Difference between revisions of "2020 AMC 10B Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
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We can first consider the equation without a floor function: | We can first consider the equation without a floor function: | ||
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<cmath>n^2 - 2900n + 1000000 = 0</cmath> | <cmath>n^2 - 2900n + 1000000 = 0</cmath> | ||
− | Now we can | + | Now we can determine the factors: |
<cmath>(n-400)(n-2500) = 0</cmath> | <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. | 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 | + | 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>, | + | 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>, | + | For <math>n = 400</math>, <math>L=20</math> and <math>R=20</math> |
− | For <math>n = 470</math>, | + | For <math>n = 470</math>, <math>L=21</math>, <math>R=21</math> |
− | For <math>n = 540</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> | Now we move to <math>n = 2500</math> | ||
− | For <math>n = 2430</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>, | + | 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>, | + | 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>, | + | 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>, | + | For <math>n = 2500</math>, <math>L=50</math> and <math>R=50</math> |
− | For <math>n = 2570</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> | Therefore we have 6 total solutions, <math>n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}</math> | ||
− | ==Solution | + | ==Solution 2== |
This is my first solution here, so please forgive me for any errors. | This is my first solution here, so please forgive me for any errors. | ||
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-Solution By Qqqwerw | -Solution By Qqqwerw | ||
− | ==Solution | + | ==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 <math>2+4 = \boxed{\textbf{(C)} 6}</math>. | + | 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 | ~Rekt4 | ||
− | ==Solution | + | ==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>. | 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>. | ||
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~the_jake314 | ~the_jake314 | ||
− | ==Solution | + | ==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>. | |
− | <math>\ | ||
− | <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> | ||
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− | + | ~KingRavi | |
− | + | ==Solution 6 == | |
− | <math>\ | + | 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*} | |
+ | 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 <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 Solutions== |
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.
Contents
[hide]Problem
How many positive integers satisfy (Recall that is the greatest integer not exceeding .)
Solution 1
We can first consider the equation without a floor function:
Multiplying both sides by 70 and then squaring:
Moving all terms to the left:
Now we can determine the factors:
This means that for and , 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 .
For , but so
For , and
For , ,
For , but so
Now we move to
For , and so
For , and so
For , and so
For , but so
For , and
For , but so
Therefore we have 6 total solutions,
Solution 2
This is my first solution here, so please forgive me for any errors.
We are given that
must be an integer, which means that is divisible by . As , this means that , so we can write for .
Therefore,
Also, we can say that and
Squaring the second inequality, we get .
Similarly solving the first inequality gives us or
is larger than and smaller than , so instead, we can say or .
Combining this with , we get are all solutions for that give a valid solution for , meaning that our answer is . -Solution By Qqqwerw
Solution 3
We start with the given equationFrom there, we can start with the general inequality that . This means thatSolving each inequality separately gives us two inequalities:Simplifying and approximating decimals yields 2 solutions for one inequality and 4 for the other. Hence, the answer is .
~Rekt4
Solution 4
Let be uniquely of the form where . Then, Rearranging and completeing the square gives This gives us Solving the left inequality shows that . Combing this with the right inequality gives that which implies either or . By directly computing the cases for using , it follows that only yield and invalid from . Since each corresponds to one and thus to one (from and the original form), there must be 6 such .
~the_jake314
Solution 5
Since the right-hand-side is an integer, so must be the left-hand-side. Therefore, we must have ; let . The given equation becomes
Since for all real , we can take with to get We can square the inequality to get The left inequality simplifies to , which yields The right inequality simplifies to , which yields
Solving , and , we get , for values .
Solving , and , we get , for values .
Thus, our answer is
~KingRavi
Solution 6
Set in the given equation and solve for to get . Set ; since , we get The left inequality simplifies to , which yields The right inequality simplifies to , which yields Solving , and , we get , for values .
Solving , and , we get , for values .
Thus, our answer is
Solution 7
If is a perfect square, we can write for a positive integer , so The given equation turns into
so or , so
If is not square, then we can say that, for a positive integer , we have
To solve this inequality, we take the intersection of the two solution sets to each of the two inequalities and . To solve the first one, we have
The second inequality yields
Notation wise, we need all integers such that
or
For the first one, since our uppoer bound is a little less than , the that works is . For the second, our lower bound is a little more than , so the that work are and .
total solutions for , since each value of corresponds to exactly one value of .
-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
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.