# Computation without a computer

In the number $(1+\sqrt{3})^{2015}$, what is the 224th digit after the decimal point?

You may NOT use a calculator, computer, or any electronic aid to answer this question. Only pen(cil), paper, and brain are allowed.

This should be relatively easy, but it's a puzzle of a kind I haven't seen on this site before, and hopefully it'll set a trend for puzzles that are the opposite of ones!

• Is the answer 0? – ghosts_in_the_code Mar 29 '15 at 14:47
• @ghosts_in_the_code There are only 10 possibilities, I don't think he would answer a question like that! – leoll2 Mar 29 '15 at 14:57
• brain is allowed ? great! i have digital one hahaha nice opening for nice set of newfashioned tagged puzzles. wonder why this was been downvoted – Abr001am Mar 29 '15 at 16:51
• It's hard, but the community here are pretty bright! It has to be at least as hard as that, or it'll probably get solved within minutes and then attacked as too easy. Even so it was solved pretty fast. – Rand al'Thor Mar 29 '15 at 19:25
• I think this is a good example of a problem that looks textbook but is nonetheless a math puzzle. In the surface it looks like a hopeless calculation, but with a clever insight, the solution can be found with little calculation. – xnor Mar 29 '15 at 21:22

Consider the auxiliary value $N=(1+\sqrt{3})^{2015}+(1-\sqrt{3})^{2015}$. In the binomial expansion of $N$ all terms with odd powers of $\sqrt3$ cancel out, so that $N$ is an integer.
The real number $M=1-\sqrt{3}$ is negative with $|M|\approx0.732$. Then $|M|<0.74$ and $|M|^8<0.74^8<10^{-1}$. Then $|M|^{2015}<(|M|^8)^{250}<10^{-250}$. Then $(1-\sqrt{3})^{2015}$ is a negative real number between $-10^{-250}$ and $0$, and the first 249 digits after the decimal point are 0s. Subtracting it from the integer $N$ gives an integer plus an astronomically small real number, so that the first 249 digits after the decimal point all are 0.
• Nice approach. However, since $M$ is negative, we have $N - M = N + |M|$, which leaves the first 249 digits after the decimal point $0$. – Lawrence Mar 29 '15 at 21:19
• I like the answer very much, except the part $0.74^8<10^{-1}$. This is too annoying to calculate by hand, so I suggest this approach (which gives a bit less sharp bound, but requires almost no calculation at all): $|M| < \frac 3 4 \implies |M|^5 < \frac{3^5}{2^{10}} < \frac{3^5}{1000} = 0.243 < \frac 14 \implies |M|^{25} < \frac 1 {2^{10}} < 10^{-3}$ $\implies |M|^{2015} < |M|^{2000}<10^{-240}$. This requires only to calculate $3^5 = 243$ and knowing $2^{10} = 1024$, which is common knowledge in this digital age. – Ennar Mar 30 '15 at 12:30