OFFICIAL SOLUTION

We are told that the smallest angle measures 136 degrees--this is the first term in the consecutive set. If the polygon has S sides, then the largest angle--the last term in the consecutive set--will be (S - 1) more than 136 degrees.

The sum of consecutive integers = (Average Term) * (# of Term)= \(\frac{First + Last}{2}\) * (# of Terms).

Given that there are S terms in the set, we can plug in for the first and last term as follows:

\(\frac{136 + 136 + (S-1)}{2} * S\) = sum of the angles in the polygon.

We also know that the sum of the angles in a polygon = 180 (S-2) where S represents the number of sides.

Therefore: \(180(S-2) = \frac{136+136+(S-1)}{2} * S\). We can solve

for S by cross-multiplying and simplifying as follows:

\(2(180)(S-2) = [272 + (S-1)] S\)
\(360S - 720 = (271 + S)S\)
\(360S - 720 = 271S + S^2\)
\(S^2 - 89S + 720 = 0\)

A look at the answer choices tells you to try (S - 8), (S - 9), or (S - 10) in factoring.

As it turns out (S-9)(S-80)=0, which means S can be 9 or 80. However S cannot be 80 because this creates a polygon with angles greater than 180.

Therefore S equals 9; there are 9 sides in the polygon.

The correct answer is B.
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I still think that the fastest (though may not be the best way as expounded extensively in this post) to solve this question is to make use of the fact that 180(n-2) will always have Unit Digit = 0 (try n=3,4,...etc)

The fact is that we are given a starting number of 136, and we know for SURE that all angles are consecutive numbers. That would mean (pun intended) that the mean * n must = UD of 0.

We can narrow down to 9 or 10, but a surer bet would be to use the fact that the MEAN would always have a UD of zero i.e. only 140 is the nearest possible answer we have here. Then, multiplying by number of sides would always yield UD zero.

Hence (B).

Why use quadratic equation when we can simply look at unit digit and options to answer this question

Check video solution here.

Answer: Option B



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Their solution seems to miss the point. It's not inherently problematic that some angles in the hypothetical 80-side polygon are greater than 180 degrees. What is a problem is that one angle in such a polygon would need to be exactly 180 degrees, and you can't have a 180-degree angle in a polygon, because then you have a straight line and do not have a vertex.

There's nothing in the GMAT instructions that tells you to assume all "polygons" are convex (the Math Review section in the OG says something to that effect, but that Math Review is not a substitute for the test directions), but as a practical matter, if you did assume that when a GMAT question mentions a "polygon" that it is a "convex polygon", I think you'll always get the right answer -- I can't imagine seeing a real GMAT question about polygons where the answer would be different if you considered concave polygons separately from convex ones. If you see a polygon with interior angles greater than 180 degrees on the GMAT, the question will include a diagram so you can see what kind of polygon you're dealing with. Q10 in the diagnostic test at the start of the OG is one example (the star shape is concave if you erase the edges of the pentagon).
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