<p>Consider the equation \(\dfrac{2}{x} + \dfrac{5}{y} = \dfrac{1}{3}\), where \(x, y \in \mathbb{N}\). Find the number of solutions of the equation.</p>
Step-by-Step Solution
Key Concept: Rearrange the equation to isolate one variable, then apply divisibility constraints on natural numbers. The denominator condition forces specific factorizations of the numerator.
<p><strong>Step 1:</strong> Rearrange the equation:</p><p>$$\frac{2}{x} + \frac{5}{y} = \frac{1}{3}$$</p><p>$$\frac{2}{x} = \frac{1}{3} - \frac{5}{y} = \frac{y-15}{3y}$$</p><p><strong>Step 2:</strong> Cross-multiply to get:</p><p>$$6y = x(y-15)$$</p><p>$$x = \frac{6y}{y-15}$$</p><p><strong>Step 3:</strong> For x to be a natural number, (y-15) must divide 6y. Since (y-15) divides 6y and (y-15) divides (y-15)·6, we have (y-15) divides 6y - 6(y-15) = 90.</p><p><strong>Step 4:</strong> So (y-15) must be a divisor of 90. The divisors of 90 are: ±1, ±2, ±3, ±5, ±6, ±9, ±10, ±15, ±18, ±30, ±45, ±90.</p><p><strong>Step 5:</strong> Since y ∈ ℕ, we need y > 0. Also, for x > 0: $$x = \frac{6y}{y-15} > 0$$ requires y > 15 (since y must be positive).</p><p><strong>Step 6:</strong> For y > 15, we need (y-15) to be a positive divisor of 90:</p><p>y - 15 ∈ {1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90}</p><p>This gives: y ∈ {16, 17, 18, 20, 21, 24, 25, 30, 33, 45, 60, 105}</p><p><strong>Step 7:</strong> Verify each gives natural number x:</p><p>y=16: x=96 ✓ | y=17: x=51 ✓ | y=18: x=36 ✓ | y=20: x=20 ✓ | y=21: x=18 ✓ | y=24: x=16 ✓ | y=25: x=15 ✓ | y=30: x=12 ✓ | y=33: x=11 ✓ | y=45: x=9 ✓ | y=60: x=8 ✓ | y=105: x=7 ✓</p><p>∴ Answer: <strong>12</strong></p>
Correct Answer: 12