Difference between revisions of "2017 AIME I Problems"
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==Problem 4== | ==Problem 4== | ||
+ | A pyramid has a triangular base with side lengths <math>20</math>, <math>20</math>, and <math>24</math>. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length <math>25</math>. The volume of the pyramid is <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. | ||
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[[2017 AIME I Problems/Problem 4 | Solution]] | [[2017 AIME I Problems/Problem 4 | Solution]] | ||
Revision as of 14:30, 8 March 2017
2017 AIME I (Answer Key) | AoPS Contest Collections • PDF | ||
Instructions
| ||
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |
Contents
[hide]Problem 1
Fifteen distinct points are designated on : the 3 vertices
,
, and
;
other points on side
;
other points on side
; and
other points on side
. Find the number of triangles with positive area whose vertices are among these
points.
Problem 2
When each of 702, 787, and 855 is divided by the positive integer , the remainder is always the positive integer
. When each of 412, 722, and 815 is divided by the positive integer
, the remainder is always the positive integer
. Fine
.
Problem 3
For a positive integer , let
be the units digit of
. Find the remainder when
is divided by
.
Problem 4
A pyramid has a triangular base with side lengths ,
, and
. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length
. The volume of the pyramid is
, where
and
are positive integers, and
is not divisible by the square of any prime. Find
.
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by 2016 AIME II |
Followed by 2017 AIME II | |
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.