On the most recent Seton Hall Joseph W. Andrushkiw Competition, the final question was as follows:
Let $A = (\sqrt{3}+\sqrt{2})^{2016}$. When A is written in decimal
form, what is its $31^{st}$ digit after the decimal point?
Brute forcing it via wolfram alpha reveals that the answer is [edit: I found the 31st number from the start, not the 31st after the decimal point] zero, yet this competition does not allow the use of a calculator. It seems to me that as irrational numbers are in the base of the exponent, there should not be an identifiable pattern in the digits.
Searching this site has made me think that perhaps the answer has something to do with the Euler phi function (something which I will admit up front I have never been acquainted with), but I can't find anything which I understand enough to give me a concrete way to start to approach this. Any help on this frustrating problem would be appreciated. Thanks!
Best Answer
Hmm. Pretty sure that the answer is $9$. The key observation to this problem is noticing that $(\sqrt{3}-\sqrt{2})^{2016}+(\sqrt{3}+\sqrt{2})^{2016}$ is an integer.
The proof of this is expansion using Binomial Theorem. The odd powers of the square roots get canceled out.
Now we have $(\sqrt{3}+\sqrt{2})^{2016} = N - (\sqrt{3}-\sqrt{2})^{2016}$, where $N$ is a positive integer.
Now this is easy. Since $(\sqrt{3}-\sqrt{2})^{2016} < (0.4)^{2016} = (0.064)^{\frac{2016}{3}} < (0.1)^{\frac{2016}{3}} < (0.1)^{600}$, we have $(\sqrt{3}+\sqrt{2})^{2016}= (N-1)+0.99\cdots 99$, and there are at least $500$ $9$'s there. The answer is $\boxed{9}$.