, 2025

We will begin by making the substitution $A=2^x$, $B=2^y$, and $C=3^z$.

With these substitutions, the three equations become $$\begin{align*} A+B+\frac{1}{3}C &= 2259 \tag{1} \\ AB+C &= 7073 \tag{2} \\ A+B+C &= 6633 \tag{3}\end{align*}$$ Subtracting Equation
$(1)$ from Equation $(3)$ gives $\dfrac{2}{3}C = 6633-2259 = 4374$.

Hence, $C=\dfrac{3}{2}\cdot4374=6561$.

Substituting $C=6561$ into Equation
$(2)$ gives $AB+6561 = 7073$ or $AB=512$.

Substituting $C=6561$ into Equation
$(3)$ gives $A+B+6561=6633$ or $A+B=72$.

Multiplying both sides of $A+B=72$ by
$A$ (since $A\neq0$) gives $A^2+AB=72A$, into which we can substitute
$AB=512$ to get $A^2+512=72A$.

Rearranging gives $A^2-72A+512=0$
which can be factored to get $(A-64)(A-8)=0$.

If $A=64$, then either $AB=512$ or $A+B=72$ can be used to deduce that $B=8$, and if $A=8$, then we could similarly deduce that
$B=64$.

Thus, we have that $A$ and $B$ are $8$ and $64$, but we cannot determine the order.
Since $64=2^6$ and $8=2^3$, we conclude that $x$ and $y$ are $6$ and $3$, though we cannot determine with
certainty which is which.

Returning to $C=6561$, we have $3^z=6561=3^8$ so $z=8$.

Regardless of which of $x$ and $y$ is $8$ and which is $64$, we get that $xyz=3\times6\times8=144$. The integer
formed by the rightmost two digits of $144$ is $44$.