Difference between revisions of "2017 AMC 12A Problems/Problem 11"
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Because the sum of the interior angles is a multiple of <math>180</math>, we know that the sum of the angles in a polygon is <math>0 \mod 9</math>. <math>2017</math> is congruent to <math>1 \mod 9</math>, so the answer has to be <math>-1 \mod 9</math>. The only answer that is congruent to <math>-1 \mod 9</math> is <math>143</math>. | Because the sum of the interior angles is a multiple of <math>180</math>, we know that the sum of the angles in a polygon is <math>0 \mod 9</math>. <math>2017</math> is congruent to <math>1 \mod 9</math>, so the answer has to be <math>-1 \mod 9</math>. The only answer that is congruent to <math>-1 \mod 9</math> is <math>143</math>. | ||
-harsha12345 | -harsha12345 | ||
| + | |||
| + | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
| + | https://youtu.be/4mVtT62THpA | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=10|num-a=12}} | {{AMC12 box|year=2017|ab=A|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:56, 10 June 2023
Contents
Problem
Claire adds the degree measures of the interior angles of a convex polygon and arrives at a sum of 
. She then discovers that she forgot to include one angle. What is the degree measure of the forgotten angle?
Solution 1
We know that the sum of the interior angles of the polygon is a multiple of 
. Note that 
and 
, so the angle Claire forgot is 
. Since the polygon is convex, the angle is 
, so the answer is 
.
Solution 2 (fast with answer choices)
Because the sum of the interior angles is a multiple of , we know that the sum of the angles in a polygon is 
. is congruent to 
, so the answer has to be 
. The only answer that is congruent to is 
.
-harsha12345
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
See Also
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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. 
