# Thread: quite a difficult question

1. ## quite a difficult question

hello , I encounter a quite difficult maths question

Each interior angle of a regular polygon, when it rounds off to the nearest degree, is "n", how many possible values of n are there?

I really have no idea how to solve this problem, all information is provided here, nothing misses.

Thanks all.

2. Originally Posted by kenny1999
hello , I encounter a quite difficult maths question Each interior angle of a regular polygon, when it rounds off to the nearest degree, is "n", how many possible values of n are there?
I really have no idea how to solve this problem, all information is provided here, nothing misses.
Thanks all.
The easiest way is just to count them.
However.
There are 360 degrees in a circle.
If you divide the number of sides of a polygon into that you get the value for "n"
EXCLUDE polygons with 0, 1 & 2 sides.
Code:
#of
sides n_value
3 120
4 90
5 72
You will notice that the angle is getting smaller with the increase in the number of sides.
When do you suppose the angle "n" will be smaller than the number of sides?
Try the square root of 360.
You will have 19 different values BUT you must exclude the "polygons" of 1 and 2 sides so you are left with 17.
There will also be 17 different values of n for a polygon of 19 sides up to a polygon of some number of sides.

Code:

360/239 = 1.506 rounds to 2
360/240 = 1.500 rounds to 2
360/241 = 1.494 rounds to 1
...
360/718 = 0.5014 rounds to 1
360/719 = 0.5007 rounds to 1
360/720 = 0.5000 rounds to 1
360/721 = 0.4993 rounds to 0
ALL polygons with sides greater that 720 will have zero for the rounded value of n.

You can create a spread sheet and determine the exact number that you are looking for very quickly.

Hope that helps.

.Actually, the answer is provided.

3. Originally Posted by aidan
The easiest way is just to count them.
However.
There are 360 degrees in a circle.
If you divide the number of sides of a polygon into that you get the value for "n"
EXCLUDE polygons with 0, 1 & 2 sides.
Code:
#of
sides n_value
3 120
4 90
5 72
You will notice that the angle is getting smaller with the increase in the number of sides.
When do you suppose the angle "n" will be smaller than the number of sides?
Try the square root of 360.
You will have 19 different values BUT you must exclude the "polygons" of 1 and 2 sides so you are left with 17.
There will also be 17 different values of n for a polygon of 19 sides up to a polygon of some number of sides.

Code:

360/239 = 1.506 rounds to 2
360/240 = 1.500 rounds to 2
360/241 = 1.494 rounds to 1
...
360/718 = 0.5014 rounds to 1
360/719 = 0.5007 rounds to 1
360/720 = 0.5000 rounds to 1
360/721 = 0.4993 rounds to 0
ALL polygons with sides greater that 720 will have zero for the rounded value of n.

You can create a spread sheet and determine the exact number that you are looking for very quickly.

Hope that helps.

.Actually, the answer is provided.
thanks

it's amazing...

but I want to know that

how to calculate it much more faster??

because it is a question from a maths competition, in which calculator cannot be used and the time allowed is very limited

4. Originally Posted by aidan
The easiest way is just to count them.
However.
There are 360 degrees in a circle.
If you divide the number of sides of a polygon into that you get the value for "n"
EXCLUDE polygons with 0, 1 & 2 sides.
Code:
#of
sides n_value
3 120
4 90
5 72
The interior angles increase with increasing number of sides, from 60 for a triangle, 90 for a square, ....

CB

5. Basically, what you want to do is find how big k has to be so that (interior angles of a k-gon) - (interior angles of a k+1-gon) ≤ 1

180(k-2)/k - 180(k-1)/(k+1) ≤ 1

(k-2)/k - (k-1)/(k+1) ≤ 1/180

( (k^2-k-2) - (k^2 - k) )/(k^2 + k) ≤ 1/180

k^2 + k ≥ 180(2)
k^2 + k - 360 ≥ 0
k ≥ -1/2+sqrt(360*4)/2 = -1/2 + 6sqrt(10) = -.5 + about 6*3.1 = 18.6 - .5 > 18

the smallest k=19
Let's check that:

180*16/18 = 10*16 = 160
180*17/19 = 9.5ish*17 = 161.5 (actually less b/c 9.5ish = 9.48ish, this will round down)
180*18/20 = 9*18 = 162

So now we know this: since we round n, n can be any integer between 162 and 180; 180-162 = 38
(This is the reason we checked for ≤1, because we know now that every increase in k will be ≤ 1 increase in the angle, and the angle is allowed to get arbitrarily close to 180 degrees)

It can also be any of the angles corresponding to k-gons up to 19. BE CAREFUL YOU DO NOT DOUBLE COUNT/LEAVE OUT 19*. This is the easiest place to screw up. We didn't count it before because (k=19) rounds down, and we didn't include 161 as an n value before.

k can't be 1 or 2, but it can be 3,4,5...19 = 17 different numbers.

17+38 = 55. There are 55 possible values for n.

This may be a little longer than the time allows, but it's probably on the top 3 fastest methods and doesn't require a calculator, just a little guesswork for the square root (which btw allows a lot of error: even if I had said sqrt(10)≤3.25 it still would get you k=19) and one long division (180/19).

* EDIT: The original said "17," which was me getting ahead of myself.