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Difference between revisions of "2020 AMC 12A Problems/Problem 25"

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<cmath>nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}</cmath>
 
<cmath>nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}</cmath>
 
We're looking at the solution with smaller <math>x</math>, which is <math>x=\frac{n-n\sqrt{1-4a}}{2a}=\frac{n}{2a}\left(1-\sqrt{1-4a}\right)</math>. Note that if <math>\lfloor x\rfloor=n</math> is the greatest <math>n</math> such that <math>\lfloor x\rfloor \cdot \{x\}=ax^2</math> has a solution, the sum of all these solutions is slightly over <math>\sum_{k=1}^{n}k=\frac{n(n+1)}{2}</math>, which is <math>406</math> when <math>n=28</math>, just under <math>420</math>. Checking this gives  
 
We're looking at the solution with smaller <math>x</math>, which is <math>x=\frac{n-n\sqrt{1-4a}}{2a}=\frac{n}{2a}\left(1-\sqrt{1-4a}\right)</math>. Note that if <math>\lfloor x\rfloor=n</math> is the greatest <math>n</math> such that <math>\lfloor x\rfloor \cdot \{x\}=ax^2</math> has a solution, the sum of all these solutions is slightly over <math>\sum_{k=1}^{n}k=\frac{n(n+1)}{2}</math>, which is <math>406</math> when <math>n=28</math>, just under <math>420</math>. Checking this gives  
−
<cmath>\sum_{k=1}^{28}\frac{n}{2a}\left(1-\sqrt{1-4a}\right)=\frac{1-\sqrt{1-4a}}{2a}\cdot 406=420</cmath>
+
<cmath>\sum_{k=1}^{28}\frac{k}{2a}\left(1-\sqrt{1-4a}\right)=\frac{1-\sqrt{1-4a}}{2a}\cdot 406=420</cmath>
 
<cmath>\frac{1-\sqrt{1-4a}}{2a}=\frac{420}{406}=\frac{30}{29}</cmath>
 
<cmath>\frac{1-\sqrt{1-4a}}{2a}=\frac{420}{406}=\frac{30}{29}</cmath>
 
<cmath>29-29\sqrt{1-4a}=60a</cmath>
 
<cmath>29-29\sqrt{1-4a}=60a</cmath>

Revision as of 18:13, 2 February 2020

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 1

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 $a=\frac{29 \cdot 1}{30^2} = \frac{29}{900} \implies \boxed{\textbf{C}}$.

Solution 2

First note that $\lfloor x\rfloor \cdot \{x\}<0$ when $x<0$ while $ax^2\ge 0\forall x\in \mathbb{R}$. Thus we only need to look at positive solutions ($x=0$ doesn't affect the sum of the solutions). Next, we breakdown $\lfloor x\rfloor\cdot \{x\}$ down for each interval $[n,n+1)$, where $n$ is a positive integer. Assume $\lfloor x\rfloor=n$, then $\{x\}=x-n$. This means that when $x\in [n,n+1)$, $\lfloor x\rfloor \cdot \{x\}=n(x-n)=nx-n^2$. Setting this equal to $ax^2$ gives \[nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}\] We're looking at the solution with smaller $x$, which is $x=\frac{n-n\sqrt{1-4a}}{2a}=\frac{n}{2a}\left(1-\sqrt{1-4a}\right)$. Note that if $\lfloor x\rfloor=n$ is the greatest $n$ such that $\lfloor x\rfloor \cdot \{x\}=ax^2$ has a solution, the sum of all these solutions is slightly over $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$, which is $406$ when $n=28$, just under $420$. Checking this gives \[\sum_{k=1}^{28}\frac{k}{2a}\left(1-\sqrt{1-4a}\right)=\frac{1-\sqrt{1-4a}}{2a}\cdot 406=420\] \[\frac{1-\sqrt{1-4a}}{2a}=\frac{420}{406}=\frac{30}{29}\] \[29-29\sqrt{1-4a}=60a\] \[29\sqrt{1-4a}=29-60a\] \[29^2-4\cdot 29^2a=29^2+3600a^2-120\cdot 29a\] \[3600a^2=116a\] \[a=\frac{116}{3600}=\frac{29}{900} \implies \boxed{\textbf{(C) }929}\] ~ktong

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
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