Difference between revisions of "2022 AIME II Problems/Problem 5"
m (→Solution 1) |
(→Solution) |
||
Line 15: | Line 15: | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | Note: I believe that you also need <math>20 >= a</math>, so you cannot just simply do 18*4. In addition, it is possible for <cmath>b - c = 2</cmath>, and <cmath>a - b = p_1</cmath>. You would need to do casework to make sure <math>a</math> does not go above 20. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=II|num-b=4|num-a=6}} | {{AIME box|year=2022|n=II|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:44, 30 August 2022
Problem
Twenty distinct points are marked on a circle and labeled through
in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original
points.
Solution
Let ,
, and
be the vertex of a triangle that satisfies this problem, where
.
. Because
is the sum of two primes,
and
,
or
must be
. Let
, then
. There are only
primes less than
:
. Only
plus
equals another prime.
.
Once is determined,
and
. There are
values of
where
, and
values of
. Therefore the answer is
Note: I believe that you also need , so you cannot just simply do 18*4. In addition, it is possible for
, and
. You would need to do casework to make sure
does not go above 20.
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.