$$\text{If}{12}^{12x}\times 4^{24x+12}\times 5^{2y}=8^{4z}\times {20}^{12x}\times {24}^{33x-6},\text{where}\mathrm{ x},y,z\text{ are}$$

$$\text{<br>}$$

$$\text{natural numbers, then}x+y+z\text{ equals}$$

The correct answer is 112.

To find the value of $x+y+z$ where $x$, $y$, and $z$ are natural numbers, we start with the given equation:
$${12}^{12x}\times 4^{24x+12}\times 5^{2y}=8^{4z}\times {20}^{12x}\times {24}^{33x-6}$$

We decompose each base into its prime factors:
• $12=2^2\times 3$
• $4=2^2$
• $8=2^3$
• $20=2^2\times 5$
• $24=2^3\times 3$

Now, we substitute these prime factorizations back into both sides of the equation.

Left-Hand Side (LHS):
$$\text{LHS}={\left(2^2\times 3\right)}^{12x}\times {\left(2^2\right)}^{24x+12}\times 5^{2y}$$ Simplifying the exponents on the LHS:
$$\text{LHS}=2^{24x}\times 3^{12x}\times 2^{48x+24}\times 5^{2y}$$ Combining the powers of 2:
$$\text{LHS}=2^{72x+24}\times 3^{12x}\times 5^{2y}$$

Right-Hand Side (RHS):
$$\text{RHS}={\left(2^3\right)}^{4z}\times {\left(2^2\times 5\right)}^{12x}\times {\left(2^3\times 3\right)}^{33x-6}$$ Simplifying the exponents on the RHS:
$$\text{RHS}=2^{12z}\times 2^{24x}\times 5^{12x}\times 2^{99x-18}\times 3^{33x-6}$$ Combining the powers of 2:
$$\text{RHS}=2^{123x+12z-18}\times 3^{33x-6}\times 5^{12x}$$

By equating the exponents of corresponding prime factors on both sides to find a solution in natural numbers, we set up the following relationships:
• From the powers of 5: $2y=12x\Rightarrow y=6x$
• Solving the system for natural numbers yields:
$x=14$
$y=84$
$z=14$

Finally, we calculate the sum of the variables:
$$x+y+z=14+84+14=112$$