Given GCD = 30.

Given GCD $= 30$.

$$30 = 2 \times 3 \times 5$$

Given LCM $= 900$.

$$900 = 9 \times 100 = 3^2 \times 10^2 = 3^2 \times (2 \times 5)^2 = 2^2 \times 3^2 \times 5^2$$

The two given numbers are 60 and 150.

$$60 = 2^2 \times 3 \times 5$$ $$150 = 2 \times 3 \times 5^2$$

Let the third number be $C$. We can express $C$ in its prime factorization form as $C = 2^x \times 3^y \times 5^z$.

For the GCD of three numbers, the exponent of each prime factor is the minimum of the exponents of that prime factor in the three numbers.
For the LCM of three numbers, the exponent of each prime factor is the maximum of the exponents of that prime factor in the three numbers.

Let's analyze each prime factor:

For the prime factor 2:
The exponents in 60, 150, and $C$ are 2, 1, and $x$ respectively.
From GCD: $\min(2, 1, x) = 1$ (since $30 = 2^1 \times 3^1 \times 5^1$). This implies $x \ge 1$.
From LCM: $\max(2, 1, x) = 2$ (since $900 = 2^2 \times 3^2 \times 5^2$). This implies $x \le 2$.
To find the least possible value of $C$, we choose the smallest possible exponent for 2, which is $x=1$.

For the prime factor 3:
The exponents in 60, 150, and $C$ are 1, 1, and $y$ respectively.
From GCD: $\min(1, 1, y) = 1$. This implies $y \ge 1$.
From LCM: $\max(1, 1, y) = 2$. This implies $y$ must be 2. If $y$ were 1, $\max(1,1,1)$ would be 1, not 2.
So, for $y$, we must choose $y=2$.

For the prime factor 5:
The exponents in 60, 150, and $C$ are 1, 2, and $z$ respectively.
From GCD: $\min(1, 2, z) = 1$. This implies $z \ge 1$.
From LCM: $\max(1, 2, z) = 2$. This implies $z \le 2$.
To find the least possible value of $C$, we choose the smallest possible exponent for 5, which is $z=1$.

Using the chosen exponents $x=1$, $y=2$, and $z=1$:

$$C = 2^1 \times 3^2 \times 5^1$$ $$C = 2 \times 9 \times 5$$ $$C = 10 \times 9$$ $$C = 90$$

Numbers: $60 = 2^2 \times 3^1 \times 5^1$, $150 = 2^1 \times 3^1 \times 5^2$, $90 = 2^1 \times 3^2 \times 5^1$.
GCD: $2^{\min(2,1,1)} \times 3^{\min(1,1,2)} \times 5^{\min(1,2,1)} = 2^1 \times 3^1 \times 5^1 = 30$. (Correct)
LCM: $2^{\max(2,1,1)} \times 3^{\max(1,1,2)} \times 5^{\max(1,2,1)} = 2^2 \times 3^2 \times 5^2 = 4 \times 9 \times 25 = 900$. (Correct)

The least possible value of the third number is $\text{}90\text{}$.