[Number Theory]
Problem: Compute the sum of all real numbers $x$ such that

$$
\left\lfloor\frac{x}{2}\right\rfloor-\left\lfloor\frac{x}{3}\right\rfloor=\frac{x}{7}
$$

Solution: Because the quantity on the left side is the difference of two integers, $x / 7$ must be an integer, hence $x$ is an integer (in fact a multiple of 7). Because the denominators on the left side are 2 and 3 , it is convenient to write $x=6 q+r$, where $0 \leq r \leq 5$, so that $\lfloor x / 2\rfloor=3 q+\lfloor r / 2\rfloor$ and $\lfloor x / 3\rfloor=2 q+\lfloor r / 3\rfloor$. Then for $r=0,1, \ldots, 5$ these expressions can be simplified as shown in the table below.

| $r$ | 0 | 1 | 2 | 3 | 4 | 5 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\left\lfloor\frac{x}{2}\right\rfloor$ | $3 q$ | $3 q$ | $3 q+1$ | $3 q+1$ | $3 q+2$ | $3 q+2$ |
| $\left\lfloor\frac{x}{3}\right\rfloor$ | $2 q$ | $2 q$ | $2 q$ | $2 q+1$ | $2 q+1$ | $2 q+1$ |
| $\left\lfloor\frac{x}{2}\right\rfloor-\left\lfloor\frac{x}{3}\right\rfloor$ | $q$ | $q$ | $q+1$ | $q$ | $q+1$ | $q+1$ |

Now proceed by cases:

$r=0:$ Then $q=x / 6$. But from the statement of the problem, $q=x / 7$, so $x=0$.

$r=1: \quad$ Then $q=(x-1) / 6=x / 7 \Rightarrow x=7$.

$r=2: \quad$ Then $q=(x-2) / 6$ and $q+1=x / 7$, so $(x+4) / 6=x / 7$, and $x=-28$.

$r=3$ : Then $q=(x-3) / 6$ and $q=x / 7$, so $x=21$.

$r=4: \quad$ Then $q=(x-4) / 6$ and $q+1=x / 7$, so $(x+2) / 6=x / 7$, and $x=-14$.

$r=5$ : Then $q=(x-5) / 6$ and $q+1=x / 7$, so $(x+1) / 6=x / 7$, and $x=-7$.

The sum of these values is $0+7+-28+21+-14+-7=\mathbf{- 2 1}$.

Final answer: -21