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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