Find the first 1000 digits after the decimal point for the numbers a) $(6+\sqrt{35})^{1999}$; b) $(6+\sqrt{37})^{1999}$; c) $(6+\sqrt{37})^{2000}$.

#
Add the given numbers with their conjugates.

## Solution

a) Let $(6+\sqrt{35})^{1999}=a+b \sqrt{35}$, then $(6-\sqrt{35})^{1999}=a-b \sqrt{35}$, and

$(6+\sqrt{35})^{1999}+(6-\sqrt{35})^{1999}=2 a$. Therefore, $(6+\sqrt{35})^{1999}=2 a-(6-\sqrt{35})^{1999}$. It remains to note that $(6-\sqrt{35})^{1999}=\frac{1}{(6+\sqrt{35})^{1999}}<\frac{1}{10^{1999}}<\frac{1}{10^{1000}}$.

b), c) Similarly to a), taking into account that the number $6-\sqrt{37}$ is negative.

## Answer

a), c) all nines; b) all zeros.