Let $a$ be a positive integer that is a multiple of 5 and such that $a+1$ is a multiple of 7, $a+2$ is a multiple of 9, and $a+3$ is a multiple of 11. Determine the smallest possible value of $a$.
Solution 1: Note that if a positive integer $m$ is a multiple of an integer $p$, then $2m - p$ is a multiple of $p$. Indeed, if $m = np$, then $2m - p = 2np - p = (2n - 1)p$. In particular, taking $m$ equal to $a, a+1, a+2$, and $a+3$, we establish that $2a - 5$ is a multiple of 5, $2(a+1) - 7$ is a multiple of 7, $2(a+2) - 9$ is a multiple of 9, and $2(a+3) - 11$ is a multiple of 11. However, it happens that $2a - 5 = 2(a+1) - 7 = 2(a+2) - 9 = 2(a+3) - 11$, so $2a - 5$ is a multiple of 5, 7, 9, and 11, and thus is a multiple of $5 \times 7 \times 9 \times 11 = 3465$. Therefore, the smallest possible value of $a$ satisfies $2a - 5 = 3465$, that is, $a = 1735$.

Solution 2: Since $a$ is a multiple of 5, we have $a = 5i$, for some positive integer $i$. But $a + 1 = 1 + 5i$ must be a multiple of 7. By trials, we verify that the smallest value of $i$ for which this happens is $i_0 = 4$. For any other value of $i$ that makes $1 + 5i$ a multiple of 7, we must have $(1 + 5i) - (1 + 5i_0) = 5(i - i_0)$ a multiple of 7, and therefore, $i - i_0$ a multiple of 7. Thus, $i = i_0 + 7j = 4 + 7j$ and $a = 5i = 5(4 + 7j) = 20 + 35j$, with $j \geq 0$ an integer. Similarly, $a + 2 = 22 + 35j$ must be a multiple of 9. The smallest value of $j$ for which this happens is $j_0 = 4$, since $22 + 35 \times 4 = 162$ is a multiple of 9. For any other value of $j$ that makes $22 + 35j$ a multiple of 9, we must have $(22 + 35j) - (22 + 35j_0) = 35(j - j_0)$ a multiple of 9, and therefore, $j - j_0$ a multiple of 9. Thus, $j = j_0 + 9k = 4 + 9k$ and $a = 20 + 35j = 20 + 35(4 + 9k) = 160 + 315k$, with $k \geq 0$ an integer. Finally, 11 must divide

$$
\begin{aligned}
a + 3 & = 163 + 315k = 14 \times 11 + 9 + (28 \times 11 + 7)k \\
& = (14 + 28k) \times 11 + (9 + 7k)
\end{aligned}
$$

Therefore, 11 divides $9 + 7k$. Again, by trials, it is easy to verify that the smallest value of $k$ for which this happens is $k_0 = 5$. Any other value of $k$ for which this happens is of the form $k = 5 + 11l$, with $l \geq 0$ an integer.

We conclude that the positive integers $a$ for which 5 divides $a$, 7 divides $a + 1$, 9 divides $a + 2$, and 11 divides $a + 3$ are exactly the integers of the form $a = 160 + 315(5 + 11l) = 1735 + 3465l$, with $l \geq 0$ an integer. In particular, the smallest value of $a$ is 1,735, obtained with $l = 0$.