Search any question & find its solution
Question:
Answered & Verified by Expert
The remainder of $n^4-2 n^3-n^2+2 n-26$ when divided by 24 , is
Options:
Solution:
1639 Upvotes
Verified Answer
The correct answer is:
22
Let $\begin{aligned} f(n) & =n^4-2 n^3-n^2+2 n-26 \\ & =n^3(n-2)-n(n-2)-26 \\ & =(n-2)\left(n^3-n\right)-26 \\ & =(n-2) n\left(n^2-1\right)-26 \\ & =(n-2)(n-1) n(n+1)-26\end{aligned}$
$$
=24 k-48+22
$$
$[\because$ product of four consecutive natural numbers is divisible by 24]
$$
=24[k-2]+22
$$
Hence, remainder is 22.
$$
=24 k-48+22
$$
$[\because$ product of four consecutive natural numbers is divisible by 24]
$$
=24[k-2]+22
$$
Hence, remainder is 22.
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.