Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
When 21505 is divided by 9, the remainder is
Question Error Report
Question is incomplete/wrong
Question not belongs to this Chapter
Answer is wrong
Solution is wrong
Answer & Solution is not matching
Spelling mistake
Image missing
Website not working properly
Other (not listed above)
Error description
Thank you for reporting, we will resolve it shortly
Back to Question
Thank you for reporting, we will resolve it shortly
Q. When $2^{1505}$ is divided by $9$, the remainder is
KEAM
KEAM 2012
Binomial Theorem
A
8
B
7
C
5
D
6
E
1
Solution:
$2^{1505} =\left(2^{6}\right)^{25} \times 2^{5} $
$=(64)^{25} \times 32$
$=(1+63)^{25} \times 32 $
$=\left[{ }^{25} C_{0}(63)^{0}+{ }^{25} C_{1}(63)^{1}+{ }^{25} C_{2}(63)^{2}\right.$
$\left.\ldots .+{ }^{25} C_{25}(63)^{25}\right] 32 $
$=\left[1+(63)+{ }^{25} C_{2}(63)^{2}+\ldots+(63)^{25}\right] 32$
$=32+63\left[1+{ }^{25} C_{2}(63)+\ldots+(63)^{24}\right] 32 $
The remainder when $2^{1505}$ is divided by
9 $=$ The remainder when 32 divided by 9
$=5$