Problem (Algebra):
If $\lceil u\rceil$ denotes the least integer greater than or equal to $u$, and $\lfloor u\rfloor$ denotes the greatest integer less than or equal to $u$, compute the largest solution $x$ to the equation

$$
\left\lfloor\frac{x}{3}\right\rfloor+\lceil 3 x\rceil=\sqrt{11} \cdot x
$$

Solution:
Let $f(x)=\left\lfloor\frac{x}{3}\right\rfloor+\lceil 3 x\rceil$. Observe that $f(x+3)=f(x)+1+9=f(x)+10$. Let $g(x)=f(x)-\frac{10}{3} x$. Then $g$ is periodic, because $g(x+3)=f(x)+10-\frac{10 x}{3}-\frac{10 \cdot 3}{3}=g(x)$. The graph of $g$ is shown below:



Because $g(x)$ is the (vertical) distance between the graph of $y=f(x)$ and the line $y=\frac{10}{3} x$, the fact that $g$ is periodic implies that $f$ always stays within some fixed distance $D$ of the line $y=\frac{10}{3} x$. On the other hand, because $\frac{10}{3}>\sqrt{11}$, the graph of $y=\frac{10}{3} x$ gets further and further away from the graph of $y=\sqrt{11} x$ as $x$ increases. Because the graph of $y=f(x)$ remains near $y=\frac{10}{3} x$, the graph of $y=f(x)$ drifts upward from the line $y=\sqrt{11} x$.

For each integer $n$, define the open interval $I_{n}=\left(\frac{n-1}{3}, \frac{n}{3}\right)$. In fact, $f$ is constant on $I_{n}$, as the following argument shows. For $x \in I_{n}, \frac{n}{9}-\frac{1}{9}<\frac{x}{3}<\frac{n}{9}$. Because $n$ is an integer, there are no integers between $\frac{n}{9}-\frac{1}{9}$ and $\frac{n}{9}$, so $\left\lfloor\frac{x}{3}\right\rfloor$ is constant; similarly, $\lceil 3 x\rceil$ is constant on the same intervals. Let $l_{n}$ be the value of $f$ on the interval $I_{n}$, and let $L_{n}=f\left(\frac{n}{3}\right)$, the value at the right end of the interval $I_{n}$. If $n$ is not a multiple of 9 , then $l_{n}=L_{n}$, because as $x$ increases from $n-\varepsilon$ to $n$, the floor function does not increase. This means that $f$ is actually constant on the half-closed interval $\left(\frac{n-1}{3}, \frac{n}{3}\right]$. If neither $n$ nor $n+1$ are multiples of 9 , then $l_{n+1}=l_{n}+1$. However if $n$ is a multiple of 9 , then $L_{n}=l_{n}+1$ and $l_{n+1}=L_{n}+1$. (The value of $f(x)$ increases when $x$ increases from $n-\varepsilon$ to $n$, as well as going from $n$ to $n+\varepsilon$.)

Hence on each interval of the form $(3 n-3,3 n)$, the graph of $f$ looks like 9 steps of height 1 and width $\frac{1}{3}$, all open on the left and closed on the right except for the last step, which is open on both ends. Between the intervals $(3 n-3,3 n)$ and $(3 n, 3 n+3), f(x)$ increases by 2 , with $f(3 n)$ halfway between steps. This graph is shown below:





On each interval $(3 n-3,3 n)$, the average rate of change is $3<\sqrt{11}$, so the steps move down relative $y=\sqrt{11} x$ within each interval. At the end of each interval, the graph of $f$ rises relative to $y=\sqrt{11} x$. Thus the last intersection point between $f(x)$ and $\sqrt{11} x$ will be on the ninth step of one of these intervals. Suppose this intersection point lies in the interval $(3 k-3,3 k)$. The ninth step is of height $10 k-1$. Set $x=3 k-r$, where $r<\frac{1}{3}$. Then the solution is the largest $k$ for which

$$
\begin{aligned}
10 k-1 & =\sqrt{11}(3 k-r) \quad\left(0<r<\frac{1}{3}\right) \\
k(10-3 \sqrt{11}) & =1-\sqrt{11} r<1 \\
k & <\frac{1}{10-3 \sqrt{11}}=10+3 \sqrt{11}<20 .
\end{aligned}
$$

Because $0<19(10-3 \sqrt{11})<1, k=19$ implies a value of $r$ between 0 and $\frac{1}{\sqrt{11}}$. And because $\frac{1}{\sqrt{11}}<\frac{1}{3}$

$$
x=3 k-r=\frac{10 k-1}{\sqrt{11}}=\frac{\mathbf{1 8 9} \sqrt{\mathbf{1 1}}}{\mathbf{1 1}}
$$

is the largest solution to $f(x)=\sqrt{11} x$.

Let $x$ be the largest real number for which $\left\lfloor\frac{x}{3}\right\rfloor+\lceil 3 x\rceil=\sqrt{11} x$. Because the left-hand side of this equation is an integer, it is simpler to work with $n=\sqrt{11} x$ instead of $x$. The equation becomes

$$
\left\lfloor\frac{n}{3 \sqrt{11}}\right\rfloor+\left\lceil\frac{3 n}{\sqrt{11}}\right\rceil=n
$$



A little bit of computation shows that $\frac{1}{3 \sqrt{11}}+\frac{3}{\sqrt{11}}>1$, so the equation cannot hold for large values of $n$. To make this explicit, write

$$
\left\lfloor\frac{n}{3 \sqrt{11}}\right\rfloor=\frac{n}{3 \sqrt{11}}-r \quad \text { and } \quad\left\lceil\frac{3 n}{\sqrt{11}}\right\rceil=\frac{3 n}{\sqrt{11}}+s
$$

where $r$ and $s$ are real numbers between 0 and 1. (If $n \neq 0$, then $r$ and $s$ are strictly between 0 and 1.) Then

$$
\begin{aligned}
1>r-s & =\left(\frac{n}{3 \sqrt{11}}-\left\lfloor\frac{n}{3 \sqrt{11}}\right\rfloor\right)-\left(\left\lceil\frac{3 n}{\sqrt{11}}\right\rceil-\frac{3 n}{\sqrt{11}}\right) \\
& =\left(\frac{n}{3 \sqrt{11}}+\frac{3 n}{\sqrt{11}}\right)-\left(\left\lfloor\frac{n}{3 \sqrt{11}}\right\rfloor+\left\lceil\frac{3 n}{\sqrt{11}}\right\rceil\right) \\
& =n\left(\frac{1}{3 \sqrt{11}}+\frac{3}{\sqrt{11}}-1\right),
\end{aligned}
$$

so $n<1 /\left(\frac{1}{3 \sqrt{11}}+\frac{3}{\sqrt{11}}-1\right)=99+30 \sqrt{11}=198.45 \ldots$

Use trial and error with $n=198,197,196, \ldots$, to find the value of $n$ that works. Computing the first row of the following table to three decimal digits, and computing both $\frac{1}{3 \sqrt{11}}$ and $\frac{3}{\sqrt{11}}$ to the same degree of accuracy, allows one to calculate the remaining rows with acceptable round-off errors.

| $n$ | $n /(3 \sqrt{11})$ | $3 n / \sqrt{11}$ |
| :---: | :---: | :---: |
|  |  |  |
| 198 | 19.900 | 179.098 |
| 197 | 19.799 | 178.193 |
| 196 | 19.699 | 177.289 |
| 195 | 19.598 | 176.384 |
| 194 | 19.498 | 175.480 |
| 193 | 19.397 | 174.575 |
| 192 | 19.297 | 173.671 |
| 191 | 19.196 | 172.766 |
| 190 | 19.096 | 171.861 |
| 189 | 18.995 | 170.957 |

Because $n=189=18+171$, the final answer is $x=\frac{\mathbf{1 8 9} \sqrt{\mathbf{1 1}}}{\mathbf{1 1}}$.

Final answer: $\frac{189 \sqrt{11}}{11}$