2020 AMC 12A Problems/Problem 25

Revision as of 16:27, 2 February 2020 by Vedadehhc (talk | contribs) (Solution 1)

Problem 25

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

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

Solution

Let $1<k<2$ be the unique solution in this range. Note that $ck$ is also a solution as long as $ck < c+1$, hence all our solutions are $k, 2k, ..., bk$ for some $b$. This sum $420$ must be between $\frac{b(b+1)}{2}$ and $\frac{(b+1)(b+2)}{2}$, which gives $b=28$ and $k=\frac{420}{406}=\frac{30}{29}$. Plugging this back in gives $a=\frac{29 \cdot 1}{30^2} = \frac{29}{900} \implies \boxed{\textbf{C}}$.

See Also

2020 AMC 12A (ProblemsAnswer KeyResources)
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