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$Given$:

D, E, V, I, L, G, N, O, T are distinct digits (1 to 9).

$D+E$ brings good luck

$I+L$ is unlucky

$N+O-T$ is unlucky too.

$DDDD$ = $(D+E)^V$ + $(I+L)^G$ + $(N+O)^T$

Fully Figure out this unusual unique devil.both Rep Digit and Pan Digital.

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    $\begingroup$ whether a number is lucky and unlucky varies greatly in cultures... is this too broad? $\endgroup$ – Omega Krypton Jun 1 '19 at 11:20
  • $\begingroup$ True..for puzzles sake...think them as clues ..leading to what if scenarios..leading to restricted choices for the digits..ultimately to final resolution $\endgroup$ – Uvc Jun 1 '19 at 13:19
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It is:

$6666 = (6+1)^4 + 13^2 + 16^3$, with $I,L$ and $N,O$ being indeterminate.

because:

$D+E=7, I+L=13, N+O-T=13$. We know $I,L,N,O,T,V,G\gt1$, so $D$ or $E$ equals $1$. (and $6666$ looks like the obvious choice!). $V,G,T\in\{2,3,4\}$ as $7^5\gt6666$. $N+O\gt14$ (as $T\gt1$), and considering $15^2, 16^3$ ($17^4$ is way too big), and noting that $13^4\gt6666$ gives only two options, leaving $7^4=2401$ as the first addend, and $D=6$.

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  • $\begingroup$ Excellent!!.... $\endgroup$ – Uvc Jun 1 '19 at 13:19

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