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!


2 Answers 2


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 $\boxed{9}$.

  • 5
    $\begingroup$ Very nice. +1 . $\endgroup$
    – Shailesh
    Commented Nov 24, 2015 at 15:26
  • 1
    $\begingroup$ As a historical note, the key observation above derived via the binomial expansion, is also a corollary to proposition 9 book 2 of Euclid's Elements. It takes a lot more work there, but gives a nice geometrical interpretation very different in feeling (for me at least) from the binomial way to see it. $\endgroup$
    – user77970
    Commented Jan 2, 2016 at 17:48
  • $\begingroup$ I don't get the last part. You didn't prove that $(\sqrt{3} - \sqrt{2} )^{2016} = 0.9999$. You have only shown that it is less than $0.1^{600}$. Does that mean that eg. $0.05^{600}$, which is less than $0.1^{600}$ also has so many nines? No, only if it is in some form of: $10^{-\text{many}}$. I really don't get how you concluded that, with these inequalities. Elaborate more? $\endgroup$
    – KeyC0de
    Commented Feb 13, 2017 at 18:44

Compare $(\sqrt{3}+\sqrt{2})^n$ and $(\sqrt{3}-\sqrt{2})^n$
Can show their sum is an integer when $n$ is even, and the second number is very small?


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