If $12^{12x}\times4^{24x+12}\times5^{2y}=8^{4z}\times20^{12x}\times243^{3x-6}$, where $x$, $y$ and $z$ are natural numbers, then $x+y+z$ equals

Every base is expressed as a product of primes ($2, 3, 5$):

$12^{12x} = 2^{24x} \times 3^{12x}$

$4^{24x+12} = 2^{48x+24}$

$8^{4z} = 2^{12z}$

$20^{12x} = 2^{24x} \times 5^{12x}$

$243^{3x-6} = 3^{15x-30}$

---

LHS $= 2^{24x} \times 3^{12x} \times 2^{48x+24} \times 5^{2y}$

$= 2^{72x+24} \times 3^{12x} \times 5^{2y}$

RHS $= 2^{12z} \times 2^{24x} \times 5^{12x} \times 3^{15x-30}$

$= 2^{12z+24x} \times 3^{15x-30} \times 5^{12x}$

---

Since both sides are products of the same primes, the power of each prime must match on both sides.

---

Comparing powers of $3$:

$12x = 15x - 30$

$-3x = -30$

$x = 10$

---

Comparing powers of $5$:

$2y = 12x = 12(10)$

$y = 60$

---

Comparing powers of $2$:

$72x + 24 = 12z + 24x$

$48x + 24 = 12z$

$48(10) + 24 = 12z$

$504 = 12z$

$z = 42$

---

$x + y + z = 10 + 60 + 42 = 112$