# Executive Editor

• The editor gets the numbers on the entrance and transforms it. The editor can perform two teams, both teams v and w mark the symbol chains

• If I understood algorithm correctly, the code looks like:

``````i = 300
while True:
i += 1
text = '5' * i

while '555' in text or '888' in text:
index = text.find('555')
text = text if index == -1 else (text[:index] + '8' + text[index + 3:])

index = text.find('888')
text = text if index == -1 else (text[:index] + '55' + text[index + 3:])

count5 = text.count('5')
count8 = text.count('8')

if count5 &gt; count8:
print(i)
break

``````

and response 303

It is theoretically possible to decide again:

It is clear that, with a change only on the left to the right, we can only get the following options:

``````855
8
885
85
88
8855
55
``````

(no versions of type 858... because we're changing numbers on the left to the right and there's no way `'5'` - There's nowhere to build it. `'888'` Always changing strictly. `'55'`)

Of which, by condition, we're only welcome. `855` and `55` (G8 and 5 from 0 to 2 are defined precisely by the fact that we replace three of these numbers, i.e. model 3).

How hard to understand all the lines above will be repeated with the period 7

Now that we have a periodicity to start from scratch, we'll look at the lines of one symbol and above:

line `'5'` We don't like it, it's out of seven options.

We' the next line. `'55'` - It's size 2 now.

``````2 + 7 * x > 300
``````

from where

``````x = 47
2 + 7 * 47 = 303
``````

Target accomplished without programming

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