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The measures of the interior angles in a polygon are consecutive integ

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The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 03 Feb 2011, 15:45
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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
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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 03 Feb 2011, 16:02
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rxs0005 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.

Answer: B.
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The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 11 Feb 2012, 16:10
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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 17 Apr 2013, 02:02
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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

Please help.

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: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 03 Feb 2011, 21:05
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rxs0005 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


When I look at this question, I say, "I know how to find the Interior angle of a regular polygon. But this is not a regular polygon since it has angles 136, 137, 138, 139, 140, 141, 142 ....etc."

Also,
Interior angle of 8 sided regular polygon = 180*6/8 = 135
Interior angle of 9 sided regular polygon = 180*7/9 = 140
Interior angle of 10 sided regular polygon = 180*8/10 = 144
etc

The average of the given angles can only match 140 (such that effectively, all the angles are 140) Hence, the polygon must have 9 sides.

Remember, capitalize on what you know.
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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 04 Feb 2011, 10:47
Karishma,
Can you explain in more details when you say "The average of the given angles can only match 140 (such that effectively, all the angles are 140)". I did not get your point. Thanks
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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 04 Feb 2011, 11:20
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Sum of interior angles for n sides: (n-2)*180
And each interior angle are increasing by 1.
This can be written as : 136+137+138+139+... = 136+(136+1)+(136+2)+(136+3)+...
For n sides

136n+(1+2+3+4....+n-1)
136n+(n-1)n/2 --- Sum of n-1 natural number

Equate:
136n+(n-1)n/2 = (n-2)*180
n^2-89n+720=0
n^2-80n-9n+720=0
n(n-80)-9(n-80)=0
(n-9)(n-80)=0

n=9
n=80

Since 9 is one of the options.

Ans: "B"
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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 04 Feb 2011, 12:56
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ajit257 wrote:
Karishma,
Can you explain in more details when you say "The average of the given angles can only match 140 (such that effectively, all the angles are 140)". I did not get your point. Thanks


Sum of interior angles of a polygon = (n-2)*180 (not necessarily regular polygon)

Why? See the figure below:

Attachment:
Ques2.jpg
Ques2.jpg [ 5.14 KiB | Viewed 41013 times ]

A 6 sided polygon can be split into 4 triangles each of which has a sum of interior angles 180 degrees.
An n sided polygon can be split into n - 2 triangles.

When the polygon is regular, each angle is same so the sum is divided by the number of sides to get the measure of each angle e.g. in a regular hexagon, each interior angle = 4*180/6 = 120 degrees.

Now if I have a hexagon whose angles are 115, 117, 119, 121, 123 and x, what will be the angle x?
We can see it in two ways -
1. The sum of all angles should be 4*180 = 720
So 115 + 117 + 119 + 121 + 123 + x = 720
or x = 125


2. The average of the angles should be 120. (Since the sum of the angles is 720 and there are 6 sides)
119 and 121 average out as 120. (119 is 1 less than 120 and 121 is 1 more than 120)
117 and 123 average out as 120.
So 115 and x should average out as 120 too. Therefore, x should be 125.

In the question, the average of the given angles of the polygon can only be 140. So it must have 9 sides. To confirm,
136, 137, 138, 139, 140, 141, 142, 143, 144 - 9 angles with average 140. So the polygon must have 9 sides.

(It cannot be 144 or anything else because 10 angles (136, 137, 138, 139, 140, 141, 142, 143, 144, 145) will not give an average of 144)
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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 05 Feb 2011, 09:01
tough one. simple solution, but its not an ez one.
am i right?
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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 06 Feb 2011, 07:25
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144144 wrote:
tough one. simple solution, but its not an ez one.
am i right?


Actually, I wouldn't say it is very tough if we know that the sum of interior angles of a polygon is 180(n -2) (I explained above why it is so). We can also use a very straight forward but long approach.

Interior angles of the given polygon: 136, 137, 138, 139, 140, 141, 142, 143....

Using options:

If the polygon had 8 sides, it would have had 8 interior angles. Their sum: (8-2)180 = 1080
Sum of 8 angles: 136 +137 + 138 + 139 + 140 + 141 + 142 + 143 is more than 1080 hence this polygon does not have 8 sides.

If the polygon had 9 sides, it would have had 9 interior angles. Their sum: (9-2)180 = 1260
Sum of 9 angles: 136 +137 + 138 + 139 + 140 + 141 + 142 + 143 + 144 is 1260. Hence this polygon does have 9 sides.
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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 06 Feb 2011, 07:45
144144 wrote:
tough one. simple solution, but its not an ez one.
am i right?


I wouldn't say it's very hard either, though I wouldn't recommend trial and error in this cases:

...
Calculating the sum of the interior angles of a polygon with 7 sides then checking whether 7 consecutive integers starting from 136 add up to that value;
Calculating the sum of the interior angles of a polygon with 8 sides then checking whether 8 consecutive integers starting from 136 add up to that value;
Calculating the sum of the interior angles of a polygon with 9 sides then checking whether 9 consecutive integers starting from 136 add up to that value;

Whereas equating two formulas and working on answer choices should give an answer in less time: \(180(n-2)=\frac{271+n}{2}*n\) --> \(360(n-2)=(271+n)*n\) --> \(n=9\).
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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 12 Feb 2012, 16:09
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 consecutive integ  [#permalink]

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New post 12 Feb 2012, 20:35
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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New post 13 Feb 2012, 04:34
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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 09 Aug 2012, 10:08
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Hey guys...please let me know whether this method is correct or not.
the stem says that the angles are consecutive integers, 136 being the smallest one. Also going by the formula of sum of interior angles, we know atleast this much that sum of angles cant be number like 729, 847, 653, 542 but it can be a number of the form xx0. So going by this knowledge, if the smallest angle is 136 then the sum of angles can only be of the form xx0 when the largest angle is 144. That is 9 sides.
Yeah the largest angle could be 154, 164 but in that case the number of sides must be 19 and 29 respectively.
Since the largest option is 13, hence the answer is 9.
Please correct me if I am wrong.
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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 17 Apr 2013, 02:36
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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 consecutive integ  [#permalink]

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New post 24 Apr 2013, 23:54
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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 consecutive integ  [#permalink]

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New post 14 Nov 2013, 21:06
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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 consecutive integ  [#permalink]

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New post 03 Nov 2014, 08:36
Bunuel wrote:
rxs0005 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.

Answer: B.



Hi Bunnel,

In above post, i could not get " 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'.
Why are you saying that LHS is 0.

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Re: The measures of the interior angles in a polygon are consecutive integ  [#permalink]

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New post 03 Nov 2014, 08:39
ammuseeru wrote:
Bunuel wrote:
rxs0005 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.

Answer: B.



Hi Bunnel,

In above post, i could not get " 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'.
Why are you saying that LHS is 0.

Regards,
Ammu


Because the left hand side (LHS) is 360(n-2) = 10*(36(n-2)) = 10*integer = something with the units digit of 0.

Does this make sense?
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Re: The measures of the interior angles in a polygon are consecutive integ   [#permalink] 03 Nov 2014, 08:39

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