Executive Editor

[Trenton]

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

[jonetta]

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 > 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