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Sample Question 25:
Step 1: Start with the given equation \(a \cdot x^2 = \lfloor x \rfloor \cdot \{x\}\).
Step 2: Rewrite this equation by let \(w = \lfloor x \rfloor\) (the whole part of \(x\)) and \(f = \{x\}\) (the fractional part of \(x\)). Hence, we obtain \(w \cdot f = a \cdot (w + f)^2\).
Step 3: Rearrange this equation as \(a \cdot f^2 + (2a - 1)w \cdot f + a \cdot w^2 = 0\).
Step 4: Solve this quadratic equation for \(f\) (assuming \(w\) is fixed) using the Quadratic Formula: \(f = w \left(\frac{1 - 2a ± \sqrt{1 - 4a}}{2a}\right)\).
Step 5: Consider the range of solutions for \(f\). Since the equation is quadratic, we need both roots to be positive, which requires \(0 < a \leq \frac{1}{4}\).
Step 6: More closely analyze the resulting solution: \(f = w \left(\frac{1}{2a} - 1 - \frac{\sqrt{1 - 4a}}{2a}\right)\). To ensure \(0 \leq f < 1\), derive a constant \(k = \frac{1}{2a} - 1 - \frac{\sqrt{1 - 4a}}{2a} > 0\), such that \(f = w \cdot k\), and \(k < \frac{1}{w}\).
Step 7: Sum up the solutions: the sum of all solutions to the original equation is \((1 + k)\cdot \frac{W(W + 1)}{2}\), where \(W\) is the largest \(w\) that produces solutions. This sum equals 420, and hence \(1 + k\) is slightly above 1, making \(\frac{W(W + 1)}{2}\) slightly below 840. This leads us to deduce \(W = 28\), and thus \(k = \frac{1}{29}\).
Step 8: Solve for \(a\) in \(k = \frac{1}{2a} - 1 - \frac{\sqrt{1 - 4a}}{2a}\) leads to \(a = \frac{29}{900}\).
Step 9: Therefore, \(p + q = 29+900 = \boxed{\textbf{(C) } 929}\).
The number \(a = \tfrac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers, has the property that the sum of all real numbers \(x\) satisfying\(\lfloor x \rfloor \cdot \{x\} = a \cdot x^2\)is \(420\), where \(\lfloor x \rfloor\) denotes the greatest integer less than or equal to \(x\) and \(\{x\} = x - \lfloor x \rfloor\) denotes the fractional part of \(x\). What is \(p + q?\)
\(\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332\)
Answer Keys
Question 25: CSolutions
Question 25Step 1: Start with the given equation \(a \cdot x^2 = \lfloor x \rfloor \cdot \{x\}\).
Step 2: Rewrite this equation by let \(w = \lfloor x \rfloor\) (the whole part of \(x\)) and \(f = \{x\}\) (the fractional part of \(x\)). Hence, we obtain \(w \cdot f = a \cdot (w + f)^2\).
Step 3: Rearrange this equation as \(a \cdot f^2 + (2a - 1)w \cdot f + a \cdot w^2 = 0\).
Step 4: Solve this quadratic equation for \(f\) (assuming \(w\) is fixed) using the Quadratic Formula: \(f = w \left(\frac{1 - 2a ± \sqrt{1 - 4a}}{2a}\right)\).
Step 5: Consider the range of solutions for \(f\). Since the equation is quadratic, we need both roots to be positive, which requires \(0 < a \leq \frac{1}{4}\).
Step 6: More closely analyze the resulting solution: \(f = w \left(\frac{1}{2a} - 1 - \frac{\sqrt{1 - 4a}}{2a}\right)\). To ensure \(0 \leq f < 1\), derive a constant \(k = \frac{1}{2a} - 1 - \frac{\sqrt{1 - 4a}}{2a} > 0\), such that \(f = w \cdot k\), and \(k < \frac{1}{w}\).
Step 7: Sum up the solutions: the sum of all solutions to the original equation is \((1 + k)\cdot \frac{W(W + 1)}{2}\), where \(W\) is the largest \(w\) that produces solutions. This sum equals 420, and hence \(1 + k\) is slightly above 1, making \(\frac{W(W + 1)}{2}\) slightly below 840. This leads us to deduce \(W = 28\), and thus \(k = \frac{1}{29}\).
Step 8: Solve for \(a\) in \(k = \frac{1}{2a} - 1 - \frac{\sqrt{1 - 4a}}{2a}\) leads to \(a = \frac{29}{900}\).
Step 9: Therefore, \(p + q = 29+900 = \boxed{\textbf{(C) } 929}\).