Question

If $\forall m\in N$, then ${11}^{m+2}+{12}^{2m-1}$ is divisible by

• A. $121$
• B. $132$
• C. $133$
• D. None of these

Answer 1

The correct option is D None of these

To find the divisor of ${11}^{m+2}+{12}^{2m-1}$ by mathematic induction, the first step is to check for the smallest natural number, i.e; for $m=1$. So, this reduces to ${11}^3+{12}^1$ or ${11}^4+1$.

So, the number when divided by $11$ leaves remainder 1.

So, we can knock out options $A$ and $B$ as $121$ as well as $132$ are both divisible by 11 and hence their multiples will always be divisible by 11.

Now, we have to check the divisibility of ${11}^{m+2}+{12}^{2m-1}$ by $133$. For $m=1$, ${11}^4+1$ is not divisible by $133$.

So, we can knock out option $C$.

Hence, $D$ is correct.