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Byju's Answer
Standard XII
Mathematics
Difference of Two Sets
Let n=p1α1,...
Question
Let
n
=
p
α
1
1
,
p
α
2
2
,
p
α
3
3
.
.
.
p
α
k
k
where (
p
1
,
p
2
,
.
.
.
p
k
are primes and
α
1
,
α
2
,
α
3
.
.
.
,
α
k
ϵ
N
)
then find the number of ways in which
n
can be expressed as the product of two factors which are prime to one another.
Open in App
Solution
x
=
P
α
1
1
,
P
α
2
2
,
P
α
3
3
.
.
.
.
.
P
α
k
k
L
e
t
t
h
e
p
r
i
m
e
n
u
m
b
e
r
⇒
P
1
P
2
P
3
.
.
.
.
.
.
.
P
k
n
=
A
×
B
w
h
e
r
e
H
C
F
(
A
,
B
)
=
0
The power of
P
1
either A & B.
A
B
p
o
s
s
i
b
l
e
w
a
y
s
p
r
i
m
e
n
o
.
→
0
k
1
1
k
−
1
k
c
1
2
k
−
2
k
c
2
1
1
k
0
k
c
k
⇒
1
2
∑
K
r
=
0
K
c
r
[
∵
E
v
e
r
y
c
a
s
e
R
e
p
e
a
t
e
d
b
y
t
w
o
T
i
m
e
s
s
o
d
i
v
i
d
e
b
y
2
]
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0
Similar questions
Q.
If "n" is a natural number such that
n
=
p
α
1
1
.
p
α
2
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
p
k
α
2
;
p
1
,
p
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
,
p
k
are distinct primes, then
Q.
Assertion :The sum of divisors of
n
=
2
10
3
2
5
3
7
2
11
2
13
3
is
1
5760
(
2
11
−
1
)
(
3
3
−
1
)
(
5
4
−
1
)
(
7
3
−
1
)
(
11
3
−
1
)
(
13
4
−
1
)
Reason: The number of divisor of
m
=
p
1
α
1
p
2
α
2
.
.
.
p
r
α
r
where
p
1
,
p
2
,
.
.
.
,
p
r
are distinct primes and
α
1
,
α
2
,
.
.
.
,
α
r
are natural number is
(
α
1
+
1
)
(
α
2
+
1
)
.
.
.
(
α
r
+
1
)
Q.
Let
N
=
2
101
×
∫
1
0
x
50
(
1
−
x
)
50
d
x
∫
1
0
x
50
(
1
−
x
102
)
50
d
x
then the number of ways in which N can be resolved into two factors which are relatively prime numbers is
Q.
Let
α
1
and
α
2
be the roots of the equation
x
2
−
4
x
+
P
1
=
0
, and
α
3
and
α
4
be the roots of the equation
x
2
−
36
x
+
P
2
=
0
. If
α
1
<
α
2
<
α
3
<
α
4
and
α
1
,
α
2
,
α
3
,
α
4
are in G.P., then the product
P
1
P
2
equals
Q.
If a number N can be represented as
P
a
1
×
P
b
2
×
P
c
3
where
P
1
,
P
2
,
P
3
are prime factors of N then number of factors of N can be found as
(
a
+
1
)
×
(
b
+
1
)
×
(
c
+
1
)
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