Fibonacci Cycles Back…Figure out the Fibonacci Number FBN

$$Given$$:

$$F$$, $$B$$, $$N$$, $$U$$, $$V$$ 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 $$FBN$$?

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.

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