Search any question & find its solution
Question:
Answered & Verified by Expert
$7^{9}+9^{7}$ is divisible by
Options:
Solution:
2307 Upvotes
Verified Answer
The correct answer is:
64
Now, $7^{9}=(8-1)^{9}=-1(1-8)^{9}$
$=-1+{ }^{9} C_{1} 8-{ }^{9} C_{2} 8^{2}+\ldots+{ }^{9} C_{9} 8^{9}$
and $\quad 9^{7}=(1+8)^{7}=1+{ }^{7} C_{1} 8+{ }^{7} C_{2} 8^{2}$
$+{ }^{7} C_{3} 8^{3}+\ldots .+{ }^{7} C_{7} 8^{7}$
$\therefore 7^{9}+9^{7}=8\left({ }^{9} C_{1}+{ }^{7} C_{1}\right)+8^{2}\left({ }^{7} C_{2}-{ }^{9} C_{2}\right)+\ldots$
$=8(9+7)+8^{2}(21-36)+\ldots$
$=64 \times 2+64(-15)+\ldots$
Hence, it is divisible by 64 .
$=-1+{ }^{9} C_{1} 8-{ }^{9} C_{2} 8^{2}+\ldots+{ }^{9} C_{9} 8^{9}$
and $\quad 9^{7}=(1+8)^{7}=1+{ }^{7} C_{1} 8+{ }^{7} C_{2} 8^{2}$
$+{ }^{7} C_{3} 8^{3}+\ldots .+{ }^{7} C_{7} 8^{7}$
$\therefore 7^{9}+9^{7}=8\left({ }^{9} C_{1}+{ }^{7} C_{1}\right)+8^{2}\left({ }^{7} C_{2}-{ }^{9} C_{2}\right)+\ldots$
$=8(9+7)+8^{2}(21-36)+\ldots$
$=64 \times 2+64(-15)+\ldots$
Hence, it is divisible by 64 .
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.