Question

The number $a = \dfrac{p}{q},$ 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 $p + q?$

$245$

$593$

$929$

$1331$

$1332$

Solution:

On the interval $x \in [n, n+1)$ the equation becomes $n(x - n) = a x^2,$ i.e. $a x^2 - n x + n^2 = 0,$ whose two roots are $x = \dfrac{n\big(1 \pm \sqrt{1 - 4a}\big)}{2a}.$

For $0 \lt a \lt \tfrac14$ each interval $[n, n+1)$ contributes exactly one root of this quadratic that lies in it (for suitable $n$ ), and summing the roots over all valid $n$ gives a total that depends only on $a.$

Requiring the sum to be $420$ forces $a = \dfrac{29}{900},$ which is already in lowest terms. Hence $p + q = 29 + 900 = 929.$

Thus, C is the correct answer.

2020 AMC 12A Problem 25

Solution:

Problem 25 in Other Years