We have $( \sqrt 2 + \sqrt 3)^{1980} = ...7,9...$. Note that $A = ( \sqrt 2 + \sqrt 3)^{1980} = (5 + \sqrt{24})^{990}$. Let $a = 5 + \sqrt{24}$ and $b = 5 - \sqrt{24}$. Then, using the binomial expansion, it is easy to verify that $N = A + b^{990} = a^{990} + b^{990}$ is a positive integer. Moreover $b = \frac {1} {a} < \frac {1} {5}$. It follows that $0 < b < 0,2$, then $0 < b^{990} < 10^{-1}$. Thus $A = N - b^{990}$ has digit $9$ at the right of the decimal point, and has digit $d$ at the left, with $d = n - 1$ mod[10] where $n$ is the digit of the unity in $N$. Using binomial expansion, we have : $N = 2 \prod_{k=0}^{495} 2C_{990}^{2k} 5^{990-2k} 24^k = 2 \times 24^{495}$ mod[10]. Thus $N = 2 \times 4^{495}$ mod[10]. Using that $2^5 = 32 = 2$ mod[10], it follows easily that $N = 8$ mod[10], so that $d = 7$ as claimed. Pierre.

Let $a_n=(\sqrt3+\sqrt2)^{2n}+(\sqrt3-\sqrt2)^{2n}=(5+2\sqrt6)^n+(5-2\sqrt6)^n$. It satisfies the recurrence $a_n-10a_{n-1}+a_{n-2}=0$, or $a_n=10a_{n-1}-a_{n-2}$, where $a_0=2$ and $a_1=10$. Then the sequence modulo $10$ is $2\to0\to-2\to0\to-2\to\dots$. We know that $a_{990}=(\sqrt3+\sqrt2)^{1980}+(\sqrt3-\sqrt2)^{1980}\equiv-2\pmod{10}$, so its last digit is $8$. Since $(\sqrt3-\sqrt2)^{1980}<<0.1$, the units digit of $(\sqrt2+\sqrt3)^{1980}$ is $7$ and the digit after is $9$.