Question

The remainder of $n^{4}-2 n^{3}-n^{2}+2 n-26$ when divided by $24$ , is

• A. 20
• B. 21
• C. 22
• D. 23

Answer 1

Answer: (C)

Let $ 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 $
$=(n-2)(n-1) n(n+1)-26$
$=24 k-48+22$
$[\because$ product of four consecutive natural numbers is divisible by 24$]$
$=24[k-2]+22$
Hence, remainder is $22 .$