Officials acquired two anonymized self texts by the G Gang member known as i, and hope these will help you identify the boss, B.
$\kern166mu$G=iNtErMsOfG$\kern199mu$
.G=i/B.i
Investigators have thus far determined that each gangster (i, N, t, E, r, M, s, O, f and B) must secretly be one of the foremost citizens of Whollywood (0, 1, 2, 3, 4, 5, 6, 7, 8 or 9) and that foul mathplay is surely afoot.
Just which digit do you finger as B?
( No leading zeros in Whollywood. G is not an 11th digit. =, . and / are not in disguise. )
Answer
The boss B is
8
Since G is defined in terms of itself, it stands to reason that
G is an infinitely repeating sequence of digits
Since B is missing from G, we can figure that .G is
$.G = 1 / 8.1 =\ .\overline{123456790}$
(Could be figured out via fractions of repeated 9s, as in:
1/9=.111...,12/99=.121212...,475/999=.475475475...,iNtErMsOf/999999999 = .iNtErMsOf iNtErMsOf iNtErMsOf .... The divisors of 999999999 are easy because it is so divisible by 3: 999999999=3×3×3×3×12345679.)
and G is then
$\overline{123456790}$
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