# How many iterations can I have in a year

I am no math wiz at all.

I have an 8 day set iteration that I want to run as many times in a year as I can. I can run 2 at a time an I can start 2 more every 3rd day.

So day 1 I start 2 iterations, day 3 I start 2 more, Day 9 I start a 3rd, day 11 I start a 4th. How many can I run per year?

EDIT to clarify and add context-

I'll try to make it more of a word problem. I brew 10 gallons of beer at a time and I have 4 fermenters. Each batch ferments for 6 days. After the beer ferments it has to clear in another tank of which I have 2 and the beer has to sit in this clearing tank for 2 days.

so I have 4 beers in fermentation but I can only move 2 into clearing tanks for 2 days, then after that I can move the other two into the clearing tanks.

At the point where I empty a fermenter I want to refill it.

How many times can I do all this in a year, and whats the equation?

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Is it a leap year? If you start on day 1, does it end on day 8 or day 9? – Ben Voigt Nov 1 '12 at 20:34
I don't want to worry about leap year. – TJ Sherrill Nov 1 '12 at 21:49

If you can only run $2$ at a time, how can you start more on day $3$? You have four running on day $4$. It sounds like you start $2$ on days $1,9,17,25 \ldots$, which are the days of the form $8k+1$. You have to start them before day $357$. What is the greatest $8k+1$ that is less than $357$?
Added: if you can have four batches in process but have only two clearing tanks, you should start two each on day $1, 3, 9, 11, \ldots , 8k+1, 8k+3.$ They finish on day $8k+8$ and $8k+10$. So find the highest $k$ that each series is done within the year. In this case, for each series it is $44$, so you can make $4 \cdot 44=176$ batches in a year.
+1 Ross, in part for helping me understand the problem statement! Just curious: would there be any advantage to starting $1$ iteration each $8K+1^{\text{st}}$ day ($k\geq 0$), and another every $8k+3^{\text{rd}}$ day? I guess I'd need to "do the math"! – amWhy Nov 1 '12 at 21:52