Find the last two digits of $3^{45}$ 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.
 A: 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, $\displaystyle3^{45}\equiv3^5\pmod{100}$
A: ${\rm mod}\ 50\!:\ \color{#0a0}{3^{\large 5}\!\equiv -7}\,\Rightarrow\,\color{#c00}{3^{\large 10}}\!\equiv 49\equiv \color{#c00}{-1}\,\Rightarrow\,3^{\large 45}\!\equiv \color{#0a0}{3^{\large 5}}(\color{#c00}{3^{\large 10}})^{\large 4}\!\equiv \color{#0a0}{-7}(\color{#c00}{-1})^{\large 4}\!\equiv -7,\,$ so $\!\bmod 100\,$ either $\,3^{\large 45}\!\equiv -7\,$ or $\,50\!-\!7\equiv\color{#90f}{43}.\,$ It must be $\,\color{#90f}{43},\,$ by ${\rm mod}\ 4\!:\ 3^{\large 45}\!\equiv (-1)^{\large 45}\!\equiv -1\equiv\color{#90f}{ 43}.$
A: A simple method: If you write down the first 24 powers of 3, keeping just the last 2 digits, you get
1   3   9   27
81  43  29  87
61  83  49  47
41  23  69  7
21  63  89  67
1   3   9   27

After a few powers you realize that you only need the last 2 digits of some power of 3 to get the last 2 digits of the next power.
There is a clear pattern that it cycles every 20th power (note that the last digit cycles every 4th power: 1-3-9-7...). Therefore the last two digits of $3^{45}$ are the same as for $3^{25}$ and for $3^5=243$, i.e. 43.
A: 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,
$\quad 3^{45} = (3^4)^{11} \times 3 \equiv 81^{11} \times 3 \equiv 81 \times 3 \equiv 43 \pmod{100}$
