The measures of the interior angles in a polygon

guerrero wrote:The measures of the interior angles in a polygon are consecutive integers. The largest angle measures 110 degrees. How many sides does this polygon have?

Plugging answer choices is a good strategy for this problem.
As that has been discussed, I'll post another two methods to solve this problem.

Tricky Approach:
If a polygon has n sides, the sum of the interior angles of the polygon = (n - 2)*180°
Hence, the sum must be a multiple of 10, i.e. units digit of the sum must be 0.

As the largest angle measures 110° and the angles are consecutive integers, the next angles will be 109°, 108°, ... etc

The 1st time we get zero as units digit when we start adding the units place of these measures is (0 + 9 + 8 + 7 + 6) ---> Hence, 5 angles --> n = 5

The 2nd time we get zero as units digit when we start adding the units place of these measures is (0 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 + ...) ---> Hence, more than 11 angles angles --> n > 11

Hence, the polygon has 5 sides.

The correct answer is A.

Algebraic Approach:
Say, the number of sides of the polygon is n.
Hence, sum of the interior angles of the polygon = (n - 2)*180°

Largest angle is 110° and all the interior angles are consecutive integers.
Hence, the measures of the interior angles of the polygon in degrees are 110, (110 - 1), (110 - 2), ..., and (110 - (n - 1)).

So, sum of the angles = [110 + (110 - 1) + (110 - 2) + ... + (110 - (n - 1))]
= (110 + 110 + ... n times) - (1 + 2 + ... + (n - 1))
= 110n - sum of first (n - 1) integers
= 110n - n(n - 1)/2

So, 110n - n(n - 1)/2 = (n - 2)*180
--> 220n - n² + n = 360n - 720
--> n² + 139n - 720 = 0
--> n² - 5n + 144n - 720 = 0
--> (n - 5)(n + 144) = 0

As n cannot be negative, only possible value of n is 5.

The correct answer is A.