2021 WSMO Accuracy Round Problems/Problem 10

Problem

The largest value of $x$ that satisfies the equation $5x^2-7\lfloor x\rfloor\{x\}=\frac{26\lfloor x\rfloor^2}{5}$ can be expressed as $\frac{a+b\sqrt{c}}{d},$ where $c$ is not divisible by the square of any prime and $\gcd(a,b,d)=1.$ Find $a+b+c+d.$ ( $\{x\}$ denotes the fractional part of $x$ , or $x-\lfloor x\rfloor$ .)

Solution

Let $\lfloor x \rfloor = n$ and $\{x\} = r$ . Plugging this into our equation yields: $5(n+r)^2-7nr=\frac{26n^2}{5}$ $\Rightarrow 25r^2 + 15rn - n^2 = 0$ $\Rightarrow r = \frac{n(\pm \sqrt{13}-3)}{10}$

We are finding the largest $x$ , so we try to maximize $n$ . Assume $n>0$ . Since $r=\{x\}\geq 0$ , we take $r = \frac{n(\sqrt{13}-3)}{10}$ . Note that: $0 \leq r < 1$ $\Rightarrow 0 \leq \frac{n(\sqrt{13}-3)}{10} < 1$ $\Rightarrow 0 \leq n < \frac{5(\sqrt{13}+3)}{2}$

Notice that $\frac{5(\sqrt{9}+3)}{2} < \frac{5(\sqrt{13}+3)}{2} < \frac{5(\sqrt{16}+3)}{2}$ $\Rightarrow 15 < \frac{5(\sqrt{13}+3)}{2} < 17.5$ . We start by checking $n=17$ : $r = \frac{n(\sqrt{13}-3)}{10}$ $\Rightarrow r = \frac{17(\sqrt{13}-3)}{10}$

This solution is valid if $0 \leq r < 1$ : $\frac{17(\sqrt{13}-3)}{10} < 1$ $\Rightarrow 17\sqrt{13} < 61$ $\Rightarrow 17^2 \cdot 13 < 61^2$ $\Rightarrow 3757 < 3721$

This is clearly false, so we now check $n=16$ . Using the same method as before, $r = \frac{16(\sqrt{13}-3)}{10} < 1$ $\Rightarrow 16\sqrt{13} < 58$ $\Rightarrow 3328 < 3364$

This is true, so $x = n+r = 16 + \frac{16(\sqrt{13}-3)}{10} = \frac{56 + 8\sqrt{13}}{5}$ , giving us an answer of $56+8+13+5 = \boxed{82}$ .

~BigKahuna227

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2021 WSMO Accuracy Round Problems/Problem 10

Problem

Solution