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> | ||
− | |||
− | + | ~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 10:26, 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
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 that
Solving 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
\begin{align*} \frac{k^2 + 1000}{70} &= k \\ k^2 - 70k + 1000 &= 0 \\ (k-20)(k-50) &= 0, \end{align*}
so or
, so
If is not square, then we can say that, for a positive integer
, 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
and
. To solve the first one, we have
\begin{align*}
k^2 - 70k + 1000 &< 0 \\
(k-20)(k-50) &< 0\\
\end{align*}
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 and
(they are roughly equal, but this is to ensure that we do not miss any solutions).
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.