5
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$D$, $E$, $F$, $G$, $H$, $S$, $T$, $U$, $V$ are distinct digits and can vary from 0 to 9.

$DE$, $FGH$, $ESDE$, $SS$, $ST$, $SU$ are all concatenated Numbers.

From the given Relations below, deduce all the digits easily.

$DE$ = $S^T$ + $T^T$

$FGH$ = $S^T$ + $T^T$ + $U^T$ + $V^T$

$ESDE$ = $S^T$ + $T^T$ + $U^T$ + $V^T$ + $G^T$ + $SS^T$ + $ST^T$ + $SU^T$

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  • $\begingroup$ Are leading zeros allowed? E.g. could $D$, $F$, or $E$ be zero? $\endgroup$ – Rand al'Thor Jun 22 at 10:35
  • $\begingroup$ We have 9 letters but 10 (0-9) digits, so how can we deduce $V$ if it is not in the relations? $\endgroup$ – TheSimpliFire Jun 22 at 10:37
  • $\begingroup$ Leading zeroes not allowed.. $\endgroup$ – Uvc Jun 22 at 10:42
  • $\begingroup$ I apologize..second t^t should be v^t..will edit..thx for spotting the error $\endgroup$ – Uvc Jun 22 at 10:44
  • $\begingroup$ Odds are tilted very heavily to one side $\endgroup$ – Uvc Jun 22 at 11:00
4
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First thing to notice

In the third relation,

the powers on the right-hand side must be small enough to give a four-digit number in total. So there are only two options: either $T=2$, or $T=3$ and $S=1$. (Clearly $T$ isn't 0 or 1, as that would give too many leading zeros among the distinct digits.)

Option 1

If $T=2$, then in the third relation we must have $S\leq5$ otherwise the right-hand side would be more than four digits. So $S$ is one of 1,3,4,5.

The first relation is

$DE=S^2+4$. But $1^2+4=5$ (not two digits), $3^2+4=13$ (reusing digit 3), $4^2+4=20$ (reusing digit 2), $5^2+4=29$ (reusing digit 2). Contradiction!

Option 2

If $T=3$ and $S=1$, then the first relation is $1^3+3^3=28$, so $D=2$ and $E=8$.

The second relation is

$FGH=28+U^3+V^3$, where we cannot reuse the digits $1,2,3,8$. The only possibility is $28+5^3+7^3=496$, so we have $F=4,G=9,H=6$, and $\{U,V\}=\{5,7\}$ in some order.

Now the third relation is

$8128=496+9^3+11^3+13^3+1U^3$, which works with $U=5$.

Final solution

$S=1,D=2,T=3,F=4,U=5,H=6,V=7,E=8,G=9$.

$$28=1^3+3^3$$
$$496=1^3+3^3+5^3+7^3$$
$$8128=1^3+3^3+5^3+7^3+9^3+11^3+13^3+15^3$$

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  • $\begingroup$ @Uvc I use iPad all the time (on it right now in fact) and deletion of comments works fine from either the web interface or the stackexchange app. For the latter which I assume you’re using, click on the comment in question and the options to upvote / reply / flag / delete / edit etc appear below it. $\endgroup$ – Rubio Jun 22 at 14:59
  • $\begingroup$ Will try and figure it out...thx $\endgroup$ – Uvc Jun 22 at 15:19
  • $\begingroup$ Did you notice that 28, 496 and 8128 are all perfect numbers? (very clever title, @Uvc!) $\endgroup$ – TheSimpliFire Jun 22 at 18:42
  • $\begingroup$ @TheSimpliFire Wow, these relations are actually amazing! :-O $\endgroup$ – Rand al'Thor Jun 22 at 18:50
  • 1
    $\begingroup$ @OmegaKrypton 6 is a special case as it is too small to fit into the formula. For further details you may wish to take a look at: arxiv.org/pdf/1504.07322.pdf (page 2 gives the theorem) $\endgroup$ – TheSimpliFire Jun 23 at 7:48

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