SOLVED EXAMPLES

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1 Prelimiaries Chapter PELIMINAIES Cocept of Divisibility: A o-zero iteger t is said to be a divisor of a iteger s if there is a iteger u such that s tu I this case we write t s (i) 6 as ca be writte as = 6 So 6 divides (ii) 5 7, sice there is o u, st 7 = 5u Hece 5 does ot divide 7 Prime Number : A prime umber is a positive iteger greater tha whose oly positive divisors are ad itself eg 7, 3, etc 3 Divisio Algorithm : Let a ad b itegers with b > 0 The there exist uique iteger is q ad r with the property that a bq r where 0 r b, q is called the quotiet ad r is remaider eg for a = 7 ad b = 5 divisio algorithm gives 7 = for a = 3, b = 6, 3 = 6( ) + GCD (Greatest Commo Divisor): The greatest commo divisor of two o-zero itegers a ad b is the largest of all commo divisors of a ad b We deote this iteger by gcd(a, b) emarks : If a, b {0}, the there exist x, y such that gcd ( a, b) ax by (i) gcd (5, ) =, sice is the oly umber which divides 5 ad both (ii) gcd (, ) = is greatest commo divisor as ad (iii) gcd (0, 5) = 5 as 5 0 ad 5 5 but o other umber divisors of both which is also greater tha 5 (i) What is the smallest positive positive iteger i the set {x 60y 000 z x, y, z } (a) (b) (c) 6 (d) [CSI-NET-03-(II)] Sol gcd (, 60, 000) Hece, correct optio is (b) (ii) For positive iteger m ad Let F m ad Gm Which of the followig statemet are true? (a) F divides G m wheever m (b) gcd ( F, Gm) wheever m (c) gcd ( F, Fm ) wheever m (d) G m divides F wheever m [CSI-NET-0-(I)] Sol Take m 3 ; optio (b) cotradicted

2 Prelimiaries Take m ; 3 optio (d) cotradicted So, optio (a) ad (c) are correct (iii) True or False The equatio 63x 70y 5z 00 has a itegral solutio [TIF-0] Sol By defiitio of gcd (63, 70, 5) there exist x, y, z such that gcd (63, 70,5) 63x 70y 5z, the 00 63(00 x) 70(00 y) 5(00 z) 00 63x 70y 5z So we get x, y, z are itegral solutio of the equatio Hece this is true statemet 5 elatively Prime Itegers or Coprime iteger : Whe gcd( a, b), we say a ad b are relatively prime or coprime iteger eg gcd(9,0), gcd(5,8) gcd (0, 0) = 0 as 0 is the greatest commo divisor of (0, 0) so, they are ot relatively prime 6 Composite umber: which are ot prime umbers ie (other the prime umber iteger all umber is composite) eg To compute gcd( a, b ) by Euclidea Algorithm Exp: To compute gcd(38, ) = + 6 = = = + = + 0 gcd(38, ) gcd(50, 5) 50 = = = + = gcd(50, 5) 7 Euclid s Lemma: If p is a prime umber that divides ab, the p divides a or p divides b ie p ab p a or p b Exp Whe p is ot a prime, the Euclid s Lemma may fail 6 3 but 6 ad Fudametal Theorem of Arithmetic: Every iteger greater tha is a prime or a product of primes This product is uique, except for the order i which the factors appear Thus, if p p pr ad q q qs where p s ad q s are primes, the r = s ad, after re-umberig the q s, we have pi qi for all i (i) 30 = 3 5, Thus, p, p 3, p3 5 (ii) 5 = 3 5, p 3, p 5 (iii) = 3 7, p, p 3, p3 7 These expressio are uique No other primes exists other tha these which gives same umber i each example give above 9 Least Commo Multiple: LCM (a, b): The least commo multiple of two o-zero itegers a ad b is the smallest positive iteger that is a multiple of both a ad b eg lcm(, 6) =

3 Prelimiaries (i) (i) lcm (5, 8) = 0 (ii) lcm (0, 0) = 0 emarks : a b lcm ( a, b)gcd ( a, b), so, if gcd (a, b) =, the lcm (a, b) = a b If a m ad b m the l m [where l is lcm (a, b) True or False Give ay iteger we ca always fid a iteger m such that each of cosecutive itegers m, m 3,, m are composite [TIF-0] As True Sol Take m lcm {, 3,,, }, we ca take commo, 3,,, from each term which will show that umbers are composite 0 Modular Arithmetic: Modular arithmetic is a abstractio of a method of coutig that you ofte use for example, if it is ow Jauary, what will be 5 th moth from ow? of course February 5 = +, whe a q r, where q is the quotiet ad r is the remaider upo dividig a by, we write (i) (ii) amod r or a r mod Thus 3mod sice 3= +, 6 mod 0 sice 6 = mod 85 6 sice 6 = More geerally, if a ad b are itegers ad is a positive iteger, we ofte write ( a b) ( ab) mod (( a mod )( b mod )) mod, ( a b) mod (( a mod ) ( b mod )) mod (7 3) mod0 ((7 mod0) (3 mod0)) mod0 = (7 3) mod0 = 0mod0 0 mod (7) meas 7 ( ) 7 This gives the remaider after dividig by 7 to ( ) Some properties of modular Arithmetic if a b (mod ) The, a + c b + c mod () a c b c mod () (iii) p p a b mod ( ) (iv) p( a) p( b) (mod ) where p( a ) polyomial i a but coverse of the property (ii), (iii) ad (iv) does ot hold as, 9 6 mod (3) also 33 3 mod (3) but 3 mod (3) ot hold Similarly for others 3 a b mod wheever

4 (i) The last digit of 80 is (a) (b) (c) 6 (d) 8 Sol Usig modular arithmetic ; 6 mod 0 Use property (iii) if (mod ) ad P P a b a b (mod ) So we get ( ) 0 = 6 0 (mod 0) 80 = 6 (mod 0) last digit of 80 is 6 Hece, correct optio is (c) Prelimiaries [TIF-00] (ii) Which of the followig statemet is FALSE? (a) There exist a atural umber which whe divided by 3, leaves remaider ad which whe divided by, leaves remaider 0 (b) There exist a atural umber which whe divided by 6, leaves remaider ad which whe divided by 9, leaves remaider (c) There exist a atural umber which whe divided by 7, leaves remaider ad which whe divided by, leaves remaider 3 (d) There exist a atural umber which whe divided by, leaves remaider 7 ad which whe divided by 8, leaves remaider 3 [TIF-00] Sol Apply modular arithmetic, check it by optio (b) assume such atural umber exist say x the x mod 6 ad x mod 9 x 6q ad x 9q 6q 9q (6q ) mod 9 0 (*) use ( a b) mod ( a (mod ) b (mod )) mod the (6q ) mod 9 (6 q (mod 9) (mod 9)) mod 9 (0 ) mod 9 or (3 ) mod 9 0 or (6 ) mod 9 7 which is cotradictio to (*), hece optio (b) is ot true Hece, correct optio is (b) (iii) What is the last digit of 97 03? (a) (b) 3 (c) 7 (d) 9 Sol By modular arithmetic (97) ((97) ) mod 0 mod 0 by property (iii) mod 0 97 mod 0 by property (ii) 7 mod 0 Hece, correct optio is (c) [TIF-0] (iv) Which of the followig primes satisfy the cogruece a (6a ) mod 3

5 Prelimiaries (a) (b) 7 (c) 67 (d) 83 Sol () (6 ) mod 3 (6 ) mod 3 (8) mod 3 mod 3 (v) ( ) mod 3 ( ) Use Euler Theorem gcd ( a, ), the a mod, a ; 3 the a mod 3 Hece, correct optios are (a) ad (c) The last two digit of 7 8 are (a) 07 (b) 7 (c) 37 (d) 7 8 Sol 7 r mod 00 Fid r (vi) 7 0(mod 00) (mod 00) r 07 Hece, correct optio is (a) 0 80 (7 ) 7 7 0(mod 00) use modular arithmatic property The last digit of (38) 0 is (a) 6 (b) (c) (d) 8 Sol 38 8 (mod 0) (vii) (38) (8) (mod 0) (6 mod 0) 50 ((38) ) 6 (mod 0) 008 (38) 6 (mod 0) 009 (38) 8 (mod 0) 00 (38) (mod 0) 0 (38) (mod 0) last digit Hece, correct optio is (b) The uit digit of 00 is (a) (b) (c) 6 (d) 8 Sol 6 (mod 0) 5 5 ( ) (6) (mod 0) 6 (mod 0) uit digit Hece, correct optio is (c) 5 [CSI-NET-05-(I)] [CSI-NET-0-(II)] [CSI-NET-0-(I)] [CSI-NET-0-(I)] Fuctios (Mappigs) ad elatios Fuctio (Mappig): A fuctio (or mappig) f from a set X to a set Y is a rule which assigs to each elemet x of X exactly oe elemet y of Y f X Domai Y Co-domai

6 6 Prelimiaries -- ' f : X Y ' a A f B a (i) a a a A f B a (ii) a a a A f 3 B a (iii) a a By the defiitio of fuctio that a rule that assigs to each elemet x of X exactly oe elemet y of Y Thus i above examples f is ot a fuctio sice a maps to a ad a both which cotradicts the defiitio of a fuctio Compositio of Fuctios: Let f : X Y ad g : Y from X to Z defied by ( gof )( x) g( f ( x)) for all x X f Z The compositio gof (or gf ) is the mappig g x f( x) g( f( x)) X Y Z Let f ( x) x, g( x) x fog( x) f g( x) f x The ( x ) x but gof ( x) g f ( x) g x g x x x ( ) ( ) Thus, x x

7 Prelimiaries 7 Hece compositio of two fuctio eed ot be commutative ie, fog ( x) gof ( x) 3 Oe to oe Fuctio: A fuctio f : X Y is called oe to oe if f ( x ) f ( x) x x or equivaletly x x f ( x ) f ( x) Oe-oe Not oe-oe A B A B a f a f a (i) a a (ii) a a emarks : (i) If domai has more elemets tha codomai fuctio ca ot be oe-oe (by defiitio of a fuctio) (ii) If domai has oly oe elemet the ay fuctio which is defied o to that domai always be oe-oe (by defiitio of a fuctio) (iii) If x m, y ; m the umber of oe-oe map is! ( m)! (i) Let m be atural umber The umber of ijective maps from a set of cardiality m to a set of cardiality is (a) m! (b)! (c) ( m)! (d) Noe [TIF-0] Sol By remark (iii) Number of oe to oe fuctio is Hece, correct optio is (d) p m! ( m)! ONTO Fuctios: A fuctio f : X Y is said to be oto if each elemet of Y is the image of atleast oe elemet of X p m ONTO Not ONTO emarks : (i) Similarly if codomai has more elemet tha domai of a fuctio fuctio ca ot be oto (ii) If codomai of a fuctio with o empty domai the it will be always oto meas for each y Y ; atleast oe x X such that f ( x) y (iii) If x m, y ; m the umber of oto map is ( ) ( ) ( 3) ( ) ( k) m m m m k m C C C3 Ck

8 (i) 8 Sol Number of surjective maps from a set of elemets to a set of 3 elemet is (a) 36 (b) 6 (c) 69 (d) 8 Use remark (iii) m m m m c ( ) ( ) ( 3) c c 3 m ; c (3 ) 3 (3 ) 3 (3 3) c c 3 8 3() 3() Hece, correct optio is (a) Prelimiaries [CSI-NET-0-(II)] 5 Properties of Fuctios: Give fuctios f : X Y, g : Y Z ad h : Z W (i) h( gf ) ( hg) f (associativity) (ii) If f ad g are oe-to-oe, the gof ad fog is oe-to-oe (iii) If f ad g are oto, the gof ad fog is oto (iv) If f is oe-to-oe ad oto, the there exist a fuctio f The from Y oto X such that ( f of )( x) x x X ad ( fof )( y) y y Y 6 elatio: Let A ad B be two sets A relatio from A to B is a subset of A B where A B is cartesia product of sets A ad B ie A B is a set of ordered pairs (a, b) such that,, 3,,,, The A B ( x, y) : x A ad y B A B a b c a A ad b B The power set of A B defied as relatios o A B Ifact, each elemet of power set of A B defies a relatio ad each relatio defied o A B gives a subset of A B as, N N (, ), (, ), (, 3), (, ), (, ), (, 3), (3,), (3, ), (3, 3), The if we take all order pairs ( a, b ) st a ad 3 is a subset of N N elatio o a set A: A relatio o a set A is subset of A A b The (, 3), (, 3), (3, 3), (, 3) which 7 Types of relatio o a set: (i) eflexive elatio: Let be a relatio o a set A ie let be a subset of A A, the is called a reflexive relatio if ( a, a), a A

9 Prelimiaries ie is reflexive if we have, aa, a A A relatio o a set A is NOT EFLEXIVE if there is atleast oe elemet a A, such that ( a, a) (i) eflexive : A,, 3 9 The (,), (, ) 3 (, ), (, ), (3, 3), (3, ) (,), (, ), (3, 3) (, ), (, ), (, 3), (3, ), (, 3), (3, ) ad are ot reflexive sice they have ot all possible pairs of ( a, a) a A I, 3 they are reflexive sice (, ), (, ), (3, 3) Similarly ; {(, ), (, ), (3, 3)} (ii) Symmetric relatio: Let be a relatio o a set A ie let be a subset of A A The is said to be a symmetric relatio if ( a, b) ( b, a) Thus is symmetric if we have ba wheever we have ab A relatio o a set A is ot symmetric if there exist two distict elemets a, b A, such that ab but ba ie ( a, b) but ( b, a) Set A = {,, 3, } 3 (, ), (, ) (, ), (, 3), (3,), (, ) (,), (, ), (3, 3), (, ) is ot symmetric, sice (, ) but (, ) (iii) Ati-symmetric relatio: Let be a relatio o a set A ie let be a subset of A A The is a b ad ( b, a) said to be a ati-symmetric relatio if (, ) ie ( a, b), ad ( b, a) a b or a b ( a, b) or ( b, a) or ( a, b) ( b, a) implies a b is ot ati-symmetric if there exist elemets a b such that ( a, b) as well as ( b, a) (iv) Trasitive relatio: Let be a relatio o a set A ie let be subset of A A The is said to be a trasitive relatio if ( a, b) ad ( b, c) ( a, c) A relatio o a set A is ot trasitive if there exist elemets a, b ad c i A, ot ecessarily distict, such that ( a, b),( b, c) but ( a, c) A,, 3,

10 0 3 (, ), (, ) (3, 3), (3, ) (, ), (, ), (, ) ad are trasitive but 3 is ot sice (, ) 3 ad (, ) 3 but (, ) 3 Prelimiaries 7 Equivalece elatio: Let be a relatio o a set A The is said to be a equivalece relatio iff Ex the followig three coditios hold simultaeously a A (i) is reflexive ie a A ( a, a) (ii) is symmetric ie ( a, b) ( b, a) (iii) is trasitive ie ( a, b) ad ( b, c) ( a, c) (i) Set of parallel lies (a) eflexive:-- Each lie is parallel to itself (b) Symmetric:-- If L L the L L Hece symmetric hold (c) Trasitivity:-- If L L ad L L3 L L3 Hece set of parallel lies defies a equivalece relatio Ex (ii) Set of all male huma beig, relatio defies brotherhood :-- (a) eflexivity :-- Each male huma beig is brother to itself (i mathematics it is hold) (b) Symmetric :-- If A is brother of A the A is also brother of A (Sice we have the set of all male huma beig) (c) Trasitive :-- Also hold Hece a equivalece relatio 8 Examples: which is ot equivalece relatio (i) set of perpedicular lies (a) ot reflexive as a lie caot perpedicular to itself (b) it is symmetric (c) ot trasitive Hece ot a equivalece relatio 9 Equivalece Class: If ~ is a equivalece relatio o a set A ad a A, the the set [ a] { x A : x ~ a} is called the equivalece class of A cotaiig a [a] is a subset of A 0 Properties of Equivalece Classes: Let A be a o-empty set ad let be a equivalece relatio o A Let a ad b be arbitrary elemets of A The (i) a [ a] (ii) If b [ a], the [ b] [ a] (iii) [ a] [ b] iff ab ie ( a, b) (iv) Either [ a] [ b] or [ a] [ b] ie two equivalece classes are either disjoit or idetical Proof: (i) Sice is reflexive, we have aa But [ a] { x x A ad aa} Hece a ad aa imply a [ a] (ii) Let b [ a] ba Now if, x be ay arbitrary elemet of [b] The x [ b] xb But is trasitive, therefore xb ad ba xa x [ a] Thus x [ b] x [ a] Therefore [ b] [ a] Agai, let y be ay arbitrary elemet of [a] The y [ a] ya Sice is symmetric, therefore ba ab Now ya ad ab yb y [ b]

11 Prelimiaries Thus y [ a] y [ b] Therefore [ a] [ b] Fially [ a] [ b] ad [ b] [ a] [ a] [ b] (iii) First Part: [ a] [ b] ab We have [ a] [ b] Sice is reflexive, therefore we have aa Now aa a [ a] a [ b] [ [ a] [ b]] ab Thus [ a] [ b] ab Coverse part: Suppose that ab, the to prove that [ a] [ b] Let x be ay arbitrary elemet of [a] The xa But it is give that ab Therefore, xa ad ab xb [ is trasitive] x [ b] Thus x [ a] x [ b] Therefore [ a] [ b] Agai let y be ay arbitrary elemet of [b] The y [ b] yb Now, we are give that ab From this we have ba sice is symmetric Now, yb ad ba ya y [ a] Thus y [ b] y [ a], therefore [ b] [ a] Hece, [ a] [ b] ad [ b] [ a] [ a] [ b] Fially sice [a] = [b] ab ad ab [ a] [ b] Therefore, [ a] [ b] iff ab (iv) If [ a] [ b], the we have othig to prove So let us suppose that [ a] [ b] therefore there exist a elemet x A such that x [ a] [ b] x [ a] ad x [ b] xa ad xb ax ad xb [ is symmetric] ab [ is trasitive] [ a] [ b] Thus, [ a] [ b] [ a] [ b] Partitio: A partitio of a set S is a collectio of o-empty disjoit subsets of S whose uio is S Cosider the set S {,,3, }, the {, },{3},{} is a partitio of S Let Z be the set of all itegers we kow that x y(mod 3) is a equivalece relatio o Z Cosider the set of three equivalece classes [0] {, 6, 3,0,3,6,} [] {, 7,,,,7,} [] {, 8, 5,,5,8,} We observe that (i) The sets [0], [] ad [] are o-empty (ii) The sets [0], [] ad [] are pairwise disjoit (iii) Z [0] [] [] Hece, {[0],[],[]} is a partitio of Z, uder the relatio x y (mod 3)

12 Prelimiaries elatio iduced by a partitio of a set: Correspodig to ay partitio of a set S, we ca defie a relatio o S by the requiremet that xy iff x ad y belog to the same subset of S belogig to the partitio The relatio is the said to be iduced by the partitio Example : Cosider the set S {,,9,0} ad its subsets B {,3}, B {7,8,0}, B {,5,6}, B {,9} 3 The set p { B, B, B3, B} is such that (i) B, B, B3, B are all o-empty subsets of S (ii) 3 B B B B S, ad (iii) For ay sets B, either Bi Bj or Bi B j i Hece the set { B, B, B3, B } is a partitio of S 3 Fudametal theorem o equivalece relatio: A equivalece relatio o a o-empty set S determie a partitio of S ad coversely a partitio of S defies a equivalece relatio o S O The equivalece classes of a equivalece relatio o a set S costitute a partitio of S coversely, for ay partitio P of S, there is a equivalece relatio o S whose equivalece classes are the elemets of P Proof: Let be a equivalece relatio i S Let A be the set of equivalece classes of S with respect to ie let A {[ a]: a S} where [ a] { x : x S ad xa} Sice is a equivalece relatio, therefore a S, we have aa Hece a [ a] ad thus [ a] Further every elemet a of S is a elemet of the equivalece class [a] i A From this we coclude that S [ a] as Fially, if [a] ad [b] are two equivalece classes the either [ a] [ b] or [ a] [ b] Hece A is a partitio of S Thus we see that a equivalece relatio i S decomposes the set S ito equivalece classes ay two of which are either equal or mutually disjoit T T T p q S Coverse Let P {,,,} be ay partitio of S If, a b c iff there is a T i i the partitio such that p, q Ti, let us defie a relatio i S by pq Now S T T T Therefore x S, there exists Ti P such that x Ti for some i Hece x a b c Ti ad x Ti meas xx Thus x S, we have xx ad thus is reflexive Agai if we have xy, the there exists Ti But x Ti ad y Ti y Ti Therefore, is symmetric ad x Ti yx P such that x Ti ad y Ti Fially suppose xy ad yz The by the defiitio of there exist subsets T j ad T k (ot ecessarily distict) such that x, y Tj ad y, z Tk Sice y Tj ad also y Tk, therefore T j Tk But Tj ad T k belog to a partitio of S Therefore T j Tk implies T j Tk Now T j Tk implies x, z Tj

13 Prelimiaries ad cosequetly we have xz Thus is trasitive Sice is reflexive, symmetric ad trasitive, therefore is a equivalece relatio SOME IMPOTANT FUNCTIONS fuctio : It is called Euler phi-fuctio ad defied as ( ) = umber of positive itegers co-prime to ad less tha or equal to Ex (i) (0) = as, 3, 7, 9 (total ) itegers which are co-prime to 0 Method for fidig umber of positive itegers co-prime to :-- a b c If p q r, where p, q, r, are prime ad a, b, c, are atural umber 3 The ( ) p q r Ex (ii) (00) () () 0 (i) For a positive iteger m, let ( m) deote the umber of itegers k such that k m ad gcd ( k, m) The which of the followig statemet are ecessarily true (a) ( ) divides for every positive iteger Sol (b) divides ( a ) for all positive iteger a ad (c) divides ( a ) for all positive iteger a ad such that gcd ( a, ) (d) a divides ( a ) for all positive iteger a ad such that gcd ( a, ) [CSI-NET-0-(II)] (b) ad (c) For (a) take = 3 ad For (d) take a = 3, =, the both (a) ad (d) are cotradicted (ii) The umber of elemet i the set { m : m 000, m ad 000 are relatively prime} is (a) 00 (b) 50 (c) 300 (d) 00 [CSI-NET-0-(I)] 3 Sol (000) (0 ) 0 00 By property 5 Hece, correct optio is (d) 98 (iii) What is the cardiality of the set { z z ad z for ay 0 98}? (a) 0 (b) (c) (d) 9 [CSI-NET-05-(II)] Sol Give group is isomorphic to cyclic group of order 98 ie, 98 the umber of elemets i 98 which is coprime to 98 is (98) Hece, correct optio is (c)

14 (): ((tau) fuctio):-- It gives total umber of divisors of, icludig ad both a b c If p q r, where p, q, r, are distict primes ad a, b, c The ( ) ( a ) ( b ) ( c ) Note: It icludes ad ad the umber itself (00) 5 ( ) ( ) (3) (3) 9 Prelimiaries (iii) ( ) :-- (Sigma fuctio) :-- It gives the sum of all divisors It too icludes ad the umber itself a b c If p q r, p, q, r primes a, b, c a b c p q r The ( ) p q r Ex (0) (i) The umber of positive divisor of 50,000 is (a) 0 (b) 30 (c) 0 (d) 50 5 Sol 50, ( ) ( ) (5 ) Hece, correct optio is (b) Properties : I particular whe is prime say p, ( p) p, ( p), ( p) p If gcd (m, ) = the (i) ( m) ( m) ( ) (ii) ( m ) ( m) ( ) (iii) ( m ) ( m) ( ) [CSI-NET-0-(I)] 3 A umber theoratic fuctio f is said to be multiplicative if f ( m, ) f ( m) f ( ) Wheever gcd (m, ) = hece,, are multiplicative fuctios For >, ( ) is always a eve iteger 5 (0 ) 0 Importat Theorems : (I) Gauss : For each positive iteger ( d ) where sum beig exteded over all positive divisors d of d / (II) Theorem : For >, the sum of the positive itegers less tha ad relatively prime to is ( K, ) K K ( )

15 Prelimiaries 5 p (III) Fermat s Theorem : If p is prime, the a a (mod p) p (IV) Fermat s little Theorem : If p is prime, the a (mod p) coverse of Fermat s theorem is eed ot to be true ( ) (V) Euler Theorem : If is a positive iteger ad gcd (a, ) =, the a (mod ), ( ) is Euler phi fuctio Euler Theorem is geeralizatio of Fermat s Theorem O k ie if x a (mod ) ad gcd ( k, ( )) d ; is prime # ( )/ d a (mod ), where ( ) is Euler fuctio (VI) Wilso s Theorem : If p is a prime, the ( p )! (mod p) or ( p )! 0 (mod p) (VII) Pseudo Prime : A composite iteger is called pseudo prime if it satisfies the cogruece equatios (mod ) (VIII) Chiese emaider Theorem Let x a mod m : i,,, for which m i are pairwise relatively prime the solutio of the set of cogruece i i (i) Sol m m m x a b a b mod m, where m m m m ad bi satisfy bi (mod mi ) m m m is If m, m,, m are pairwise relatively prime positive itegers ad if a, a,, a are ay itegers the simultaeous cogruece x a mod m ; i,, have a solutio ad solutio is uique modulo m, where m m m m The equatio x (mod 3) has (a) o solutio (c) exactly oe solutio Use Euler theorem (v) i i (b) 3 solutio (d) solutio k ; a ; 3 ; d gcd (, (3)) gcd (, ), the (3)/ d / (mod ) (mod 3) (mod 3) i [TIF-0] which is cotradictio Hece o solutio exist Hece, correct optio is (a) (ii) If is the positive iteger such that the sum of all positive iteger a satisfyig a ad gcd ( a, ) is Sol equal to 0 the umber of summad amely ( ) is (a) 0 (b) (c) 0 (d) 80 Use theorem II ( k, ) k k ( ) 0 ( ) ( ) 80 Hece, correct optio is (d) [CSI-NET-0-(I)]