That's . . . . . 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17
Added to . . 17 + 16 + 15 + 14 + 13 + 12 + 11 + 10
The pairs 10 with 17, 11 with 16, and so on to 17 with 10, all make a sum of 27, and there are 8 pairs in all doing that.
Using this same method to find the sum of : 10, 11, 12, . . . . . . . . . 79, 80?
71 pairs each with a sum of 90, makes 6390, and half of that is 3195
The first number is s so the last number is s + n - 1
The pairs all have a sum of 2s + n - 1 and there are n pairs like that.
Finally the sum we want is half of that, which in algebra is . . . . n (2s + n -1) / 2
16 and 32 are pure powers of 2 : they are $2^4 and 2^5$ respectively
We are looking for a run of consecutive numbers that totals to 16 :
Whether s is an odd or an even number, 2s will be even , so n - 1 must also be even , because the whole factor (2s + n - 1) has to be some power of two.
When n -1 is even, n itself will be odd , but 16 does not have any odd factors, so there is no value, odd or even, from which to start a run of consecutive numbers whose sum will be 16.
We can also see that the same reasoning would apply to making 32 and any other pure power of 2 .
Try this reasoning, based on odd-ness and even-ness, while looking at the rectangle of dots on the main problem page.
Do you prefer to reason from the algebra or from the image, or do you find that taking the algebra and the image together is somehow the best way to increase your confidence in the validity of your argument ?