My Digit Sum is 12, Who Can I Be?

If the number is $\overline{AB},$ we have $(10B+A) - (10A+B) = 9(B-A) = 54,$ so $B-A=6.$ We are given that $B+A = 12,$ so $B=9$ and $A=3.$ Thus, the number is $39.$

I just tested the options.

Let $t$ be the ten's digit and $u$ be the unit;s digit, then

$t+u=12$ $\implies$ $\boxed{1}$

$10u+t+54=10t+u$

$9u-9t=-54$

$u-t=-6$ $\implies$ $\boxed{2}$

Adding $\boxed{1}$ and $\boxed{2}$ , we get

$2u=6$

$u=3$

It follows that

$t=12-3=9$

So the number is $\boxed{39}$ .