Question

Prove that if n is a positive integer, then 133 divides ${11}^{n+1}+{12}^{2n-1}$

Short answer

133 divides ${11}^{n+1}+{12}^{2n-1}$ whenever n is a positive integer

Step-by-step solution

Step: 1

If n=1,

$\begin{array}{r}{11}^{1+1}+{12}^{21-1}\\ =121+12\\ =133\end{array}$

Step: 2

Let P(k) be true.

${11}^{k+1}+{12}^{2k-1}$

We need to prove that P(k+1) is true.

Step: 3

$\begin{array}{r}{11}^{(k+1)+1}+{12}^{2(k+1)-1}\\ =11\cdot {11}^{k+1}+144\cdot {12}^{2k-1}\\ =11\cdot {11}^{k+1}+(133+11)\cdot {12}^{2k-1}\\ =11\cdot \left({11}^{k+1}+{12}^{2k-1}\right)+133\cdot {12}^{2k-1}\end{array}$

It is divisible by 133.

It is true for P(k+1) is true.