2
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$Given$:

$F$, $B$, $N$, $U$, $V$ are all digits that can vary from 0 to 9..but not necessarily distinct.

$FBN$, $NBF$, $UV$, $VU$ are all concatenated numbers.

From information given below, what is $FBN$?

enter image description here

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3
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One possible solution

$F=U=1$, $V=2$, $N=B=4$

Reasoning

$144 = 12^2$ and $441=21^2$ and $FBN=144$ is a Fibonacci number.

Alternative,

It's easy to show that the above is the only solution with $F>0$ as there are only five Fibonacci numbers between $100$ and $1000$ and only one is a square. If we also allow $F=0$ then there are two other possible solutions.
$$F=B=U=0, V=N=1$$ $$F=B=U=V=N=0 $$ These would satisfy the constraints given that the reversal strictly implies a $3$-digit reversal at the top and a $2$-digit reversal at the bottom.

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  • $\begingroup$ 121 is not a fibonacci $\endgroup$ – Uvc Jun 16 at 15:02
  • $\begingroup$ @Uvc sorry I missed that requirement, updated now. $\endgroup$ – hexomino Jun 16 at 15:06
  • $\begingroup$ That’s much better... $\endgroup$ – Uvc Jun 16 at 15:07

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