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Cosets and Lagrange's Theorem · Easy problem

By Fermat, 210≡1(mod11)2^{10}\equiv1\pmod{11}, so ord⁡(2)\operatorname{ord}(2) in U(11)U(11) must divide 1010. By checking the divisors of 1010, find the actual (smallest) order of 22 mod 1111.