Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
10n+3(4n+2)+5 is divisible by (n∈ N) :
Q.
10
n
+
3
(
4
n
+
2
)
+
5
is divisible by
(
n
∈
N
)
:
1864
197
KEAM
KEAM 2005
Report Error
A
7
B
5
C
9
D
17
E
13
Solution:
For
n
=
1
,
10
n
+
3.4
n
+
2
+
5
=
10
+
3.4
3
+
5
=
10
+
192
+
5
=
207
which is divisible by 9.
∴
By Induction, the result is divisible by 9.