# Prime Divisibility

**Number Theory**Level 4

For coprime positive integers \(n\) and \(k\), we define \(\text{ord}_n (k) \) to be the smallest **positive** integer such that \(k^{\text{ord}_n (k)} \equiv 1 \pmod{n} \).

Let \(L= \text{lcm} \left \{ \text{ord}_{36097} (1), \text{ord}_{36097} (2) , \cdots , \text{ord}_{36097} (36096) \right \}. \) Find the last three digits of \(L+1\).

**Details and assumptions**

To put it in words, \(L\) is the smallest positive integer which is a multiple of all of the following numbers: \[\text{ord}_{36097}(1), \text{ord}_{36097} (2) , \cdots , \text{ord}_{36097} (36096) .\]

You may use the fact that \(36097\) is a prime.

Even though \(1^0 \equiv 1 \pmod{36097}\), \(\text{ord}_{36097} (1) \neq 0\), since the order is defined to be positive. In this case, \(\text{ord}_{36097} (1) = 1\).

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