Let $ 0 < a_1 < a_2 < a_3 < ... < a_{10000} < 20000$ be integers such that $ gcd(a_i,a_j) < a_i, \forall i < j$ ; is $ 500 < a_1$ [i](always)[/i] true ?

[i](own)[/i]  :lol:
To determine whether \(500 < a_1\) is always true given the conditions \(0 < a_1 < a_2 < a_3 < \ldots < a_{10000} < 20000\) and \(\gcd(a_i, a_j) < a_i\) for all \(i < j\), we will proceed with the following steps:

1. **Define the Sets \(X\) and \(Y\)**:
   Let \(X = \{a_i \mid 1 \leq i \leq 10000\}\) and \(Y = \{x \mid 10000 \leq x < 20000 \text{ and } x \in \mathbb{N}\}\).

2. **Mapping Function \(f(x)\)**:
   For a number \(x < 20000\), there exists a unique non-negative integer \(t\) such that \(x \cdot 2^t \in Y\). Define \(f(x)\) as this unique positive integer \(x \cdot 2^t\). For example:
   \[
   f(12345) = 12345, \quad f(4321) = 17284, \quad f(700) = 11200, \quad f(1) = 16384
   \]

3. **Injectivity of \(f(x)\)**:
   If \(f(x) = f(y)\), then \(\frac{x}{y} = 2^k\) for some integer \(k\), implying one of \(x\) or \(y\) divides the other. Therefore, \(f(x) \neq f(y)\) for all \(x, y \in X\) with \(x \neq y\). This shows that \(f\) is an injective map from \(X\) to \(Y\).

4. **Cardinality and Bijectivity**:
   Since \(|X| = |Y| = 10000\), \(f\) is a bijective map from \(X\) to \(Y\).

5. **Odd Integers and Function \(g(x)\)**:
   For some odd \(x \in (0, 20000)\), there exists a unique \(t\) such that \(x \cdot 2^t \in Y\). Let \(g(x)\) denote this unique \(t\). Define \(h(x) = x \cdot 2^{g(x)}\). Similarly, \(h\) is a bijective map between the set of odd integers less than 20000 and \(X\).

6. **Claim**:
   We claim that \(x \cdot 2^{g(x)} > 500\) for all odd \(x \in (0, 20000)\).

7. **Proof by Contradiction**:
   For odd \(x, y \in (0, 20000)\) such that \(y \neq 1\), if \(g(x) \leq g(xy)\), then \(x \cdot 2^{g(x)} \mid xy \cdot 2^{g(xy)}\). Since both \(x \cdot 2^{g(x)}\) and \(xy \cdot 2^{g(xy)}\) are in \(X\) and distinct, this leads to a contradiction. Thus, \(g(x) > g(xy)\) or \(g(x) \geq g(xy) + 1\).

8. **Inductive Steps**:
   - For \(g(1)\):
     \[
     g(1) \geq g(3) + 1 \geq \cdots \geq g(3^9) + 9 \geq 9 \quad (\text{since } 3^9 < 20000)
     \]
     \[
     1 \cdot 2^{g(1)} \geq 512 > 500
     \]
   - For \(g(3)\):
     \[
     g(3) \geq g(3^2) + 1 \geq \cdots \geq g(3^9) + 8 \geq 8
     \]
     \[
     3 \cdot 2^{g(3)} \geq 768 > 500
     \]
   - For \(g(5)\):
     \[
     g(5) \geq g(5 \cdot 3) + 1 \geq g(5 \cdot 3^2) + 2 \geq \cdots \geq g(5 \cdot 3^7) + 7 \geq 7 \quad (\text{since } 5 \cdot 3^7 < 20000)
     \]
     \[
     5 \cdot 2^{g(5)} \geq 640 > 500
     \]
   - For \(g(7)\):
     \[
     g(7) \geq g(7 \cdot 3) + 1 \geq \cdots \geq g(7 \cdot 3^7) + 7 \geq 7 \quad (\text{since } 7 \cdot 3^7 < 20000)
     \]
     \[
     7 \cdot 2^{g(7)} \geq 896 > 500
     \]

9. **General Case**:
   Suppose \(t\) is an arbitrary odd integer \(\geq 9\) and \(< 20000\). Let \(m\) be the unique non-negative integer such that \(3^{m+2} \leq t < 3^{m+3}\):
   \[
   t \cdot 3^{6 - m} < 3^{m+3} \cdot 3^{6 - m} = 3^9 < 20000
   \]
   \[
   g(t) \geq g(t \cdot 3) + 1 \geq \cdots \geq g(t \cdot 3^{6 - m}) + 6 - m \geq 6 - m
   \]
   \[
   t \cdot 2^{g(t)} \geq 3^{m+2} \cdot 2^{6 - m} = 9 \cdot 3^m \cdot 2^{6 - m} \geq 9 \cdot 2^m \cdot 2^{6 - m} = 9 \cdot 2^6 = 576 > 500
   \]

Thus, we have shown that \(a_1 > 500\) is always true.

The final answer is \( \boxed{ 500 < a_1 } \).