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UPSC 2015 CDS I Math Paper
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© examsnet.com
Question : 35
Total: 100
If n is a natural number and
n
=
p
1
x
1
p
2
x
2
p
3
x
3
where
p
1
,
p
2
,
p
3
are distinct prime factors, then the number of prime factors for n is
[2015 CDS-I]
x
1
+
x
2
+
x
3
x
1
x
2
x
3
(
x
1
+
1
)
(
x
2
+
1
)
(
x
3
+
1
)
None of the above
Validate
Solution:
n
=
p
1
x
1
p
2
x
2
p
3
x
3
where
p
1
,
p
2
,
p
3
aredistinct prime factors.
The prime factors of n are
p
1
,
p
2
and
p
3
.
Number of prime factors of
n
=
(
x
1
+
1
)
(
x
2
+
1
)
(
x
3
+
1
)
© examsnet.com
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