Question

What is ${\mathbf{2}}^{{\mathbf{2}}^{\mathbf{2006}}}\mathbf{\left(}\mathbf{mod}\mathbf{3}\mathbf{\right)}$?

Short answer

The solution is $2^{2^{2006}}\left(\mathrm{mod}3\right)=1$.

Step-by-step solution

Introduction

A function that returns a number or variable's absolute value is known as a modulus function. The magnitude of the number of variables is produced. Another name for it is an absolute value function. No matter what input was provided to this function, a positive result is always the outcome.

Calculating 222006(mod3).

The value starts from $\mathrm{n}=0$. i.e.,

$\begin{array}{c}\mathrm{For}\mathrm{n}=1,2^1=2\left(\mathrm{mod}3\right)\\ \mathrm{For}\mathrm{n}=2,2^2=1\left(\mathrm{mod}3\right)\\ \mathrm{For}\mathrm{n}=3,2^3=2\left(\mathrm{mod}3\right)\\ \mathrm{For}\mathrm{n}=4,2^4=1\left(\mathrm{mod}3\right)\end{array}$

The even value of$\mathrm{n}$ remainder is$1$ and for odd it is $2$. i.e.,

$\begin{array}{c}{2}^{2\mathrm{n}}=1\left(\mathrm{mod}3\right)\\ 2^{2\mathrm{n}+1}=2\left(\mathrm{mod}3\right)\end{array}$

Therefore, the solution is$2^{2^{2006}}\left(\mathrm{mod}3\right)=1$ .