Difference between revisions of "2017 AMC 12A Problems/Problem 11"
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We know that the sum of the interior angles of the polygon is a multiple of <math>180</math>. Note that <math>\left\lceil\frac{2017}{180}\right\rceil = 12</math> and <math>180\cdot 12 = 2160</math>, so the angle Claire forgot is <math>\equiv 2160-2017=143\mod 180</math>. Since the polygon is convex, the angle is <math>\leq 180</math>, so the answer is <math>\boxed{(D)\ =\ 143}</math>. | We know that the sum of the interior angles of the polygon is a multiple of <math>180</math>. Note that <math>\left\lceil\frac{2017}{180}\right\rceil = 12</math> and <math>180\cdot 12 = 2160</math>, so the angle Claire forgot is <math>\equiv 2160-2017=143\mod 180</math>. Since the polygon is convex, the angle is <math>\leq 180</math>, so the answer is <math>\boxed{(D)\ =\ 143}</math>. | ||
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| + | ==Solution 2== | ||
| + | Because the sum of the interior angles is a multiple of 180, we know that the sum of the angles in a polygon is 0 mod 9. 2017 is congruent to 1 mod 9, so the answer has to be -1 mod 9. The only answer that is -1 mod 9 is 143. | ||
| + | -harsha12345 | ||
==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}} | ||
Revision as of 10:25, 21 July 2019
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
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
Because the sum of the interior angles is a multiple of 180, we know that the sum of the angles in a polygon is 0 mod 9. 2017 is congruent to 1 mod 9, so the answer has to be -1 mod 9. The only answer that is -1 mod 9 is 143. -harsha12345
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. 
