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ELEMENTS OF NUMBER THEORY & CONGRUENCES Lagrange, Legendre and Gauss M th t Mathematics ELEMENTS OF NUMBER THEORY & CONGRUENCES 1) 1) 2) 3) 4) If a≠0, b≠0 ∈Z and a/b, b/a then a=b a=1 1 b=1 a=±b M th t Mathematics Ans : is 4 known result. If a/b ⇒b=ma → (1) where m ∈Z & b/a ⇒ a=bn → (2) where n ∈Z from (1) & (2), a=(am)n=a(mn) ⇒ mn=1, mn=1 possible if m=1 & n=1 or m=-1 & n=-1. For the values of n=1 &-1 then (2)→a=± b M th t Mathematics 2)) 1) 2) 3) 4) 0a and d1a are e primes composite it numbers b neither prime nor composite none of these M th t Mathematics Ans : is 3 by defn. of prime & composite numbers its implied M th t Mathematics 3) 1) 2) 3) 4) If (ab,c) = 1 & (a, c) c)=1 1 then (b, c) c)= 1 c b none of the these M th t Mathematics Ans : is 1 known result (a, c) = 1, (b, c) = 1 ⇒ (ab,c)=1 M th t Mathematics 4)) 1) 2) 3) 4) If p is sp prime e number u be then t e p/ab ⇒ p/a p/b /b p/a or p/b none of the these M th t Mathematics A : iis 3. Ans 3 k known result lt p/ab ⇒ p/a or p/b M th t Mathematics 5) 1) 2) 3) 4) 111………1 (9 (91 ttimes) es) is s a composite number a prime i number b a surd Irrational M th t Mathematics Ans : is 1 since 91 = 7 x 13 1111..1 = 1111111 . 1111111 1111111----(13 (13 factors) 91 times 7 times 7 times & ∴ it is diviible by 1111111. (7 times) ∴ It is a composite number. M th t Mathematics 6) 1) 2) 3) 4) The number of positive divisors off 1400 1400, iincluding l di 1 and d itself it lf is i 18 24 22 21 M th t Mathematics Ans : is 2 1400 = 23 x 52 x 7 ∴ T(1400) = (3+1) (2+1)(+1) = 24 M th t Mathematics 7) 1) 2) 3) 4) The sum of all positive divisors off 960 excluding l di 1 and d it itself lf is i 3047 2180 2087 3087 M th t Mathematics Ans : is 3 960 = 26 x 3 x 5 S =127 x 4 x 6 = 3048 but 3048 – 960 – 1 = 2087. M th t Mathematics 8) 1) 2) 3) 4) If (a (a+b) b)3 ≡ x ((mod od a) then t e x=a2 x=b b3 x=a3 x=b2 M th t Mathematics Ans : is 2 ((a+b))3 = a3+3a2b + 3ab2 + b3 ⇒ (a+b)3 – b3 = a(a2+3ab+3b2)=ak ⇒ a / [(a+b)3 – b3] ∴ (a+b)3 ≡ b3 (mod a) M th t Mathematics 9) 1) 2) 3) 4) Which of the following statement i false is f l ? 98 ≡ –7 (mod 3) 67 2 ((mod 67≡ d 5) 123 ≡ –4 (mod 7) 240 ≡ 9 (mod 11) M th t Mathematics Ans : is 3 123 + 4 = 127 is not a multiple of 7 M th t Mathematics 10) If 100 ≡ x (mod 7), then the least positive iti value l off x iis 1) 2) 3) 4) 1 3 4 2 M th t Mathematics Ans : is 4 7 / (100 – x) when x = 2, 7 / 98 M th t Mathematics 11) When 520 is divided by 7 the remainder i d is i 1) 2) 3) 4) 1 3 4 6 M th t Mathematics Ans : is 3 53 = 125 ≡ -1 ((mod 7)) ∴ (53)6 ≡ (-1)6 (mod 7) 518.52 ≡ 1.52 (mod 7) ∴ 520 ≡25(mod 7) ≡ 4(mod 7) M th t Mathematics 12)) The e last ast d digit g t in 7291 is s 1) 2) 3) 4) 1 3 7 9 M th t Mathematics Ans : is 2 72 = 49 ≡ -1 1 (mod 10) ⇒ (72)145 ≡ (-1)145 (mod 10) 7290 ≡ -1(mod 10) also 7 ≡ -3 3 (mod 10) ∴ 7190x7 ≡ ((-1)(-3) )( ) ((mod 10)) ∴ 7291 ≡ 3(mod 10) M th t Mathematics 13) The digit in the unit place of the number b 183! + 3183 is i 1) 2) 3) 4) 7 6 3 0 M th t Mathematics Ans : is 1 Unit place in 183 ! is 0 ( 2 ≡ -1 (mod 10) (32)91 ≡ (-1)91 (mod 10)= -1 (mod 10) ∴ 3182 ≡ -1 (mod 10) also, also 3 ≡ -7 (mod 10) ∴ 3182 .3 ≡ (-1) (-7) (mod 10) ∴ 3183 ≡ 7 (mod 10) M th t Mathematics 14) If – 17 ≡ 3 (mod x), then x can t k th take the value l 1) 2) 3) 4) 7 3 5 None of these M th t Mathematics Ans : is 3 -17 – 3 = - 20 is divisible by y5 M th t Mathematics 15) The smallest positive divisor of a co composite pos te integer tege a ((>1)) does not exceed 1) a2 2) 3) a3 4) M th t Mathematics Ans : is 4 K Known result lt M th t Mathematics 16) Which following linear congruences has h no solution l ti 1) 2) 3) 4) 4x ≡ 1 (mod 3) 3 ≡ 2 ((mod 3x d 6) 5x ≡ 3 (mod 4) 2x ≡ 1 (mod 3) M th t Mathematics Ans : is 2 Since (3, 6) = 3 & 3 does not divide 2 ∴ No solution M th t Mathematics 17) The relation congruence modulo m is i 1) 2) 3) 4) Reflexive S Symmetric ti Transitive only All of these M th t Mathematics Ans : is 4 K Known result lt ≡ b (mod m) is an equivalence relation M th t Mathematics 18) The least positive integer to which c 79 9 x 101 0 x 125 5 is sd divided ded by 11 is 1) 2) 3) 4) 5 6 4 8 M th t Mathematics Ans : is 1 79 ≡ 2(mod 11), 101 ≡ 2 (mod 11) & 125 ≡ 4(mod 11) multiplying these, these 79x101x125 ≡ 2x2x4 ≡16 (mod 11) but 16 ≡ 5 (mod 11) ∴ 79 x 101 x 125 ≡ 5 (mod ( d 11) M th t Mathematics 19) 9) If p ≡ q ((mod od m)) if a and d only o y if 1) 2) 3) 4) (p – q) / m m/(p /( – q)) m/p m/q M th t Mathematics Ans : is 2 by very defn. Of congruence i.e. if a ≡ b ((mod m)) ⇒ m/(a-b) ( ) M th t Mathematics 20) When 2100 is divided by 11, the remainder i d is i 1) 2) 3) 4) 3 5 1 2 M th t Mathematics Ans : is 3 25=32 ≡ -1 1 (mod 11) ∴ (25)20 ≡ (-1) ( 1)20 (mod ( d 11) ∴ 2100 ≡ 1 (mod 11) M th t Mathematics 21) If a ≡ b (mod m) and (a, m) = 1, then th 1) 2) 3) 4) (a, b) = 1 (b m)) = 1 (b, (b, m) = a (a, b) = m M th t Mathematics Ans : is 2 Known result ((a,m)) = (b,m) ( ) =1 M th t Mathematics 22) If n ≡ 0 (mod 4) then n3 – n is di i ibl by divisible b 1) 2) 3) 4) 6 but not 24 12 b butt nott 24 24 12 & 24 M th t Mathematics A : iis 2 Ans n is a multiple of 4 if n=4,, n3 – n = 60 ∴ 12/60, 6/60 but 24 does not divided by 60 Thus 6 & 12 divide n3 – n. M th t Mathematics 23) If 195 ≡ 35 m= 1) 2) 3) 4) then 4 5 0 7 M th t Mathematics Ans : is 3 (m+2) / (195-35) ⇒ (m+2) / 160 ⇒ m+2 ≥ 2 ⇒ m+2 = 2 2, 4 4, 5, 5 8 - - - etc. etc ⇒m m= 0, 2, 3, 6 etc., ∴ (3) is the answer M th t Mathematics 24) If 28 ≡ (a+1) (mod 7) is true then a is i 1) 2) 3) 4) 3 4 0 5 M th t Mathematics Ans : is 1 26=64 ≡ 1 (mod 7) 26.2 22 ≡ 1 1.2 22 (mod ( d 7)) ∴ 28 ≡ 4 (mod 7) ⇒a a+1 1 = 4 i.e., (a=3) (a 3) M th t Mathematics 25) 5) The eu unitt digit d g t in 13 337 is s 1) 2) 3) 4) 5 2 6 3 M th t Mathematics Ans : is 4 132 = 169 ≡ -1 (mod 10) (132)18 ≡ (-1)18 (mod 10) 1336.13 13 ≡ 1.13 1 13 (mod 10) ∴ 1337 ≡ 3 (mod 10) M th t Mathematics 26) The number of incongruent solutions l ti off 24x 24 ≡ 8 ((mod d 32) iis 1) 2) 3) 4) 2 4 6 8 M th t Mathematics Ans : is 4 by y thm. (24, 32) = 8 & 8/8 ∴ the number of incongruent solutions = 8 M th t Mathematics 27) The remainder when 3100 x 250 is di id d by divided b 5 is i 1) 2) 3) 4) 3 4 1 2 M th t Mathematics Ans : is 2 32=9 ≡ -1 (mod 5) ⇒ (32)50 ≡ (-1)50 (mod 5) ∴ 3100 ≡ 1(mod 5) → (1) & 22=4≡ -1 (mod 5)⇒ (22)25 ≡ (-1)25 (mod 5) ∴ 250 ≡ -1 (mod 5) → (2) (1) x (2)→3100x250 ≡1x-1(mod 5)≡ -1 (mod 5) but -1 ≡ 4 (mod 5) ∴ 3100x250 ≡ 4 (mod 5) M th t Mathematics 28) If a and b are positive integers such that a2 – b2 is a prime number, then a2 – b2 is 1) 2) 3) 4) a+b a–b ab 1 M th t Mathematics Ans : is 1 a2 – b2 = (a (a+b) b) (a-b) (a b) is a prime. ∴ (a+b) (a-b) is divisible by 1 or its self. But a – b < a+b ∴ a-b=1 ∴ a2 – b2 = a+b M th t Mathematics 29) Which of the following is a prime i number b ? 1) 2) 3) 4) 370261 1003 73271 667 M th t Mathematics Ans : is 1 17/1003,, 11/73271 & 29/667. but none of the prime & less than 608 divides the first No. M th t Mathematics 30) Which c o of tthe e following o o g is s false a se ? 1) An odd number is relatively prime to the next even number 2) 3x ≡ 4 (mod 6) has solution 3) ax ≡ bx (mod m) ; x ≠ 0 ⇒ a ≡ b (mod m) 4) a.x + b.y = d ⇒ (a, b) = d M th t Mathematics Ans : is 2 (3 6) = 3 b (3,6) butt 3 d does nott di divides id 4 ∴ no solution. Remaining are all known results M th t Mathematics 31) For all positive values of p, q, r, and s, s will not be less than 1) 2) 3) 4) 81 91 101 111 M th t Mathematics Ans : is 1 ( ly y ∴ given expression is ≥ 3.3.3.3=81. ∴ expression cannot be less than 81. M th t Mathematics 32) If a+b)n ≡ x (mod a), then (n is a + integer) +ve i t ) 1) 2) 3) 4) x= a2 x=an x=bn none of these M th t Mathematics Ans : is 3 (a+b)n =an+nc1an-1.b+ … +ncn-1ab-1+bn ∴ ((a+b))n–bn=a [[an-1+nc1an-2.b+ … +ncn-1bn-1] (a+b)n–b bn=ak ak where k∈Z k∈Z. ∴ a/[(a+b)n – bn] ⇒ (a+b)n≡bn (mod a) ∴ x = bn M th t Mathematics 33) If 27= 189m 89 + 24n tthen e m&na are e 1) 2) 3) 4) unique nott unique i prime numbers none of these M th t Mathematics Ans : is 2 If (a,b) = d ⇒ d = ax + by where x, y ∈ Z. Here x, y are not unique unique. M th t Mathematics 34) If 2x≡3 (mod 7), then the values off x such h that th t 9 ≤ x ≤ 30 are 1) 2) 3) 4) 12, 19, 26 11 18 11, 18, 25 10, 17,24 None of these M th t Mathematics A : iis 1 Ans Th soln. The l is i x ≡ 5 ((mod d 7) ∴ Soln. S l sett is i { ….. 2, 2 5, 5 12, 12 19 19, 26, 26 33 33, …. } ∴ required i d values l off x are 12, 12 19, 19 26 26. M th t Mathematics 35) If p is a prime number and P is the product of all prime numbers less than or equal to p1 then 1) 2) 3)) 4) P – 1 is a prime P + 1 is not a prime number P + 1 is a p prime number P + 1 is a composite number M th t Mathematics Ans : is 3 Known result while proving p g the thm. The primes are infinite. M th t Mathematics 36) 4x + 9 ≡ 3 ((mod od 5) ca can be written tte as 1) 2) 3) 4) x ≡ 5 (mod 6) x ≡ 3 ((mod d 15) x ≡ 6 (mod 15) None of these M th t Mathematics Ans : is 3 when h x=6, 6 4 4.6+9 6 9 = 33 ≡ 3 (mod ( d 5) it satisfies the given congruence. Hence (3) is right answer M th t Mathematics 37)) If (3 3 (3-x)) ≡ ((2x-5) 5) ((mod od 4), ), then t e one o e of the values of x is 1) 2) 3) 4) 3 4 18 5 M th t Mathematics Ans : is 2 3-x-2x+5 = -3x+8 is divisible by 4 when x=4, -3 (4)+8 = -4 is divisible by 4. M th t Mathematics 38) The e remainder e a de when e 6 64x65x66 65 66 is s divided by 67 is 1) 2) 3) 4) 60 61 62 63 M th t Mathematics Ans : is 2 64 x 65 x 66 ≡ ((-3)) ((-2)) ((-1)) ((mod 67)) ≡ - 6 (mod 67) ≡ 61 (mod 67) M th t Mathematics GROUPS Lagrange, Legendre and Gauss M th t Mathematics 1) 1) 2) 3) 4) GROUP If x,y,z are three th elements l t off a group and then (xy-1z)-1= x-1y-1z-1 x-11yz z-1yx-1 (xy-1z)-1 M th t Mathematics Ans : is 3 since (a ¸b)-11=b-11¸a-11. Question is just extension of this property. M th t Mathematics 2)) 1) 2) 3) 4) If a¸b = , then ¸ is a binary operation on R Q+ Ro R+ M th t Mathematics Ans : is 4 if a = -1, b = 3 then , C M th t Mathematics 3) 1) 2) 3) 4) The identity element of a¸b=ab–1 is 1 0 2 –1 M th t Mathematics Ans : is 3 a¹e = a ⇒ ae-1=a ⇒ e – 1 = 1 ⇒ e=2 M th t Mathematics 4) 1) 2) 3) 4) In the group of rational numbers under d a binary bi operation ti ¸ defined d fi d by b a ¸ b = a+b–1 then identity element is 1 0 2 -1 M th t Mathematics Ans : is 1 a¹e = a ⇒ a+e-1=a ∴ e – 1 = 0 ⇒ e=1 M th t Mathematics 5) The set G={ -3, -2, -1, 0, 1, 2, 3} w.r.t. t add addition t o does not ot form o a group since. 1) The closure axiom is not satisfied 2) The Th associative i ti axiom i iis nott satisfied ti fi d 3) The commutative axiom is not satisfied 4)) Identity y axiom is not satisfied M th t Mathematics Ans : is 1 since 2, 3∈G but 2 2+3=5∉G 3 5∉G M th t Mathematics 6) If a¸b=2a – 3b on the set of i t integers. Then Th ¸ is i 1) 2) 3) 4) Associative but not commutative A Associative i ti and d commutative t ti A binary operation Commutative but not associative M th t Mathematics Ans : is 3 ∀ a, b∈Z, a¸b=2a-3b ∈Z (i.e., if a =1, b=-2 then 2 .1 -3 (-2) = 2+6= 8∈Z ) M th t Mathematics 7) 1) 2) 3) 4) In the multiplicative of cube roots oots of o unity u ty the t e inverse e se o of w99 is w 1 w2 Does not exist. M th t Mathematics Ans : is 2 W3=1 ∴ (w ( 3)33 = 1 M th t Mathematics 8) The e incorrect co ect state statement e t is s 1) In (G, .) ab ab=ac ac ⇒ b b=c, c, ∀ a, b, c ∈G 2) Cube roots of unity form an abelian group under addition 3) In a abelian group (ab)3=a3b3, ∀a, b∈G 4) In a group of even order, order there exists atleast two elements with their own inverse. M th t Mathematics Ans : is 2 Cube roots of unity; 1, 1 w, w w2 f form an abelian b li group under multiplication M th t Mathematics 9) 1) 2) 3) 4) If H & K are two subgroups of a group g oup G, then t e identify de t y the t e correct statement H∩K is a sub group H K iis a sub H∪K b group Neither H∪K nor H∩K is sub group Nothing can be said about H∪K and H∩K M th t Mathematics Ans : is 1 Let H= {{0, 2, 4}, } K={0,3} { } are sub groups of G={0, 1, 2, 3, 4, 5} under +6 i.e., H∪K = {0, 2, 3, 4} is not closed i.e., 2+3=5 ∉ H∪K M th t Mathematics 10) In the group G= {e, a, b} of order 3 a5b4 is 3, i 1) 2) 3) 4) 3 ab b a b M th t Mathematics Ans : is 3 ab=e ⇒ (ab)4 = e i a (a i.e. ( 4b4) =ae ⇒ a5b4=a M th t Mathematics 11) In a group (G, ¸), a¸x=b where a, b∈G b G has h 1) 2) 3) 4) Unique solution N solution No l ti More than one solution Infinite number of solution M th t Mathematics Ans : is 1 a¹x=b ⇒ a-1¹(a¹x)=a-1¹b ((a-1¹a)) ¹x=a-1¹b ⇒ x=a-1¹b M th t Mathematics 12) The set of (non singular) matrices at ces o of o order de 2 x 2 o over e z under matrix multiplication is 1) 2) 3) 4) Group S i group Semi Abelian group Non-abelian group M th t Mathematics Ans : is 2 M th t Mathematics 13) Which of the following is a subgroup subg oup o of G G={0, {0, 1,, 2,, 3, 4,, 5} under addition modulo 6 1) 2) 3) 4) {0, 2} {0 1} {0, {0, 4} {0, 3} M th t Mathematics Ans : is 4 2+62=4∉{0,2} etc., but 3+63=0 M th t Mathematics 14)) The e set o of integers tege s is s 1) 2) 3) 4) Finite group Additi group Additive Multiplicative group None of these M th t Mathematics Ans : is 2 M th t Mathematics 15) The set of all integers is not a group g oup under u de multiplication u t p cat o because 1) 2) 3) 4) Closure property fails A Associative i ti law l does d nott hold h ld good d There is no identity element There is no inverse M th t Mathematics Ans : is 4 Inverse 0 does not exists (also 2∈z but 2-1=½∉z) M th t Mathematics 16) A subset H of a group (G, ¸) is a subgroup b off G iff 1) 2) 3) 4) a, b∈H ⇒ a ¸ b∈H a∈H H ⇒ a-11∈H H a, b∈H ⇒ a ¸ b-1∈H H contains identity off G. M th t Mathematics A : iis 3 Ans By thm. M th t Mathematics 17) Zn= {0, 1, 2, - - - -, (n–1) } fails to be a group under multiplication modulo n because 1) 2) 3) 4) Closure property fails Cl Closure holds h ld but b t nott associativity i ti it There is no identity There is no inverse for an element of the set M th t Mathematics Ans : is 4 at least for one element ‘0’ has no inverse in Zn. M th t Mathematics 18) is an abelian group under matrix multiplication.Then multiplication Then the identity element is 1) 2) 3) 4)) M th t Mathematics Ans : is 3 M th t Mathematics 19) In the group G = {3, 6, 9, 12} under d x15, the th identity id tit iis 1) 2) 3) 4) 3 6 9 12 M th t Mathematics Ans : is 2 Since 3 x156=3, 6=3 6x156=6 9 156=9 9x 6 9 etc., t M th t Mathematics 20) The set of all 2 x 2 matrices over the real numbers is not a group under d matrix t i multiplication lti li ti because 1) 2) 3) 4) Inverse law is not satisfied A Associative i ti law l is i nott satisfied ti fi d Identity element does not exist Closure law is not satisfied M th t Mathematics Ans : is 1 If A is a singular matrix of 2 x 2 order matrix then A-1 does not exist. M th t Mathematics 21) (Z, ¸) is a group with a ¸ b = a+b+1, a b , ∀ a, b∈Z. b The e inverse e se o of a is 1) 2) 3) 4) A+2 – a+2 +2 – a –2 a–2 M th t Mathematics Ans : is 3 a¹e=a ⇒ a+e+1=a ⇒ e= -1 1 a¹a-11 = e ⇒ a+a-11+1 = - 1 ⇒ a-1 = -2-a M th t Mathematics 22) The four matrices under d multiplication lti li ti form f is i 1) 2) 3) 4) a group a semii group an abelian group infinite group M th t Mathematics Ans : is 3 Taking them as I, A, B, C then AB=C, BC=A, etc., & A.I=A etc. Also,, A.A=I ⇒ A-1=A |||ly B-1=B,, C-1=C C also AB=BA AB BA M th t Mathematics 23) In the group (G, ¸), where ∀ a, b∈ b G G. The e identity de t ty a and d inverse of 8 are respectively. 1) 2) 3) 4) M th t Mathematics Ans : is 2 a¹e=a ⇒ae/5=a⇒ e=5 & a¹a-1=e ⇒ aa-1=5 ⇒ a-1 = 25 5 a ∴8-1= 25 8 M th t Mathematics 24) The proper subgroups of the group G = {0, {0 1, 1 2, 2 3, 3 4, 4 5} under addition modulo 6 are 1) 2) 3) 4) {0, 3} and {0, 2, 4} {0 1 {0, 1, 3} and {0 {0, 1 1, 4} {0, 1} and {3, 4, 5} {0} and {0, 1, 2, 3, 4, 5} M th t Mathematics Ans : iis 1 A Since 0(G)=6 & 6=2 x 3 ∴ It has proper subgroups of orders 2&3 In (1) 3+63=0 & 2+62=4, 4+62=0 4+64 4=2 2 all in the sets M th t Mathematics 25) In the group G = {1, 3, 7, 9} under u de multiplication u t p cat o modulo odu o 10, 0, the value of is 1) 2) 3) 4) 5 3 7 9 M th t Mathematics Ans : is 4 e=1 7X103=1 ⇒ 7-1=3 ∴ 3 X103=9 M th t Mathematics 26) The incorrect statement is 1) The identity element in a group is unique 2) In a group of even order, there exists an element a≠e such that a2=e. 3) The cube roots of unity are , 4) In an abelian group (ab)2=a2b2, ∀ a, b∈ G. M th t Mathematics Ans : is 3 Cube roots of unity are M th t Mathematics 27) In the multiplicative group of fourth ou t roots oots of o unity u ty the t e inverse e se of i103 is 1) 2) 3) 4) 1 –1 i –i M th t Mathematics Ans : is 3 e=1 i103=i100 . i3 = (i4)25 . (i2).i = 1. 1 ((-1) 1) .ii = -ii ∴ inverse of –i is i. M th t Mathematics 28) Let Q1=Q – {1} be the set of all rationals except 1 and ¸ is d fi d as a¸b defined ¸b = a+b +b – ab b ∀ a, b∈Q1. The inverse of 2 is 1) 2) 3) 4) 2 1 0 –2 M th t Mathematics Ans : is 1 a¹e=a ⇒ a+e-ae=a ⇒ e(1-a)=0 ⇒ e=0 (∴a≠1∉Q1) & a¹a-1=e ⇒ a+a-1-aa-1=0 ⇒ a-1(1-a)=-a ⇒ a-1 = -a 1a 1-a (∴ 1-a ≠ 0) ∴ 2-11 = - 2 ⇒ 2-11=2 1-2 M th t Mathematics 29) In the group {Z6, + (mod 6)}, 2+4 –11 + 3–11 is i equall to t 1) 2) 3) 4) 2 1 4 3 M th t Mathematics Ans : is 2 e=0 ∴ 2+64-1+63-1=2+62+63=1 M th t Mathematics 30) Every e yg group oup of o order o de 7 is s 1) 2) 3) 4) Not abelian N t cyclic Not li Cyclic None of these M th t Mathematics Ans s : is s 3 Every group of prime order is cyclic 7 is prime M th t Mathematics 31) If g = and h = are a e ttwo o pe permutations utat o s in group g oup S4, then (h x g) (2) = 1) 2) 3) 4) 2 1 3 4 M th t Mathematics Ans : is 2 (hxg)2 = h[g(2)]=h(3)=1 M th t Mathematics 32)) If g = 3 tthen e g–1 1) 2) 3) 4) M th t Mathematics Ans : is 1 M th t Mathematics 33) In the group {1, 2, 3, 4, 5, 6} under d multiplication lti li ti modulo d l 7, 7 5x=4 has the solution x = 1) 2) 3) 4) 0.8 2 3 5 M th t Mathematics Ans : is 4 (e=1) 5x73=1 ⇒ 5-1 = 3 ∴ 5x 5x=4 4⇒x x= 5-1x74 = 3x74 4=5 5 M th t Mathematics 34) In the group G={2, 4, 6, 8} under X10, the th inverse i off 4 is i 1) 2) 3) 4) 6 8 4 2 M th t Mathematics Ans : is 3 Here e=6 since 4x106=4 etc. ∴ 4x104=6 ⇒ 4-1=4 M th t Mathematics 35) The Set { –1 , 0, 1} is not a group w.r.t. t add addition t o because itt does not satisfy 1) 2) 3) 4) Closure property A Associative i ti law l Existence of identity Existence of inverse M th t Mathematics Ans : is 1 1+1=2 ∉ the set M th t Mathematics 36) If e every e ye element e e to of a g group oup G is s its own inverse, then G is 1) 2) 3) 4) Finite I fi it Infinite Cyclic Abelian M th t Mathematics Ans : is 4 since a =a-1, b=b-1 ∀ a, a b b∈G G Now (ab)-1=ab ab (by hypothesis) ⇒ b-1a-1=ab, by y property y ⇒ ba = ab ∴ G is abelian M th t Mathematics 37) If a, b, c, are three elements of a group (G, (G ¹), ¹) and d (a¹b) ( ¹b) ¹x=c, ¹ then x= 1) 2) 3) 4) c ¹(a-1¹b-1) c ¹(b-11¹a ¹ -11) (b-1¹c-1)¹c (a-1¹b-1)¹c M th t Mathematics Ans : is 3 (a¹b)-1¹(a¹b)¹x=(a¹b)-1¹c e¹x ¹ = (b-11¹a ¹ -11)¹c )¹ M th t Mathematics 38) If { z7, x7} is sag group, oup, then t e the t e inverse of 6 is 1) 2) 3) 4) 6 4 1 3 M th t Mathematics Ans : is 1 since 6x76=36≡1 (mod 7) where h e=1 ∴ 6-1=6 M th t Mathematics