<p>Determine the sum of all the real solutions of \(\left[\dfrac{x}{2}\right] + \left[\dfrac{2x}{3}\right] = x\).</p>
Step-by-Step Solution
Key Concept: Recognize that the equation involves floor functions, so x must be an integer (since the sum of two floor functions equals x). Then systematically test integer values by analyzing when the floor function equation is satisfied.
<p><strong>Step 1: Determine constraints on x</strong></p><p>For the equation $\left[\frac{x}{2}\right] + \left[\frac{2x}{3}\right] = x$ to hold, the right side must equal an integer. Since the left side is always an integer (sum of two floor functions), x must be an integer.</p><p><strong>Step 2: Analyze for integer x</strong></p><p>Let $x = n$ where $n \in \mathbb{Z}$. Then:</p><p>$\left[\frac{n}{2}\right] + \left[\frac{2n}{3}\right] = n$</p><p><strong>Step 3: Test cases based on n mod 6</strong></p><p>Test $n = 0$: $[0] + [0] = 0$ ✓</p><p>Test $n = 1$: $[0.5] + [0.67] = 0 + 0 = 0 \neq 1$ ✗</p><p>Test $n = 2$: $[1] + [1.33] = 1 + 1 = 2$ ✓</p><p>Test $n = 3$: $[1.5] + [2] = 1 + 2 = 3$ ✓</p><p>Test $n = 4$: $[2] + [2.67] = 2 + 2 = 4$ ✓</p><p>Test $n = 5$: $[2.5] + [3.33] = 2 + 3 = 5$ ✓</p><p>Test $n = 6$: $[3] + [4] = 3 + 4 = 7 \neq 6$ ✗</p><p><strong>Step 4: Identify the pattern</strong></p><p>For $n \geq 6$, we can verify that $\left[\frac{n}{2}\right] + \left[\frac{2n}{3}\right] > n$. This is because $\frac{n}{2} + \frac{2n}{3} = \frac{7n}{6} > n$, and floor functions don't reduce this enough.</p><p>For $n < 0$, similar testing shows no additional solutions.</p><p><strong>Step 5: Calculate sum</strong></p><p>The solutions are: $x \in \{0, 2, 3, 4, 5\}$</p><p>Sum $= 0 + 2 + 3 + 4 + 5 = 14$</p><p>∴ Answer: C</p>
Correct Answer: C