Difference between revisions of "2021 AIME I Problems/Problem 15"
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==Solution== | ==Solution== | ||
| + | ===Solution 1=== | ||
Using the computer science algorithm called binary search, you can narrow down the answer. Binary search takes 10 iterations because the range is 0 to 999 and log base 2 of 999 is 10. | Using the computer science algorithm called binary search, you can narrow down the answer. Binary search takes 10 iterations because the range is 0 to 999 and log base 2 of 999 is 10. | ||
| + | |||
| + | You end up finding the bounds of 5 and 280. The sum is 285 | ||
| + | |||
| + | ~Lopkiloinm | ||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=14|after=Last problem}} | {{AIME box|year=2021|n=I|num-b=14|after=Last problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 03:23, 12 March 2021
Contents
Problem
Let 
be the set of positive integers 
such that the two parabolas![\[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\]](http://latex.artofproblemsolving.com/6/f/9/6f9e25390a9657644dc69c3cd2ca6d154cd6e9be.png)
intersect in four distinct points, and these four points lie on a circle with radius at most 
. Find the sum of the least element of and the greatest element of .
Solution
Solution 1
Using the computer science algorithm called binary search, you can narrow down the answer. Binary search takes 10 iterations because the range is 0 to 999 and log base 2 of 999 is 10.
You end up finding the bounds of 5 and 280. The sum is 285
~Lopkiloinm
See also
| 2021 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last problem | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
