\in \{33, \ldots , 36\}</math>.
+Thus, our answer is <math>2 + 4 = \boxed{\textbf{(C) }6}</math>
+~KingRavi
+==Solution 6 ==
+Set <math>x=\sqrt{n}</math> in the given equation and solve for <math>x</math> to get <math>x^2 = 70 \cdot \lfloor x \rfloor - 1000</math>. Set <math>k = \lfloor x \rfloor \ge 0</math>; since <math>\lfloor x \rfloor^2 \le x^2 < (\lfloor x \rfloor + 1)^2</math>, we get  <cmath>k^2 \le  70k - 1000 < k^2 + 2k + 1.</cmath>
+The left inequality simplifies to <math>(k-20)(k-50) \le 0</math>, which yields <cmath>20 \le k \le 50.</cmath>
+The right inequality simplifies to <math>(k-34)^2 > 155</math>, which yields <cmath>k < 34 - \sqrt{155} < 22 \quad \text{or} \quad k > 34 + \sqrt{155} > 46</cmath>
+Solving <math>k < 22</math>, and <math>20 \le k \le 50</math>, we get <math>20 \le k < 22</math>, for <math>2</math> values <math>k\in \{20, 21\}</math>.
+Solving <math>k >46</math>, and <math>20 \le k \le 50</math>, we get <math>46 < k \le 50</math>, for <math>4</math> values <math>k\in \{47, \ldots , 50\}</math>.
+Thus, our answer is <math>2 + 4 = \boxed{\textbf{(C) }6}</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
+==Solution 7 ==
+If <math>n</math> is a perfect square, we can write <math>n = k^2</math> for a positive integer <math>k</math>, so <math>\lfloor \sqrt{n} \rfloor = \sqrt{n} = k.</math> The given equation turns into
+\begin{align*}
+\frac{k^2 + 1000}{70} &= k \\
+k^2 - 70k + 1000 &= 0 \\
+(k-20)(k-50) &= 0,
+\end{align*}
+so <math>k = 20</math> or <math>k= 50</math>, so <math>n = 400, 2500.</math>
+If <math>n</math> is not square, then we can say that, for a positive integer <math>k</math>, we have
+\begin{align*}
+k^2 < &n < (k+1)^2 \\
+k^2 + 1000 < &n + 1000  = 70\lfloor \sqrt{n} \rfloor = 70k< (k+1)^2 + 1000 \\
+k^2 + 1000 < &70k < (k+1)^2 + 1000.
+\end{align*}
+To solve this inequality, we take the intersection of the two solution sets to each of the two inequalities
+<math>k^2 + 1000 < 70k</math> and <math>70k < (k+1)^2 + 1000</math>. To solve the first one, we have
+\begin{align*}
+k^2 - 70k + 1000 &< 0 \\
+(k-20)(k-50) &< 0\\
+\end{align*}
+<math>k\in (20, 50),</math> because the portion of the parabola between its two roots will be negative.
+The second inequality yields
+\begin{align*}
+k &< k^2 + 2k + 1 + 1000 \\
+&< k^2 -68k + 1001.
+\end{align*}
+This time, the inequality will hold for all portions of the parabola that are not on or between the its two roots, which are <math>34 + \sqrt{155}>46</math> and <math>34-\sqrt{155}<22</math> (they are roughly equal, but this is to ensure that we do not miss any solutions).
+Notation wise, we need all integers <math>k</math> such that
+<cmath>k \in \left(20, 50\right) \cap \left(-\infty,34 - \sqrt{155} \right)</cmath>
+or
+<cmath>k \in \left(20, 50\right) \cap \left(34 + \sqrt{155}, \infty \right).</cmath>
+For the first one, since our uppoer bound is a little less than <math>22</math>, the <math>k</math> that works is <math>21</math>. For the second, our lower bound is a little more than <math>46</math>, so the <math>k</math> that work are <math>47, 48,</math> and  <math>49</math>.
+<math>\boxed{\textbf{(C) }6}</math> total solutions for <math>n</math>, since each value of <math>k</math> corresponds to exactly one value of <math>n</math>.
+-Benedict T (countmath1)
+==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
-A in-depth graph with intersection points is linked below.
-https://www.desmos.com/calculator/e5wk9adbuk
 ==See Also==
 {{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}}

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

Latest revision as of 09:26, 9 March 2024

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Video Solutions

Video Solution 1

Video Solution 2

Video Solution 3 by the Beauty of Math

See Also