The measures of the interior angles in a polygon are consecutive odd i

I sort of used brute force (non-algebraic approach) to solve this.

Since Sum of Interior angles = (n-2) * 180, no matter how many sides the polygon has, the sum of interior angles will have unit digit of 0.

Hence, I tried listing the possible options and paid attention to the sum of unit digit, stopping when the sum ends with a unit digit of 0 -> ended up with 10.

153
151
149
147
.
.
.
xx5

The sum of the interior angles must be a MULTIPLE OF 180 and thus must be an integer with a UNITS DIGIT OF 0.
We can PLUG IN THE ANSWERS, which represent the number of sides and thus the number of angles.
For any evenly spaced set, SUM = COUNT * MEDIAN.

B: 9
Implied angles:
153, 151, 149, 147, 145, 143, 141, 139, 137
Sum = count * median = 9 * 145 = 1305
The sum does not have a units digit of 0.
Eliminate B.

If we include the next smallest angle -- 135 -- the sum will have a units digit of 0:
1305 + 135 = 1440
Check whether 1440 is a multiple of 180:
1440/180 = 144/18 = 72/9 = 8
Success!
If we include the next smallest angle -- for a total of 10 angles and thus 10 sides -- the sum of the angles is a multiple of 180.

ArvindCrackVerbal
Bunuel
daagh
VeritasKarishma
chetan2u
GMATPrepNow
ScottTargetTestPrep


I have a silly doubt here.

I am always confused how do we form equations of such kind ie 153 – 2(n-1) .
I can see that 153 – 2(n-1)=155-2n and as n starts from 1 this equation is fine here.

But in general is there any rule/logic to derive such kind of equations?

Asking me about Quants!!. I am honored, thank you

Hi,

we have the largest angle as 153, and all other angles are consecutive odd number..
To continue having odd numbers we require to subtract 2 from successive angles..
153-2*0, 153-2*1,153-2*2..153*2*(n-1)..

If I were told that some series has consecutive odd integer, then I would take the numbers as 2n-1, 2n-3,..

So, the derivation of a term in n would depend on the numbers we are dealing with

Given that 153 is the largest angle , 153=a+(n-1)2. Now the sum of interior angles in a polygon=(n-2)180 , sum of n terms in ap =n/2(2a+(n-1)2)..>n(a+n-1) now (n-2)180=n(154-n) only c)10 fits

Posted from my mobile device

Thanks a lot Chetan

You can derive most of such formulas simply by observing the first few terms for a pattern and extending the pattern for the general case.
For instance, the formula of 153 - 2(n - 1) gives you the smallest of n consecutive odd integers when the largest one is 153. Let's observe what happens for the first few values of n:

If n = 2, the consecutive odd integers are 153 and 151. The difference is 153 - 151 = 2.

If n = 3, the consecutive odd integers are 153, 151 and 149. The difference is 153 - 149 = 4.

If n = 4, the consecutive odd integers are 153, 151, 149 and 147. The difference is 153 - 147 = 6.

We can recognize a pattern here. The difference is always two less than twice the number of terms or, in other words, the largest odd integer is always 2n - 2 = 2(n - 1) greater than the smallest odd integer. Thus, we obtain the formula 153 - 2(n - 1).

As an additional example, let's consider the number of terms in an equally spaced set of integers. For instance, how can we solve a question like "how many three digit multiples of 7 are there?" without using any formula? The smallest three digit multiple of 7 is 105 and the largest three digit multiple of 7 is 994. Let's write these numbers:

105 112 119 ... 994

7*15 7*16 7*17 ... 7*142

Let's divide each term by 7. Notice that we are only interested in the number of terms and dividing each term by 7 will not change the number of terms:

15 16 17 ... 142
Let's subtract 14 from each term. Again, we are only interested in the number of terms and subtracting 14 from each term will not change the number of terms:

1 2 3 ... 128

It's not hard to see that there are exactly 128 terms above. Notice that we would get the same answer if we used the familiar formula of [(last term - first term)/(common difference)] + 1 = [(994 - 105)/7] + 1 = (889/7) + 1 = 127 + 1 = 128.

Thanks a lot Scott!

Hello axezcole,

In deriving expressions like these, I always consider smaller values and find out a pattern before applying the same logic to bigger values. You also need to keep in mind the type of numbers you are dealing with.

In this case, since the angles are odd integers, I know that any two of them will differ by 2. I also know that the largest angle is 153. This is one of the 'n' angles, which means that there are (n-1) more angles left. Therefore, the smallest angle can be obtained by subtracting the number 2 from 153 (n-1) times.

Note: I first considered a polygon with 4 angles and saw what would be the smallest angle if the largest angle was 153 and the angles were consecutive odd integers. This is what I meant when I said I established a pattern using smaller values.

That gave me,

Smallest angle = 153 - 2 - 2 -2 .... (n-1) times, which is

Smallest angle = 153 - 2(n-1)

Hope that helps!

Sure thing!!

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.