## Modular inverses

### Problem 451

Published on 22 December 2013 at 10:00 am [Server Time]

Consider the number 15.
There are eight positive numbers less than 15 which are coprime to 15: 1, 2, 4, 7, 8, 11, 13, 14.
The modular inverses of these numbers modulo 15 are: 1, 8, 4, 13, 2, 11, 7, 14
because
1*1 mod 15=1
2*8=16 mod 15=1
4*4=16 mod 15=1
7*13=91 mod 15=1
11*11=121 mod 15=1
14*14=196 mod 15=1

Let I(n) be the largest positive number m smaller than n-1 such that the modular inverse of m modulo n equals m itself.
So I(15)=11.
Also I(100)=51 and I(7)=1.

Find I(n) for 3n2·107