# Deduce the Component Digits from these three Sets of Symmetric Power Relations

Use calculator to the minimum.

Under Logical Deduction, these Daunting Relations will quickly melt away to reveal the constituent digits.

All three sets show beautiful symmetric Relations with first and second part of strings with powers on other side.

A, B, C, D, E, F, G, H are single digits..can vary from 0 to 9.

All other letter combinations are Concatenated Numbers. Leading zeroes allowed in the numbers for symmetry sake.

1st Set:

$$ABCC$$ = $$AB^B$$ + $$CC^B$$

$$DDCC$$ = $$DD^B$$ + $$CC^B$$

2nd Set:

$$EAEAEE$$ = $$EAE^B$$ + $$AEE^B$$

$$FFEAEE$$ = $$FFE^B$$ + $$AEE^B$$

3rd Set:

$$EGDDBCGC$$ = $$EGDD^B$$ + $$BCGC^B$$

$$FHABBCGC$$ = $$FHAB^B$$ + $$BCGC^B$$

Further another relationship exists

$$EGDD$$ + $$FHAB$$ = $$AE*(EAE+FAE)$$ = $$(AB+DD)*AE^B$$

• Is $AB^B$ = $(AB)^B$ or $A(B^B)$? E.g. if $A=1$ and $B=2$, then is $AB^B$ equal to $(12)^2$ or to $10 + 2^2$? Jun 3, 2019 at 5:07
• All letter combinations other than single letter used are concatenated Numbers as stated.
– Uvc
Jun 3, 2019 at 7:01

Set 1:

$$1233 = 12^2 + 33^2$$
$$8833 = 88^2 + 33^2$$

Because:

We can deduce $$B=2$$. It must be $$2$$ or $$3$$ as $$11^4\gt9999$$. If $$B=3$$, then either $$A$$ or $$C$$ is at least $$2$$, and $$22^3>9999$$.

I like this bit!

$$DDCC-ABCC = 100(DD-AB)$$.
But also $$DDCC-ABCC=DD^2-AB^2=(DD-AB)(DD+AB)$$.
Therefore $$DD+AB=100$$, and as $$B=2$$, $$D=8$$ and $$A=1$$.

Finally:

Both $$1200=12^2$$ and $$8800-88^2$$ equal $$1056$$. By inspection, we notice $$1089$$ is the next square, and so $$C=3$$.

The rest:

$$E=0$$ ($$E=5$$ is the only other option, but there is no $$F$$ that works) AND $$F=9$$. $$G+H=9$$, and only $$4,5,6,7$$ remain, and by inspection $$G=5, H=4$$.