Question

If $10^n+3\cdot 4^{n+1}+k$ is divisible by $9$ for all $n\in N$, then the least positive integral value of $k$ is

• A. $5$
• B. $3$
• C. $7$
• D. $1$

Answer 1

Answer: (A)

Let $P\left(n\right):10^n+3\cdot 4^{n+1}+k$ is divisible by $9\forall n\in N$.
For $n=1$, we get
$P\left(1\right)=10^1+3\cdot 4^{1+2}+k$ is divisible by $9$
$=202+k$ is divisible by $9$
So, the least value of $k$ is $5$.