QUANTUM COMPUTING - EXAMPLE 32.4 : A system of linear congruences consists of 3

QUANTUM COMPUTING - EXAMPLE 32.4 : A system of linear congruences consists of 3 equations : X ≡ 1 (mod 2), X ≡ 3 (mod 3), X ≡ 4 (mod 5). X has positive values. (a)(i) List the values of these equations from 1 to approximately 40. (ii) Find the first smallest value and second smallest value of X. (iii) Guess the third smallest value of X. (b) Let X ≡ Aa (mod Ma), X ≡ Ab (mod Mb), X ≡ Ac (mod Mc). According to Chinese remainder theorem, X ≡ (Aa x Ya x Md + Ab x Yb x Me + Ac x Yc x Mf) [ mod (Ma x Mb x Mc) ]. (i) Show that Ma, Mb and Mc have the greatest common divisor of Ma x Mb x Mc. (ii) Find the values of Md, Me and Mf if Md = Mb x Mc, Me = Ma x Mc and Mf = Ma x Mb. (iii) Find the values of Ya, Yb and Yc if Ya = Remainder of (Md / Ma), Yb = Remainder of (Me / Mb) and Yc = Remainder of (Mf / Mc). (iv) Use Chinese remainder theorem to find X.

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QUANTUM COMPUTING - EXAMPLE 32.4 : A system of linear congruences consists of 3 equations : X &equiv..

Answer / kangchuentat

QUANTUM COMPUTING - ANSWER 32.4 : (a)(i) 1 (mod 2) = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39. 3 (mod 3) = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39. 4 (mod 5) = 4, 9, 14, 19, 24, 29, 34, 39. (ii) By observation on X, 3 equations have common values of 9 and 39. First smallest value = 9, second smallest value = 39. (iii) Third smallest value = second smallest value + (second smallest value - first smallest value) = 39 + (39 - 9) = 69. (b)(i) Let Ma = 2, Mb = 3, Mc = 5 where they are prime numbers. Their greatest common divisor is 2 x 3 x 5 = Ma x Mb x Mc (shown). (ii) Md = Mb x Mc = 3 x 5 = 15, Me = Ma x Mc = 2 x 5 = 10, Mf = Ma x Mb = 2 x 3 = 6. (iii) Ma = 2, Mb = 3, Mc = 5, Md = 15, Me = 10, Mf = 6. Md / Ma = 15 / 2 = 7 remain 1, Ya = 1. Me / Mb = 10 / 3 = 3 remain 1, Yb = 1. Mf / Mc = 6 / 5 = 1 remain 1, Yc = 1. (iv) Let Aa = 1, Ab = 3, Ac = 4, Ya = 1, Yb = 1, Yc = 1, Ma = 2, Mb = 3, Mc = 5, Md = 15, Me = 10, Mf = 6. X ≡ (Aa x Ya x Md + Ab x Yb x Me + Ac x Yc x Mf) [ mod (Ma x Mb x Mc) ] = (1 x 1 x 15 + 3 x 1 x 10 + 4 x 1 x 6) [ mod (2 x 3 x 5) ] = 69 mod 30 = 39 mod 30 = 9 mod 30. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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