Find the last two digits of $3^{45}$

Question

I was wondering if there is a simpler way to find the last to digits of a power such as $3^{45}$. I reduced it modulo 100 to get the answer, which is 43. But I was curious if there was a simpler, or more eloquent way to get this result.

Here's my method:

Using the fact that $3^4=81\equiv -19 \pmod{100}$ and $19^5=2,476,099\equiv -1 \pmod{100}$

$$ 3^{45}\equiv 3(3^4)^{11}\equiv 3(-19)^{11}\equiv 3(-19)(19^5)^2\equiv 3(-19)(-1)^2 \equiv -57 \equiv 43 \pmod {100} $$

I just felt like this was sort of a messy way to find out that the last two digits are 43.

Answer 1

Essentially, we need to find $\displaystyle 3^{45}\pmod{100}$

Method $\#1:$

Now, $$3^{45}=3\cdot3^{44}=3(3^2)^{22}=3(10-1)^{22}=3(1-10)^{22}$$

Now, $$(1-10)^{22}\equiv1-10\cdot22\pmod{100}\equiv-19\equiv81$$

Method $\#2:$

As $(3,10)=1$

using Euler's Totient Theorem, $\displaystyle \phi(100)=40\implies 3^{40}\equiv1\pmod{100}$

or using Carmichael Function, $\displaystyle \lambda(100)=20\implies 3^{20}\equiv1\pmod{100}$

In either case, $\displaystyle3^{45}\equiv3^5\pmod{100}$

Answer 2

${\rm mod}\ 50\!:\ \color{#0a0}{3^{\large 5}\!\equiv -7}\,\Rightarrow\,\color{#c00}{3^{\large 10}}\!\equiv 49\equiv \color{#c00}{-1}\,\Rightarrow\,3^{\large 45}\!\equiv \color{#0a0}{3^{\large 5}}(\color{#c00}{3^{\large 10}})^{\large 4}\!\equiv \color{#0a0}{-7}(\color{#c00}{-1})^{\large 4}\!\equiv -7,\,$ so $\!\bmod 100\,$ either $\,3^{\large 45}\!\equiv -7\,$ or $\,50\!-\!7\equiv\color{#90f}{43}.\,$ It must be $\,\color{#90f}{43},\,$ by ${\rm mod}\ 4\!:\ 3^{\large 45}\!\equiv (-1)^{\large 45}\!\equiv -1\equiv\color{#90f}{ 43}.$

Answer 3

The background impetus for what follows is the multiplicative group

contained in $\mathbb {Z} / \text{100} \mathbb {Z}$.

and we stand ready to write

$\quad 3^4 \equiv 81 \pmod{100}$

From the (cyclic) group table we know that $81^5 \equiv 1 \pmod{100}$.

So,

$\quad 3^{45} = (3^4)^{11} \times 3 \equiv 81^{11} \times 3 \equiv 81 \times 3 \equiv 43 \pmod{100}$