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.
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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Question 62 – The names of the flow streams could be represented by : H1 for first hot stream, H2 for second hot stream, C1 for first cold stream, C2 for second cold stream. Data of supply temperature Ts in degree Celsius : 150 for H1, 170 for H2, 30 for C1, 30 for C2. Data of target temperature Tt in degree Celsius : 50 for H1, 169 for H2, 150 for C1, 40 for C2. Data of heat capacity Cp in kW / degree Celsius : 3 for H1, 360 for H2, 3 for C1, 30 for C2. (a) Find the enthalpy changes, dH for all streams of flow H1, H2, C1 and C2 in the unit of kW. Take note of the formula dH = (Cp) (Tt - Ts). (b) Match the hot streams H1 and H2 with the suitable cold streams C1 and C2 to achieve the maximum energy efficiency.
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