What is the side length of the largest square tile that can tile a room with dimensions 5.4m by 4.2m?

The length of the room is $5.4 \text{ m}$.
The width of the room is $4.2 \text{ m}$.
Since $1 \text{ m} = 100 \text{ cm}$:
Length ($L$) $= 5.4 \times 100 \text{ cm} = 540 \text{ cm}$
Width ($W$) $= 4.2 \times 100 \text{ cm} = 420 \text{ cm}$

To cover the rectangular floor with square tiles without cutting any tiles, the side length of the square tile must be a common divisor of both the length and the width of the room. To find the minimum number of tiles, we need to use the largest possible square tiles. This means the side length of the tile must be the Greatest Common Divisor (GCD) of the room's length and width.

We need to find $\text{GCD}(540, 420)$.
We can use prime factorization:

The common prime factors with their lowest powers are $2^2$, $3^1$, and $5^1$.

So, the side length of the largest square tile is $60 \text{ cm}$.

Number of tiles along the length $= \frac{\text{Length of room}}{\text{Side length of tile}} = \frac{540 \text{ cm}}{60 \text{ cm}} = 9$ tiles
Number of tiles along the width $= \frac{\text{Width of room}}{\text{Side length of tile}} = \frac{420 \text{ cm}}{60 \text{ cm}} = 7$ tiles

The total number of tiles is the product of the number of tiles along the length and the number of tiles along the width.
Total number of tiles $= 9 \times 7 = 63$

The minimum number of square tiles that can be used to cover the floor is $\boxed{\text{63}}$.