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# The measures of the interior angles in a polygon are

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The measures of the interior angles in a polygon are [#permalink]  11 Feb 2012, 15:07
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The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

[Reveal] Spoiler: OA

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Re: The measures of the interior angles in a polygon are [#permalink]  17 Apr 2013, 01:02
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Expert's post
enigma123 wrote:
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

Just another way of doing this sum:
The sum of exterior angles for any polygon = 360 degrees.Now, given that the minimum internal angle measure is 136 degrees--> the exterior angle = 180-136 = 44 degrees.
Also, we know that this value will keep decreasing like 43,42,41 etc. It is easy to see that only if there 9 terms, the middle value is 40 and we know that 40*9 =360 degrees.

B.
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Re: Sides of a Polygon. [#permalink]  11 Feb 2012, 15:10
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enigma123 wrote:
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

Sum of Interior Angles of a polygon is $$180(n-2)$$ where $$n$$ is the number of sides (so is the number of angles);

We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, $$n_{th}$$ angle, will be $$136+(n-1)$$ degrees. The sum of the $$n$$ consecutive integers (the sum of $$n$$ angles) is given by $$\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n-1)}{2}*n=\frac{271+n}{2}*n$$;

So we have that $$180(n-2)=\frac{271+n}{2}*n$$ --> $$360(n-2)=(271+n)*n$$, now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then $$n$$, out of the options listed, could be either 10 or 9, $$n=9$$ fits.

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Re: The measures of the interior angles in a polygon are [#permalink]  17 Apr 2013, 01:36
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Another way of solving the problem

sum of internal angles of a polygon = (n-2)*180. as sum is multiplied by 180, the unit degit of sum should be 0.

also looking at the options least number of sides is 8.

sum of unit degits for 8 sides is 6 (from 136) + 7+8+9+0+1+2+3 = 6.
sum of unit degist for 9 sides is 6+7+8+9+0+1+2+3+4 = 0.
Hnec option B is correct anser
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Re: The measures of the interior angles in a polygon are [#permalink]  14 Nov 2013, 20:06
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Remember that sum of angles in a polygon=(n-2)180. This means that whatever the value of n there will always be a 0 in units digit. Now, the question becomes 'How many consecutive integers we should add starting from 136 so that units digit is 0. 6+7+8+9+0=(3)0. Note that 3 is put in brackets bcos it is carried over. Now we know that sum of angles in a pentagon=540. But does 136+137+....140 equal 540. Use formula $$\frac{n}{2}$$(a+l)=$$\frac{5}{2}$$*(136+140)=690. Does not match. So proceed further. 6+7+8+9+0+1+2+3+4=(4)0. Now the series becomes 136+137+138+........+144 consisting of (144-136+1=)9 elements. Again use above formula for sum and it comes out to be 1260. Now we know that sum of angles in a nonagon is (n-2)180=1260. This choice matches. So, answer is 9.
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Re: The measures of the interior angles in a polygon are [#permalink]  12 Feb 2012, 15:09
Buneul - how did you get that largest angle will be 136+n-1 degrees?
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Re: The measures of the interior angles in a polygon are [#permalink]  12 Feb 2012, 19:35
He's using the average formula. Average uses the first and last number of an evenly spaced sequence. Since 136 is the smallest integer in the sequence it is one of the numbers. The largest number in the sequence since it's consecutive is 135+n or 136+(n-1). Lets n was one then. The answer would be 135+n which is equal to one, which is our smallest number. Let's say n (#of sides) is 4 then the largest number in the sequence will be 135+4= 139
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Re: The measures of the interior angles in a polygon are [#permalink]  13 Feb 2012, 03:34
Expert's post
enigma123 wrote:
Buneul - how did you get that largest angle will be 136+n-1 degrees?

There are n angles (sides):
The smallest angle is 136 degrees;
Next (2nd) angle is 136+1 degrees;
...
nth angle is 136+(n-1) degrees.

Hope it's clear.
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Re: The measures of the interior angles in a polygon are [#permalink]  24 Apr 2013, 22:54
sum of angles =180(n-2) where n is the number of sides
since we have consecutive angles, then median=mean

B n=9 ,then the sum of angles =180*7

the median of 9 consecutive integers is 140 (136 137 138 139 140 141 142 143 144)
the sum of consecutive integers is 140*9

180*7/140*9 =(20*9*7)/(20*7*9)= 1
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Re: The measures of the interior angles in a polygon are [#permalink]  02 Jan 2014, 09:13
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enigma123 wrote:
The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8
B) 9
C) 10
D) 11
E) 13

Note that the formula for the sum of the angles of a polygon is (n-2)*180

Therefore, the sum of the angles must end in zero

Now, let's start adding until we get units digit zero and then check if it satisfies

136+137+138+139 = We get digit zero

But if we ply (4-2)(180) = 360 this obviously doesn't satisfy cause 136+137...+139 is more than 360. So let's keep trying

+141+142+143+144...another digit 0

Let's see we now have 9 numbers so 180*7 = 1260

We then grab the median of the consecutive integers which is 140*9 = 1260

It satisfies so answer is thus B

Hope it helps
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Re: The measures of the interior angles in a polygon are [#permalink]  25 Jun 2015, 03:02
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Re: The measures of the interior angles in a polygon are   [#permalink] 25 Jun 2015, 03:02
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