# Perfect Powered Relations - Please Figure them out

$$D$$, $$E$$, $$F$$, $$G$$, $$H$$, $$S$$, $$T$$, $$U$$, $$V$$ 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.

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

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

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

• Are leading zeros allowed? E.g. could $D$, $F$, or $E$ be zero? – Rand al'Thor Jun 22 '19 at 10:35
• We have 9 letters but 10 (0-9) digits, so how can we deduce $V$ if it is not in the relations? – TheSimpliFire Jun 22 '19 at 10:37
• Leading zeroes not allowed.. – Uvc Jun 22 '19 at 10:42
• I apologize..second t^t should be v^t..will edit..thx for spotting the error – Uvc Jun 22 '19 at 10:44
• Odds are tilted very heavily to one side – Uvc Jun 22 '19 at 11:00

## 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

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

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

• @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. – Rubio Jun 22 '19 at 14:59
• Will try and figure it out...thx – Uvc Jun 22 '19 at 15:19
• Did you notice that 28, 496 and 8128 are all perfect numbers? (very clever title, @Uvc!) – TheSimpliFire Jun 22 '19 at 18:42
• @TheSimpliFire Wow, these relations are actually amazing! :-O – Rand al'Thor Jun 22 '19 at 18:50
• @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) – TheSimpliFire Jun 23 '19 at 7:48