# We are two immediate neighbors who forged our own powers to form concatenated relationship. Who are we?

Our concatenated number is $$\overline{ABAC},$$ where $$A, B, C$$ are all positive digits (1 - 9).

Our relationship is

$$\overline{ABAC} = A^A + B^B + A^A + C^C$$

Who are we?

• Having read this question which appeared in the HNQ just today made this puzzle very easy ;-) – Christoph May 9 '19 at 19:13
• @Christoph. What is HNQ? – Uvc May 9 '19 at 19:23
• Hot Network Questions, a selection of questions from the whole Stack Exchange network which are "featured" to appear in the right-hand sidebar. – Rand al'Thor May 9 '19 at 19:59

Here is the solution

$$A=3, B=4, C=5$$

$$3^3 + 4^4 + 3^3 + 5^5 = 3435$$

Reasoning

$$2 \times 4^4 = 512$$ and $$6^6=46656$$ which is more than four digits so at least one of the digits has to be $$5$$ and the others have to be less than or equal to $$5$$.
Then, we have $$5^5=3125$$ and since double this is more than $$6000$$, the digit $$5$$ can appear only once. Furthermore, since $$3125 + 2 \times 4^4 < 4000$$, the digit $$3$$ must also appear at least once.
Since $$5^5 + 3^3 = 3152$$, the other unique digit to appear must either be $$1$$ or $$4$$ (as this will necessarily be the $$2$$nd digit). This leaves a small number of possibilities to try.

• Nice, you beat me to it. I think I got a slightly slicker solution by deducing some extra conditions required on $A$ and $B$. – Rand al'Thor May 9 '19 at 16:05

## Initial bounds

• $$6^6$$ is too big as it has 5 digits.

• $$4^4$$ is only 256, too small if everything was at most that.

So at least one of $$A,B,C$$ must be

$$5$$.

## Narrowing possibilities

• $$B$$ must be even, because $$ABAC$$ and $$C^C$$ have matching parity. So $$B=2$$ or $$B=4$$.

• If $$A=5$$, then $$A^A+A^A$$ is already bigger than $$6000$$, too big. But $$5^5=3125$$ is involved somewhere, so $$A$$ must be at least $$3$$. So $$A=3$$ or $$A=4$$.

So the only option

to be $$5$$ is $$C$$.

• Trying $$A=4$$ gives $$4B45=3637+B^B$$, which is impossible since $$2^2$$ and $$4^4$$ are too small to get that high.
• Trying $$A=3$$ gives $$3B35=3179+B^B$$, which works with $$B=4$$.
$$A=3,B=4,C=5$$.