7. On the board is written the multiplication $x \times y$, where $x$ and $y$ are positive integers of three digits. Nicolò, a bit absent-minded, did not notice the multiplication symbol and copied onto his notebook the six-digit number obtained by concatenating $x$ and $y$. The teacher, walking between the desks, points out to Nicolò that the number he wrote is equal to 7 times the product $x y$. What is the sum $x+y$?
(A) 125
(B) 286
(C) 312
(D) 487
(E) 513
7. The answer is (B). The hypotheses of the problem translate to $1000 x + y = 7 x y$, and we can rewrite this equation as $y = \frac{1000 x}{7 x - 1}$. At this point, we can proceed in (at least) two ways:

(a) we have $y = \frac{1000 x}{7 x - 1} \geq \frac{1000 x}{7 x} = \frac{1000}{7} = 142.8 \ldots$, and on the other hand

$$
y = \frac{1000 x}{7 x - 1} = \frac{1000(x - 1 / 7) + 1000 / 7}{7 x - 1} = \frac{1000}{7} + \frac{1000 / 7}{7 x - 1} \leq \frac{1000}{7} + \frac{1000 / 7}{7 \cdot 100 - 1} < \frac{1000}{7} + 1,
$$

so $y$ must be the only integer between 1000/7 and 1000/7 + 1, i.e., $y = 143$. From this, we immediately find that $x = \frac{y}{7 y - 1000} = 143$ and thus $x + y = 286$.

(b) alternatively, observe that $x$ and $7 x - 1$ cannot have common prime factors (since if $p$ divides both $x$ and $7 x - 1$, then it also divides $(7 x - 1) - 7 \cdot x = -1$, but no prime number divides -1). The fact that $7 x - 1$ divides $1000 x$ implies then that $7 x - 1$ divides 1000; on the other hand, by hypothesis $x$ has 3 digits, so $7 x - 1 \geq 7 \cdot 100 - 1 = 699$. The only divisor of 1000 greater than or equal to 699 is 1000 itself, hence $7 x - 1 = 1000 \Rightarrow x = 143$ and $y = \frac{1000 \cdot x}{7 x - 1} = x = 143$. It follows that $x + y = 286$.