Question

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

• A. 20
• B. 21
• C. 22
• D. 23

Answer 1

Answer: (C)

Let $f(n)=n^4-2n^3-n^2+2n-26$
$=n^3(n-2)-n(n-2)-26$
$=(n-2)\left(n^3-n\right)-26$
$=(n-2)n\left(n^2-1\right)-26$
$=(n-2)(n-1)n(n+1)-26$
$=24k-48+22$
$[\because$ product of four consecutive natural numbers is divisible by 24$]$
$=24[k-2]+22$
Hence, remainder is $22.$