Question

If $nϵN$, then ${11}^{n+2}+{12}^{2n+1}$ is divisible by:

• A. $113$
• B. $123$
• C. $133$
• D. None of these

Answer 1

The correct option is C $133$

${11}^{n+2}+{12}^{2n+1}$

Put $n=1,{11}^3+{12}^3=1331+1728=133\left(26\right)$

${11}^{n+2}+{12}^{2n+1}$ is divisible by $133$ for $n=1$.

Now using induction, let us prove for all $n$ natural numbers.

Assure it is true for a natural number $k$.

To prove it is true for the number $k+1$.

For $k+1$ :

${11}^{(k+1)+2}+{12}^{2(k+1)+1}={11}^{(k+2)+1}+{12}^{(2k+1)+2}$

$={11}^{k+2}\times 11+{12}^{k+1}\times {12}^2$

$={11}^{k+2}\times 11+(11+133)\times {12}^{2k+1}$

$=11[{11}^{k+2}+{12}^{2k+1}]+\left(133\right)\left({12}^{2k+1}\right)$

$=11\left(133\right(m\left)\right)+\left(133\right)\left({12}^{2k+1}\right)$

$=133[11m+{12}^{2k+1}]$

${11}^{n+2}+{12}^{2n+1}$ is divisible by $133$.