Determine the smallest $n$ such that $n \equiv (a - 1)$ mod $a$ for all $a \in \{2,3,..., 10\}$.
1. We need to determine the smallest \( n \) such that \( n \equiv (a - 1) \pmod{a} \) for all \( a \in \{2, 3, \ldots, 10\} \).
2. This condition implies that \( n \equiv -1 \pmod{a} \) for each \( a \) in the set \(\{2, 3, \ldots, 10\}\).
3. To satisfy \( n \equiv -1 \pmod{a} \) for all \( a \), \( n + 1 \) must be a multiple of each \( a \) in \(\{2, 3, \ldots, 10\}\).
4. Therefore, \( n + 1 \) must be the least common multiple (LCM) of the numbers \( 2, 3, \ldots, 10 \).
5. We calculate the LCM of \( 2, 3, \ldots, 10 \):

   \[
   \text{LCM}(2, 3, 4, 5, 6, 7, 8, 9, 10)
   \]

   - The prime factorizations of the numbers are:
     \[
     \begin{aligned}
     2 &= 2, \\
     3 &= 3, \\
     4 &= 2^2, \\
     5 &= 5, \\
     6 &= 2 \cdot 3, \\
     7 &= 7, \\
     8 &= 2^3, \\
     9 &= 3^2, \\
     10 &= 2 \cdot 5.
     \end{aligned}
     \]

   - The LCM is found by taking the highest power of each prime that appears in these factorizations:
     \[
     \text{LCM} = 2^3 \cdot 3^2 \cdot 5 \cdot 7.
     \]

   - Calculating this:
     \[
     \begin{aligned}
     2^3 &= 8, \\
     3^2 &= 9, \\
     8 \cdot 9 &= 72, \\
     72 \cdot 5 &= 360, \\
     360 \cdot 7 &= 2520.
     \end{aligned}
     \]

6. Thus, the LCM of \( 2, 3, \ldots, 10 \) is \( 2520 \).
7. Since \( n + 1 = 2520 \), we have \( n = 2520 - 1 = 2519 \).

The final answer is \(\boxed{2519}\).