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?

  • 5
    $\begingroup$ Having read this question which appeared in the HNQ just today made this puzzle very easy ;-) $\endgroup$ – Christoph May 9 at 19:13
  • $\begingroup$ @Christoph. What is HNQ? $\endgroup$ – Uvc May 9 at 19:23
  • 2
    $\begingroup$ Hot Network Questions, a selection of questions from the whole Stack Exchange network which are "featured" to appear in the right-hand sidebar. $\endgroup$ – Rand al'Thor May 9 at 19:59

Here is the solution

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

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


$2 \times 4^4 = 512$ 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.

  • 1
    $\begingroup$ Nice, you beat me to it. I think I got a slightly slicker solution by deducing some extra conditions required on $A$ and $B$. $\endgroup$ – Rand al'Thor May 9 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


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 $C$.

Final answer

  • 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 $B=4$.

So the final answer is



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