Question

If $12^{12x} \times 4^{24x+12} \times 5^{2y} = 8^{4z} \times 20^{12x} \times 243^{3x-6}$, where $x$, $y$ and $z$ are natural numbers, then $x + y + z$ equals

Hint:

Whenever you see an equation involving different bases with exponents, convert all bases to prime factors. Then compare the exponents of each prime on both sides — it turns the problem into a simple system of linear equations.

Answer 1

Correct Answer: 112

Step 1: Express all bases using prime factors.
Rewrite each base in terms of primes 2, 3 and 5: \[ 12 = 2^2 \cdot 3,\quad 4 = 2^2,\quad 5 = 5,\quad 8 = 2^3,\quad 20 = 2^2 \cdot 5,\quad 243 = 3^5. \] Substitute into the given equation: \[ (2^2 \cdot 3)^{12x} \times (2^2)^{24x+12} \times 5^{2y} = (2^3)^{4z} \times (2^2 \cdot 5)^{12x} \times (3^5)^{3x-6}. \] Step 2: Simplify the exponents. Left-hand side (LHS): \[ (2^2 \cdot 3)^{12x} = 2^{24x} \cdot 3^{12x}, \] \[ (2^2)^{24x+12} = 2^{48x+24}. \] So LHS becomes: \[ 2^{24x} \cdot 3^{12x} \cdot 2^{48x+24} \cdot 5^{2y} = 2^{(24x + 48x + 24)} \cdot 3^{12x} \cdot 5^{2y} = 2^{72x + 24} \cdot 3^{12x} \cdot 5^{2y}. \] Right-hand side (RHS): \[ (2^3)^{4z} = 2^{12z}, \] \[ (2^2 \cdot 5)^{12x} = 2^{24x} \cdot 5^{12x}, \] \[ (3^5)^{3x-6} = 3^{5(3x-6)} = 3^{15x - 30}. \] So RHS becomes: \[ 2^{12z} \cdot 2^{24x} \cdot 5^{12x} \cdot 3^{15x - 30} = 2^{12z + 24x} \cdot 3^{15x - 30} \cdot 5^{12x}. \] Step 3: Equate exponents of corresponding primes. Since the expressions are equal and factorized into primes, equate exponents of 2, 3 and 5. \underline{For base 3:} \[ 12x = 15x - 30 \Rightarrow 30 = 3x \Rightarrow x = 10. \] \underline{For base 5:} \[ 2y = 12x. \] Substitute \(x = 10\): \[ 2y = 12 \cdot 10 = 120 \Rightarrow y = 60. \] \underline{For base 2:} \[ 72x + 24 = 12z + 24x. \] Substitute \(x = 10\): \[ 72 \cdot 10 + 24 = 12z + 24 \cdot 10 \Rightarrow 720 + 24 = 12z + 240 \Rightarrow 744 = 12z + 240 \Rightarrow 12z = 504 \Rightarrow z = 42. \] Step 4: Compute \(x + y + z\). \[ x + y + z = 10 + 60 + 42 = 112. \] Thus, \(x + y + z = \boxed{112}\).