MCQ
If $\forall \mathbf{m} \in \mathbf{N}$, then $11^{\mathrm{m}+2}+12^{2 \mathrm{~m}-1}$ is divisible by:
  • A
    121
  • B
    132
  • C
    133
  • None of these

Answer

Correct option: D.
None of these
  1. None of these
Solution:
To find the divisor of $11^{\mathrm{m}+2}+12^{2 \mathrm{~m}-1}$ 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 $11^{m+2}+12^{2 m-1}$ by 133 . For $m=1,11^4+1$ is not divisible by 133 .
So, we can knock out option C.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free