Try this

Twelve concentric rows.

I got 11

12 (Twelve)

[eqn]-\left(\sum_{k=1}^{\infty}k\right)^{-1}[/eqn]

8 [ n Σ ] =3216
n Σ =402
(50)+(50-3)+(50-6)+(50-9)+(50-12)+(50-15)+(50-18)+(50-21)+(50-24)+(50-27)+(50-30)+(50-33) = 402

there are twelve differences possible for one side according to the formula. Each difference represents a row, therefore there are 12 rows.

>>8945911
5/7, well meme'd anon

>>8945964
You're simply looking for how many times the differences can occur according to a certain pattern. That pattern is continuously subtracting 3 from 50 to see how many rows fufill the niche of there being 402 seats to a side

>>8945685
you probably started from row 0.

12
I just divided 3216 by 8 and then started taking away 50 every number taken away next was three less than the last number which was done until I got to zero and then I counted how many times i needed to take numbers away and that was the amount of rows

File: Screenshot_2017-05-31-14-08-05.png (987KB, 720x1280px) Image search: [Google]

987KB, 720x1280px

12 rows

>>8945651

The poorly phrased problem may be validly interpreted in at least two different ways. Due to the context of the admonishment which concludes the problem, we could validly interpret the problem to mean, for example, that the top row (all round, all eight 'sectors' of the row) has 400 seats, while the next row down (again, all eight sectors) has 397 seats, then 394, 391 and so on. The reason why this interpretation is valid and must be dispensed with, is that the language of the problem fails to unambiguously distinguish between rows and sectors of rows, yet the picture is a helpful suggestion.

Given this, it becomes clear that a simple addition will suffice to check whether this valid interpretation leads to a solution which is consistent with the rest of the problem's statement. It doesn't (hitting 3116 at one point, only to skip over 3216), and so we do reject what had been a valid initial interpretation of the problem as being spurious.

We then proceed to the other interpretation of the problem, filling in the gap, and again conclude that all that is needed is a simple addition; no more elaborate expression is required. The thing will either hit 3216 in very short order, or it won't in which case we would reject the problem as bogus in toto. But happily this other interpretation leads to the same twelve rows which have been reported by everybody else ITT, as the solution which we were meant to find, and as the answer.

>>8947746

Despite the fact that a simple brute force sum really is the most efficient method of solution, let's now have a bit of autistic fun and solve the problem using some brainlet high school math instead, with slightly more formal manipulations and comments. This will actually entail a few steps that work together nicely as an easy bit of fun.

The first task is to set up the equation properly, using a finite series. Indexing properly gives 1); rearrangement of finite series, together with the recognition of the definition of a triangular number (and its alternate expression), which every good boy should know off the top of his head, leads to 2). Ah, a quadratic in n. This leads to 3), and although this quadratic in n does have two distinct, strictly positive rational solutions, only one of them is natural, which is manifestly what is required. We therefore reject the "false root" where this problem is concerned, and again arive at 12.

[eqn] 1) \;\;\;\;\; \sum\limits_{k=1}^{n} \bigg( 400 -24(k-1) \bigg) = 3216 [/eqn]

[eqn] 2) \;\;\;\;\; \frac{53}{3} n - 134 = \sum\limits_{k=1}^{n} k = \frac{n(n+1)}{2}[/eqn]

[eqn] 3) \;\;\;\;\; \frac{1}{2} n^{2} - \frac{103}{6} n + 134 = 0 \;\;\; \rightarrow \;\;\; n = \frac{67}{3}, 12 \;\;\; \rightarrow \;\;\; n = 12 [/eqn]

Some general comments on the above: the discriminant takes care of itself very nicely (a fraction where both numerator and denominator are perfect squares), we get to do three or four different things (manipulate finite series, rearrangements, use an important fact about triangular numbers, use the quadratic formula) A cute, albeit unnecessarily complicated, solution of the problem. Notice also that the fractional form at the left in 2) immediately tells us something about n: since both n and triangular numbers themselves are manifestly natural, n had better be a multiple of 3 or the deal's off. 12 works.

It's simple, it forms an arithmetic sequence.

Sn = (n/2)(2a+(n-1)d)

3216/8 = (n/2)(2*50 - 3n + 3)
402 = (n/2)(103 - 3n)
804 = 103n - 3n^2
3n^2 - 103n + 804, put into the quadratic formula and you get n = 12 or n = 22.3333...

n = 22.33.. will probably be when there are a negative number of seats in each row, which can't happen (Un = a + (n - 1)d), so n = 12.