Sample Question 25:

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 $p + q?$

$\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332$

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Answer Keys

Question 25: C

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Solutions

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, $p + q = 29+900 = \boxed{\textbf{(C) } 929}$.