Chapte 3: Theoy of Modula Aithmetic 38 Section D Chinese Remainde Theoem By the end of this section you will be able to pove the Chinese Remainde Theoem apply this theoem to solve simultaneous linea conguences The Chinese Remainde Theoem helps us to solve the following poblem: Suppose you ae in chage of an amy and you need to count the numbe of soldies you have. You can count them one by one but this is a tedious tas and is pone to eos. An easie way to count them is to goup them into ows. Suppose the following: If you place them in ows of 3 soldies then 2 soldies ae left ove. If you place them in ows of 5 soldies then 3 soldies ae left ove. If you place them in ows of 7 soldies then 2 soldies ae left ove. We can convet this poblem into modula aithmetic which means we need to solve the following simultaneous linea conguences fo the numbe of soldies x which satisfies: x 2 mod 3 [Dividing x by 3 leaves emainde 2] x 3 mod 5 [Dividing x by 5 leaves emainde 3] x 2 mod 7 [Dividing x by 7 leaves emainde 2] The numbe of soldies x needs to satisfy all thee of these equations. This poblem is equivalent to the following. Find the intege x which leaves a emainde of 2 when divided by 3, leaves a emainde of 3 when divided by 5 and leaves a emainde of 2 when divided by 7. We will solve this poblem late in the section. D1 Solving Simultaneous Conguences Up to now we have solved a single linea conguence such as ax b mod n In this section we examine solving a set of simultaneous linea equations. We ll begin by looing at an example befoe going on to develop the geneal method. Example 21 Find the value of x which satisfies both the following equations: x 1 mod 5 1 x 4 mod 7 2 Note that we have one unnown x which satisfies both conguent equations. We need to find a value of x such that equations (1) and (2) ae tue. Let us fist use bute foce to esolve these equations. Ceating a table of values: x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 xmod 5 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 xmod 7 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 We can see in the table that the only value of x that will satisfy both ou equations is x 11: 11 1 mod 5 11 4 mod 7 and Of couse we can apply bute foce fo simple intege values. Howeve we need a systematic appoach to solve these because modulus n may be a lage numbe and we just do not have the time o willpowe to ceate a table fo lage n. Say if n 701 then this would mean we would have to ceate a table with 701 columns.
Chapte 3: Theoy of Modula Aithmetic 39 What does the fist equation x 1mod 5 in the above example mean? This means that x 1 is a multiple of 5 o x 1 5 fo some intege. Re-aanging this we have x 1 5. x 4 mod 7 means x 4 7c fo some intege c and so Similaly the othe equation x 4 7c. Equating these two equations, x 1 5 and x 4 7c, gives x 1 5 4 7c 3 7c 5 Since is an intege we need 3 7c to be a multiple of 5. If c 1 then 3 7c 3 7 10 10 and 2. 5 Substituting this, c 1, into x 4 7c gives x 4 7c 4 7 11 This is ou solution fom the above example. Example 22 Solve the simultaneous equations: x 31 mod 49 x 6 mod 20 Fom these equations we have x 31 49 implies x 31 49 x 6 20c implies x 6 20c whee and c ae integes. Equating these last two equations because both ae equal to x gives 25 49 6 20c 31 49 implies c 20 Since we want intege solutions so we ty values of such that the numeato 25 49 is a multiple of 20. (Multiplying 49 by 5 will give esults ending in 5 o 0. Only esults ending in 5 will be divisible by 20 afte adding 25.) Hence we tial multiples of 5 fo, that is 5, 10, 15,. Note that 5, 10 does not give a multiple of 20 but 15 does because 25 4915 c 38 20 Substituting c 38 into x 6 20c x 6 20 38 766. gives Chec that this x 766 satisfies both the given equations: x 31 mod 49 766 31 mod 49 x 6 mod 20 766 6 mod 20 Example 23 Find an intege x such that when divided by 2, 3 and 5 the emainde is 1. We can wite these as conguence equations:
Chapte 3: Theoy of Modula Aithmetic 40 x x x x 1 mod 2 Dividing by 2 with emainde 1 x 1 mod 3 Dividing by 3 with emainde 1 x 1 mod 5 Dividing by 5 with emainde 1 What does this mean? Means that x 1 is a multiple of 2, 3 and 5 which we can wite as: x 1 2, x 1 3 c and x 1 5m whee, c and m ae integes Which numbe is a multiple of 2, 3 and 5? gcd 2, 3, 5 1 so the smallest numbe which is a multiple of 2, 3 and 5 is Since 235 30 This means that x 1 is a multiple of 30 o x 1 30n implies x 30n 1 The geneal solution of the given thee equations is x 30n 1. Putting in vaious values of n such as 1, 2, 3, into x 30n 1 gives x 30 1 31, x 30 2 1 61, x 303 1 91, Ou solution is a multiple of 30 plus 1. You may chec that each of these solutions satisfies the given equations: x 1 mod 2, x 1 mod 3 and x 1 mod 5 We can also wite the solutions x 31, 61, 91, x 1 mod 30 in modula aithmetic as x satisfies whee 30 235 is the poduct of the thee given moduli in the simultaneous equations. In the next subsection we encounte the Chinese Remainde Theoem which povides us with a stuctued way of solving simultaneous conguence equations. D2 Poof of the Chinese Remainde Theoem In the above example we say the numbes 2, 3 and 5 ae paiwise pime. What does paiwise pime mean? It means that the only facto in common between any pai of numbes is 1. Fo example gcd 2, 3 1 gcd 2, 5 1 gcd 3, 5 1, and We say integes 2, 3 and 5 ae paiwise pime. Definition (3.21). Let n1, n2, n3,, n be integes such that any two of these numbes do not have a common facto apat fom 1, that is gcd n, n 1 fo i j This means that evey pai of numbes ae elatively pime. We say this list of integes n1, n2, n3,, n ae paiwise pime. Fo example the numbes 25, 26 and 27 ae paiwise pime because gcd 25, 26 1 gcd 26, 27 1 gcd 25, 27 1 i j, and Chinese Remainde Theoem (3.22): Let n1, n2, n3,, n be positive integes which ae paiwise pime, that is Then the simultaneous linea conguences gcd n, n 1 fo i j i j
Chapte 3: Theoy of Modula Aithmetic 41 x a mod n 1 1 x a mod n 2 2 x a mod n has a solution satisfying all these equations. Moeove the solution is unique modulo n 1 n 2 n 3 n. How do we pove this esult? We need to show two things (1) existence of solution and (2) uniqueness of solution. Poof. (1) Existence Let n n1n 2n3 n. Fo each intege 1, 2, 3,, let n n1n 2n3 n 1n n 1n N n1n 2n3 n 1n 1n Cancelling out n n n This means that We ae given that i j N is the poduct of all the moduli ni with the numbe gcd n, n 1 fo i j which implies that gcd n, N 1 whee N n1n 2 n 1n 1n Why? Because by poblem 23 of Execise 1(a) we have: If gcd a, b gcd a, c 1 then a Conside the linea conguence equation N x Does this equation have any solutions? gcd n, N 1 this equation Since gcd, bc 1 1 mod n N x 1 mod n has a unique solution. Why? Because by Coollay (3.19) of the pevious section: Let ax b mod n has unique solution povided g a n x be the unique solution of N x 1 mod n which means that N x 1 mod n ( ) x is the multiplicative invese of N mod n. Note that n missing. gcd, 1 We need to constuct a solution which satisfies all the given simultaneous conguence equations. The solution we conside is x ' a1n1 x1 a2n2x2 a3n3x3 a N x Let us see if this solution x ' satisfies the fist given equation: x a mod n 1 1 Taing the solution to modulus n1 gives x ' a N x a N x a N x a N x mod n (*) By the above definition of 1 1 1 2 2 2 3 3 3 1 N :
Chapte 3: Theoy of Modula Aithmetic 42 N n1n 2n3 n 1n 1 n [Poduct of n s with n missing] The numbes N,,, 2 N, 3 N4 N ae multiplies of n1 because n 1 is pesent in the poduct. Theefoe these numbes N,,, 2 N, 3 N4 N ae conguent to zeo modulo n 1 that is N N N N n 2 3 4 1 0 mod so we have a N x 0 mod n, a N x 0 mod n,, a N x 0 mod n 2 2 2 1 3 3 3 1 1 Substituting this into (*) gives x ' a N x 0 0 0 a N x mod n By the above ( ) we have N1x1 1 mod n1 x ' a1n1 x1 mod n1 gives x ' a1n1 x1 a1 1 a1 mod n1 Hence x ' satisfies the fist conguence equation x a mod n 1 1 1 1 1 1 1. Substituting this into the above 1 1. Aguing along vey simila lines we can show that the solution constucted x ' satisfies the emaining conguence equations. Theefoe x x '. (2) Uniqueness Suppose thee is anothe solution y which satisfies the given equations. This means we have x a y mod n fo 1, 2, 3, and Fom this conguence x y mod n fo 1, 2, 3, and n x y, n x y, n x y,, n x y 1 2 3 we have Remembe we ae given that the n s ae paiwise pime - i j Applying the esult of question 13 of Execise1(b): If a c and b c with gcd a, b 1 then ab c. To the above ( ) list gives n n n n x y 1 2 3 ( ) gcd n, n 1 fo i j. This means that x y is a multiple of n1n 2 n o in modula aithmetic we have x y mod n n n n Hence x and y ae the same solution modulo n 1 n 2 n 3 n. This completes ou poof. The poof gives us a systematic appoach on how to constuct the solution of any given linea simultaneous conguences. In the poof the solution we constucted was given by 1 2 3 (3.23) x a1n1x 1 a2n2x2 a3n3x3 a N x Note the vaious components of this fomula ae: 1. The a s ae the numbes on the ight hand side of the given equations mod n in x a mod n x a n.. 2. The lowe case n s ae the given moduli 3. The uppe case N s ae poduct of the given moduli with N n n n n n 4. x is the multiplicative invese of 1 2 1 1 N modulo n, that is n missing, x satisfies
Chapte 3: Theoy of Modula Aithmetic 43 N x 1 mod n We use this fomula to solve the emaining examples. D3 Applying the Chinese Remainde Theoem Example 24 Let us now solve the soldies poblem stated at the beginning of the section. Remembe this poblem was equivalent to solving the following simultaneous equations: x 2 mod 3, x 3 mod 5 and x 2 mod 7 Find the exact numbe of soldies given that you now thee ae between 500 and 600 soldies. How do we solve this poblem? We use the above fomula: (3.23) x a1n1x 1 a2n2x2 a3n3x3 a N x In this case 3 because we ae given 3 simultaneous equations: x a N x a N x a N x (*) 1 1 1 2 2 2 3 3 3 n Each of these N s is given by n whee n is the poduct of the given moduli: n 35 7 105 [Because we ae given moduli 3, 5 and 7] Theefoe 3 5 7 N1 35 3 35 7 N2 21 5 357 N3 15 7 We need to find the N x 1 mod n fo 1, 2 and 3 : We ae equied to find the numbe 35 2 mod 3 : x s which ae given by N1x1 35x1 1 mod 3 N2x2 21x2 1 mod 5 N3x3 15x3 1 mod 7 x which satisfies 35 1 mod 3 1 1 35x 2x 1 mod 3 implies x1 2 Similaly solving the othe two equations: 21x x 1 mod 5 implies x2 1 Since we ae given 2 2 3 3 15x x 1 mod 7 implies x3 1 x. Note that x 2 mod 3, x 3 mod 5 and x 2 mod 7 So the a s ae: a 2, a 3 and a 2 [Because x a mod n 1 2 3 1 ]
Chapte 3: Theoy of Modula Aithmetic 44 Substituting a 1 2, a 2 3, a 3 2, N1 35, N2 21, N3 15, x1 2, x2 1 and x 3 1 into (*) gives x a N x a N x a N x 1 1 1 2 2 2 3 3 3 235 2 3 211 2151 233 We have x 233 satisfying all the given simultaneous equations. Howeve thee may be a smalle numbe which also satisfies all the equations. How can we find this numbe? In the poof we ae given that the solution is unique modulo n 1 n 2 n 3 n. In ou case we have n1 n2 n3 35 7 105 Hence x 233 23 mod 105. The geneal solution x is given by x 23 105t Theefoe ou solution is a multiple of 105 plus 23. We ae given that thee ae between 500 and 600 soldies so substitute t 5 into x 23 105t : We have 548 soldies in ou ans. x 231055 548 Poposition (3.24). Let n1, n2, n3,, n be positive integes which ae paiwise pime. Also integes a s gcd a, n 1. Then the simultaneous linea conguences satisfy a x b mod n 1 1 1 a x b mod n 2 2 2 a x b mod n has a solution satisfying all these equations. Moeove the solution is unique modulo n 1 n 2 n 3 n. Poof. See Execise 3(d). Example 25 Solve the following simultaneous linea conguences: 2x 1 mod 5, 3x 9 mod 6 and 4x 1 mod 7 This time we do not have x? mod m but ax? mod m We convet them into x? mod m Multiplying the fist linea conguence 2x 1 mod 5 by 3 gives 6x x 3 mod 5 We can simplify the second 3x 9 mod 6 because 9 3 mod 6 3x 3 mod 6 By Poposition (3.10):. How do we solve these? by fist solving each of these equations sepaately. so we have
Chapte 3: Theoy of Modula Aithmetic 45 ac bc mod n a b mod n g Applying this to 3 3 mod 6 whee g gcd c, n x with gcd 3, 6 3 3 3 6 x mod 3 3 3 g means dividing by 3: implies x 1 mod 2 We multiply the thid 4x 1mod 7 by 2: 8x x 2 mod 7 We now solve the equivalent system: 3 mod 5 x, x 1mod 2 and x 2 mod 7 The solution is given by using fomula (3.23): (3.23) x a1n1x 1 a2n2x2 a3n3x3 a N x Since we ae given 3 equations so we use this fomula with 3: x a1n1 x1 a2n2x2 a3n3x3 ( ) The modulus n is the poduct of all the given moduli theefoe n n1 n2 n3 5 2 7 70 [We have moduli n1 5, n2 2 and n3 7 ] Evaluating N1, N2 and N3 gives 5 27 N1 14 5 5 2 7 N2 35 2 5 2 7 N3 10 7 We need to find the N x 1 mod n fo 1, 2 and 3 : x s which ae given by 1 1 14x1 1mod 5 2 2 35x2 1mod 2 10x 1mod 7 N x N x N x [Remembe n1 5] [Remembe n2 2 ] [Remembe n3 7 ] 3 3 3 Simplifying each of these conguences and solving gives: 14x x 1 mod 5 implies x 4 1 1 1 35x x 1 mod 2 implies x 1 2 2 2 10x 3x 1 mod 7 implies x 5 3 3 3 What othe ingedients do we need in ode to use ( )? The a s which ae a 1 3, a2 1 and a 3 2 because we ae solving x 3 mod 5, x 1 mod 2 and x 2 mod 7 Putting all these numbes a1 3, a2 1, a3 2, N1 14, N2 35, N3 10, x1 4, x2 1 and x 3 5 into ( ) gives: x a N x a N x a N x 1 1 1 2 2 2 3 3 3 We have x 303 23 mod 70. 314 4 1351 2105 303
Chapte 3: Theoy of Modula Aithmetic 46 Thus ou value of satisfies all thee of ou conguences. Checing this solution by maing sue each of the given conguences is satisfied: 2 23 2 3 1 mod 5 Hence x 23 mod 70 3 23 35 15 3 9 mod 6 4 23 4 2 1 mod 7 is the unique solution of the given equations. Example 26 Find the least positive intege x which satisfies the following simultaneous equations: 2x 1 mod 5, 3x 9 mod 6, 4x 1 mod 7 and 5x 9 mod 11 Again we do not have x? mod m but ax? mod m We convet them into x? mod m fist linea conguence 2x 1 mod 5 by 3: 6x x 3 mod 5 We can simplify the second 3x 9 mod 6 by dividing though by x 3 mod 2 1 mod 2 We multiply the thid 4x 1mod 7 by 2: 8x x 2 mod 7 Lastly we multiply the last given conguence 5x 9 mod 11 by 9: 45x x 81 4 mod 11 This last conguence is x 4 mod 11. We solve the equivalent system: 3 mod 5 1 mod 2. How do we solve these? by multiplying by an appopiate facto. Multiply the gcd 3, 6 3: x, x, x 2 mod 7 and x 4 mod 11 The solution is found by fomula (3.23): (3.23) x a1n1x 1 a2n2x2 a3n3x3 a N x We ae given 4 equations so we use this fomula with 4 : x a N x a N x a N x a N x ( ) 1 1 1 2 2 2 3 3 3 4 4 4 n1n 2 n Evaluating each of the N s which ae given by N with n 1 5, n 2 2, n 3 7 n and n4 11: 25 711 N1 154 5 2 5711 N2 385 2 25 7 11 N3 110 7 25711 N4 70 11
Chapte 3: Theoy of Modula Aithmetic 47 We need to find the x s which ae given by N x 1 mod n N1x1 154x1 1mod 5 N2x2 385x2 1mod 2 N3x3 110x3 1mod 7 N x 70x 1mod 11 fo 1, 2, 3 and 4 : [Remembe n1 5] [Remembe n2 2 ] [Remembe n3 7 ] [Remembe n4 11] 4 4 4 Solving each of these equations gives: 154x x 1 mod 5 implies x 4 1 1 1 385x x 1 mod 2 implies x2 1 2 2 The a s ae a 1 3, a2 1, a3 2 and a 4 4 110x 5x 1 mod 7 implies x 3 3 3 3 70x 4x 1 mod 11 implies x 3 4 4 4 because we ae solving x 3 mod 5, x 1 mod 2, x 2 mod 7 and x 4 mod 11 Putting all these numbes a1 3, a2 1, a3 2, a4 4, N1 154, N2 385, N3 110, N4 70, x1 4, x2 1, x3 3 and x 4 3 into ( ) gives: x a N x a N x a N x a N x 1 1 1 2 2 2 3 3 3 4 4 4 3154 4 13851 21103 4 703 3733 Remembe the modulus n is the poduct of the given moduli: n 5 2 711 770 x 3733 653 mod 770. The least positive intege is 653. We have Chec that this solution is coect. Summay To solve simultaneous conguence equations we apply the Chinese Remainde Theoem to esolve fo the unnown.