Question

Let $f\left(n\right)={10}^n+3\cdot 4^{n+2}+5$,$nϵN$. The greatest value of the integer which divides $f\left(n\right)$ for all $n$ is

• A. $27$
• B. $9$
• C. $3$
• D. none of these

Answer 1

The correct option is B $9$

$f\left(n\right)={10}^n+\left(3\right)4^{n+2}+5$
$=(1+9{)}^n+3(16\left)\right(4{)}^n+5$
$=(1+9n+9k)+48(4{)}^n+5$
$=(1+9n+9k)+48(1+3{)}^n+5$
$=(1+9n+9k)+48(1+3n+3I)+5$
$=1+9n+9\left(16n\right)+9k+9\left(16\right)I+48+5$
$=9n\left(17\right)+9(k+16I)+54$
$=9n\left(17\right)+9(k+16I)+9\left(6\right)$
$=9(n\alpha +\beta +6)$
Hence the largest integer for which $f\left(n\right)$ is divisible by $n$ is $9.$