Difference between revisions of "2020 AMC 10B Problems/Problem 24"
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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>. | 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>. | 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>. | ||
| − | 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>. | + | 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>. |
~DrJoyo | ~DrJoyo | ||
Revision as of 13:30, 8 February 2020
Problem
How many positive integers 
satisfy![\[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\]](http://latex.artofproblemsolving.com/0/d/d/0dd28ce1709d2e91e3b6ff963c8adf1cac05b9fa.png)
(Recall that 
is the greatest integer not exceeding 
.)
Solution
First notice that the graphs of 
and ![$\sqrt[]{n}$](http://latex.artofproblemsolving.com/b/f/1/bf1c18f78bd8f62980b31821dd7585c37733fed4.png)
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 
.
For the second intersection, using binary search and/or estimating from the graph can narrow down the other cases, being 
, 
, 
, and 
. 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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
