Computing the unit digit I'm asked to compute the unit digit of $(\sqrt2+\sqrt3)^{100}$. I know I'm suppose to add $(\sqrt3-\sqrt2)^{100}$ to the above number and use binomial theorem to simplify. But I can't seem to figure out how to simplify the summation. Can someone help me out?
 A: Consider $a=(\sqrt3+\sqrt2)^{100}+(\sqrt3-\sqrt2)^{100}$. If you expand both 100th powers using the binomial theorem, you find that all the terms with odd degrees in $\sqrt3$ or $\sqrt2$ cancel out, so all that's left are terms of the form $2k(\sqrt3)^{2n}(\sqrt2)^{2m}$ for some integers $k,m,n$. These terms are all integers, so $a$ is an integer too.
Since $0<(\sqrt3-\sqrt2)^{100}\ll 1$, the units digit of $(\sqrt3+\sqrt2)^{100}$ is one less than the units digit of $a$, so all we need to do is compute $a$ modulo 10.
Now all of the coefficients $2k$ in the above general term happen to be $2\binom{100}{j}$ for some $j$. It happens that $\binom{100}{j}$ is usually divisible by $5$ -- there are $24$ factors of $5$ in $100!$ and very rarely that many 5s in $j!(100-j)!$. This means that the entire term has zero units digit for most $j$s and can be ignored when we're only interested in the last digit of $a$. You only need to find the few $j$ where $5\nmid \binom{100}{j}$ and sum the last digits of the corresponding terms.
You'll need to compute one large binomial coefficient modulo 5, but since you know that the factors of $5$ cancel out, you can work modulo $5$ for that, and most of the other factors of the factorials will cancel out too.
A: Another way is to use linear recurrence theory:
We can note that $(\sqrt3 + \sqrt2)^{100} = (5+2\sqrt6)^{50}$.  Let $a_n = (5 + 2\sqrt6)^n, b_n = (5+2\sqrt6)^n+(5-2\sqrt6)^n$.  We note that
$$5-2\sqrt6 < \tfrac15 \implies (5-2\sqrt6)^{50} < \frac1{5^{50}} = \frac1{25^{25}}< \frac1{10^{25}}$$
so $a_{50}$ is marginally less than $b_{50}$ which is an integer as noted below.  So we may find the unit digit of $b_{50}$ instead and subtract one from it.
For finding the unit digit of $b_n$ easily, we will construct a recurrence equation which it satisfies. Looking at the form of $b_n$, which is $c_1 \alpha_1^n + c_2 \alpha_2^2$, it should satisfy a second order linear homogeneous recurrence equation, with characteristic equation $(r-\alpha_1)(r-\alpha_2)=0$.
Thus the characteristic equation is $\left(r-(5+2\sqrt6)\right)\left(r-(5-2\sqrt6)\right) = r^2-10r+1=0$
So the recurrence we have is $b_{n+2} = 10 b_{n+1}-b_n$.  We calculate $b_0 = 2, b_1 = 10$.  As these are integers, we are assured $b_n$ is an integer for all $n \ge 0$.
So $b_n$ are integers which satisfy the recurrence $b_{n+2} \equiv -b_n \pmod {10}$.  Hence $b_{50} \equiv -b_{48} \equiv b_{46} \dots \equiv -b_0 \equiv 8 \pmod {10}$
