A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is

To keep the total number of cylinders to a minimum, the volume of each cylinder must be maximized. Since each cylinder has the same volume, the volume of a single cylinder must be the Highest Common Factor (HCF) of the volumes of the three metals: 405 cc, 783 cc, and 351 cc.

Let's find the HCF of 405, 783, and 351:
405 = 9 × 45 = 27 × 15 = 81 × 5 = 3⁴ × 5
783 = 9 × 87 = 27 × 29 = 3³ × 29
351 = 9 × 39 = 27 × 13 = 3³ × 13
The HCF of 405, 783, and 351 is 27. Thus, the volume of each cylinder $V=27{\text{cm}}^3$.

The radius $r$ of each cylinder is 3 cm. The volume of a cylinder is:
$V=\pi r^{2}h$
$27=\pi \left(3^2\right)h$
$27=9\pi h$
$\pi h=3$
$h=\frac{3}{\pi }\text{cm}$

The total number of cylinders is:
$N=\frac{405+783+351}{27}=15+29+13=57$

The total surface area of one cylinder is:
$S=2\pi r(r+h)=2\pi rh+2\pi r^2$
Substituting $r=3$ and $\pi h=3$:
$S=2\left(\pi h\right)r+2\pi \left(3^2\right)=2\left(3\right)\left(3\right)+18\pi =18+18\pi =18(1+\pi )$

The total surface area of all 57 cylinders is:
$S_{\text{total}}=57\times 18(1+\pi )=1026(1+\pi )$

As per the options, the value is $1026(1+\pi )$, which corresponds to option 4 (written as 1026(1+π) or 1026(1)).