Last two digits

Cool. I hadn't heard of Carmichael's function before. Good to know. :)

How is this good for two digits. Should it not be only the last digit is 0?

We can use the fact that Euler's totient function is multiplicative, i.e., if $m,n$ are coprime, then $\phi(mn) = \phi(m) * \phi(n).$

So as $100 = 4*25$ and $gcd(4,25) = 1$ we know that $\phi(100) = \phi(4)*\phi(25).$

Next, we can use the fact that for $p$ prime, $\phi(p^{k}) = p^{k} - p^{k-1}.$ Then

$\phi(4) = \phi(2^{2}) = 2^{2} - 2^{1} = 2$ and $\phi(25) = \phi(5^{2}) = 5^{2} - 5^{1} = 20.$

We can thus conclude that $\phi(100) = 2*20 = 40.$

With $1000 = 2^{3}*5^{3}$ we would have

$\phi(1000) = \phi(2^{3})*\phi(5^{3}) = (2^{3} - 2^{2})(5^{3} - 5^{2}) = 4*100 = 400.$

So one method to determine $\phi(n)$ is to first decompose $n$ into its prime factors and then use the two formulas mentioned above to make the final calculation. This method can be summarized by Euler's product formula, (proof provided in the aforementioned link),

$\phi(n) = n\displaystyle\prod_{p|n} \left(1 - \dfrac{1}{p}\right).$

As an example, with $n = 1000$ this formula gives us that

$\phi(1000) = 1000*\left(1 - \dfrac{1}{2}\right)\left(1 - \dfrac{1}{5}\right) = 1000*\dfrac{1}{2}*\dfrac{4}{5} = 400,$

in agreement with our previous result.

Oh, sorry, that was a typo; it should be $43^{4} \equiv 1 \pmod{100}.$ As this is a small power we could just multiply it out by hand. Noting that $43^{2} = 1849,$ we would have that

$43^{4} \equiv 49^{2} \pmod{100} \equiv 1 \pmod{100}$ since $49^{2} = 2401.$

For this last step we could also have noted that

$49^{2} = (50 - 1)^{2} = 50^{2} - 2*50 + 1 \equiv 1 \pmod{100}.$

Or we could have directly used the binomial expansion for $43^{4}$ as follows:

$43^{4} = (50 - 7)^{4} = 50^{4} - 4*50^{3}*7 + 6*50^{2}*7^{2} - 4*50*7^{3} + 7^{4} \equiv 7^{4} \pmod{100} \equiv 1 \pmod{100}.$

So there are a number of ways to make this calculation. :)

We are free to choose powers that we think will make the calculation easiest, so there is no requirement to choose the same power in each case. Having said that, and now that you mention it, I could have gone with a binomial expansion in the following way:

$3^{20} = (3^{4})^{5} = 81^{5} = (80 + 1)^{5} = 80^{5} + 5*80^{4} + 10*80^{3} + 10*80^{2} + 5*80 + 1 \equiv 1 \pmod{100},$

since all the terms in the expansion except for $1$ are divisible by $100.$