last $2$ digit and last $3$ digit in $\displaystyle 2011^{{2012}^{2013}}$

Calculation of last $2$ digit and last $3$ digit in $\displaystyle 2011^{{2012}^{2013}}$.

$\bf{My\; Try}::$ for last $2$ digit:: which is same as when we divide $2011^{2012^{2013}}$ is divided by $100$

So $\displaystyle 2011^{2012^{2013}}\mod(100) = (11)^{2012^{2013}}\mod(100)$

Now how can i calculate after that

Help Required

Thanks

Solutions Collecting From Web of "last $2$ digit and last $3$ digit in $\displaystyle 2011^{{2012}^{2013}}$"

As pointed out in many places here on MSE, the general approach to the
calculation of modular powers is Euler’s Theorem: $$a^{\varphi(n)}
\equiv 1 \bmod n$$ for a coprime to $n$. This also shows $$a^m \equiv
a^{(m \,\bmod \varphi(n))} \bmod n.$$ So as a general principle
calculate the modulus of the exponent $\bmod \varphi(n)$.

In your case I suggest to use Chinese Remainder Theorem first: factor $100 = 4 \cdot 25$ and find solutions to $x:=11^{2012^{2013}}\bmod\,4$ and $\bmod\,25$ in order to reconstruct the congruence $\bmod \,100$ afterwards.

  • We find $11 \equiv -1 \bmod 4$. Using that $2012^{2013}$ is even, we
    find $x \equiv 1$ immediately.

  • For calculation $\bmod\; 25$ we use $11^5\equiv 1 \bmod 25$. Therefore, it is enough to consider the exponent $2012^{2013} \equiv 2^{2013}
    \bmod 5$. Using Euler’s Theorem again, we find $2^4 \equiv 1 \bmod 5$
    and thus $2^{2013} \equiv 2^1 \bmod 5$. Back to $\bmod 25$ we find
    $x \equiv 11^2 \equiv 21 \bmod 25.$

So we know that $x \equiv 21 \bmod 25$ and $x \equiv 4 \bmod 4$. As a general principle, check all numbers of the form $25k+21$ for their congruence $\bmod 4$. In this case we see immediately that $21\equiv 21 \bmod 25$ and $21 \equiv 1 \bmod 4$ is the solution.

So $2011^{2012^{2013}} \equiv 21 \bmod 100$.

For the last 3 digits do the same using $1000 = 8 \cdot 125$ and $\varphi(8) = 4$, $\varphi(125) = 100$. As a cross-check: I got $321$.