Prime Divisibility

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

Let L=lcm{ord36097(1),ord36097(2),,ord36097(36096)}.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+1L+1.

Details and assumptions

  • To put it in words, LL is the smallest positive integer which is a multiple of all of the following numbers: ord36097(1),ord36097(2),,ord36097(36096).\text{ord}_{36097}(1), \text{ord}_{36097} (2) , \cdots , \text{ord}_{36097} (36096) .

  • You may use the fact that 3609736097 is a prime.

  • Even though 101(mod36097)1^0 \equiv 1 \pmod{36097}, ord36097(1)0\text{ord}_{36097} (1) \neq 0, since the order is defined to be positive. In this case, ord36097(1)=1\text{ord}_{36097} (1) = 1.

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