Question

Problem 2.6.1. We have 7*6* 55 = 2310. Extend the technique of the Chinese remainder theorem to find the solutions in Z2310 of the equations. 2x => 3 X 564 5x 355 50 Hint: There will be 5 solutions because the last equation has 5 solutions. You may use the Mathematica command ExtendedGCD.

Answer 1

That's easy.

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Sol? 2x=3(mod 7) = x=1255(wod 7) ► x 35 (mod 7) 2 234 (mod 6) ► x=4(modo) * 57 350 mod 55) ► x 310lmodl1) ►x=10(mod 11)' Wesolve & using chinese remainder theorem, 9,=5,92=4, 92=10 m;=7m2=6, m3 = 11 :: M=1,-12-12 = 7x6 xll. : Mi = 6X11=66 → MM! =(mod 7)66M=1(mod 7)→M=5 M2 = 7X11 = 77M2M2' =/(mod6)→ 77 M2' =i[mod 6) > M=5 M3 = 7x6 =42=> M3M) El(mod 11) — 92 M3' = 1/modil) →M5=5 .: 739,M, M +42M2 M2 +93 M2M (mod M) .:X55X 66X5+ 4x77x5 + 10x42x5( mod M) 25 208(mod 462) since this is Jol? to given problem. ::83208(mod 2310) [:46212310) a=208 is a soln ot given eg? in 7.2010 .