Difference between revisions of "2020 AMC 10B Problems/Problem 24"
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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: | ||
− | For <math>n = 330</math>, left hand side <math>=19</math> but <math>18^2 < 330 < 19^2</math> so right hand side | + | 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 | + | 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 | + | 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 | + | 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> | 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 | + | 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 | + | 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 | + | 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 | + | 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 | + | 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 | + | 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> | Therefore we have 6 total solutions, <math>n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}</math> |
Revision as of 19:28, 19 February 2020
- 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
First notice that the graphs of and
intersect at 2 points. Then, notice that
must be an integer. This means that n is congruent to
.
For the first intersection, testing the first few values of (adding
to
each time and noticing the left side increases by
each time) yields
and
.
timating from the graph can narrow down the other cases, being
, d
. This results in a total of 6 cases, for an answer of
.
~DrJoyo
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 . 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
values of intersection lay closer to the left side of the stair, and
values lay closer to the right side of the stair.
With meticulous graphing, you can realize that the answer is .
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:
Multiplying both sides by 70 and then squaring:
Moving all terms to the left:
Now we can use wishful thinking to 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:
For , left hand side
but
so right hand side
For , left hand side
and right hand side
For , left hand side
and right hand side
For , left hand side
but
so right hand side
Now we move to
For , left hand side
and
so right hand side
For , left hand side
and
so right hand side
For , left hand side
and
so right hand side
For , left hand side
but
so right hand side
For , left hand side
and right hand side
For , left hand side
but
so right hand side
Therefore we have 6 total solutions,
Solution 4
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 slightly larger 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 5
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
.
~Rekt4
Video Solution
On The Spot STEM: https://youtu.be/BEJybl9TLMA
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.