The height and the base radius of a closed right circular cylinder are positive integers and its total surface area is numerically equal to its volume. If its volume is $$k \pi $$ where $$𝑘$$ is a positive integer, what is the smallest possible value of 𝑘?

The total surface area (TSA) of a closed right circular cylinder of radius $$r$$ and height $$h$$ is
$$\text{TSA}=2\pi r(r+h).$$

The volume is
$$\text{Volume}= \pi r^{2}h.$$

The condition in the problem states that these two are numerically equal:
$$2\pi r(r+h)=\pi r^{2}h.$$

Cancel $$\pi$$ from both sides:
$$2r(r+h)=r^{2}h \; \; -(1)$$

Divide by $$r\,(r\gt0)$$:
$$rh=2(r+h). \; \; -(2)$$

Rearrange $$-(2)$$ to isolate a factor of $$(r-2)$$:
$$rh-2h=2r \;\;\Longrightarrow\;\;h(r-2)=2r. \; \; -(3)$$

Let $$d=r-2$$. Because $$r$$ is a positive integer and $$r\ge3$$ (else $$h$$ would be non-positive), $$d$$ is a positive integer. Substitute $$r=d+2$$ into $$-(3)$$:
$$h\,d = 2(d+2)=2d+4.$$

Solve for $$h$$:
$$h = 2+\frac{4}{d}. \; \; -(4)$$

For $$h$$ to be an integer, $$\dfrac{4}{d}$$ must be an integer, so $$d$$ must be a positive divisor of $$4$$.
Possible $$d$$ values: $$1,2,4$$.

Compute the corresponding volumes $$V=\pi r^{2}h$$:

Case 1: $$V=\pi\,(3)^{2}(6)=54\pi$$
Case 2: $$V=\pi\,(4)^{2}(4)=64\pi$$
Case 3: $$V=\pi\,(6)^{2}(3)=108\pi$$

The smallest positive integer $$k$$ for which the volume equals $$k\pi$$ is therefore $$k=54$$.

Answer: 54