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=i N t E r M s O f G

$\kern199mu$ . G=i   /   B . i

Investigators have thus far determined that each gangster (iN, t, E, r, M, s, O, f and B) must secretly be one of the foremost citizens of Whollywood (01, 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. )

  • 2
    $\begingroup$ Wait... so G is the last few digits of itself? $\endgroup$ – BaSzAt Apr 21 '16 at 17:47
  • 3
    $\begingroup$ Not a typo, G is defined in terms of itself (and is not an 11th digit) $\endgroup$ – humn Apr 21 '16 at 18:40
  • 1
    $\begingroup$ Probably G should be some sort of expression starting with digit(s) to match all places where it is. $\endgroup$ – kamenf Apr 21 '16 at 19:13
  • $\begingroup$ Just to be clear: Every letter except G represents one of the digits from 0 to 9? The =, ., and / represent equality, a decimal point and division, respectively? $\endgroup$ – GentlePurpleRain Apr 21 '16 at 20:55
  • $\begingroup$ That the symbols represent themselves is now included in the puzzle statement $\endgroup$ – humn Apr 22 '16 at 11:07

The boss B is


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


  • $\begingroup$ Very clever! I was wondering if G could be $\infty$, but I didn't see how $.\infty$ made any sense. $\endgroup$ – GentlePurpleRain Apr 21 '16 at 21:02
  • 2
    $\begingroup$ This video may be of interest to readers. $\endgroup$ – Will Apr 21 '16 at 21:10

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