Find the digits immediately on the left and right of $(\sqrt 11 + \sqrt 10)^{2002}$. What digits are immediately on the left and right of the decimal point in $(\sqrt 11 + \sqrt 10)^{2002}$?
attempt: 
I have tried to use the binomial theorem as follows.
$(\sqrt 11 + \sqrt 10)^{2002}$ = $\Sigma_{k=0}^{2002}$$2002 \choose k$$(\sqrt(11))^k (\sqrt(10))^{2002 - k}$.
And then expanding $2002 \choose 0 $$(\sqrt(11))^0((\sqrt10))^{2002} + ....+$$2002 \choose 2002 $$(\sqrt(11))^{2002}((\sqrt10))^{0} $.
I am not really sure how to proceed or approach this problem. If someone could help , that would really help. Thank you!
 A: We are only interested in the last digit before the decimal and first after.
$(\sqrt {11}+\sqrt {10})^{2002}=$
$\sum ({2002\choose j}\sqrt {11}^j\sqrt {10}^{2002-j}=a_j )$
If $j $ is even $j <2002$ then $a_j $ is a multiple of $10$ and the last and first digit by the decimal point are zero.
If $j = 2002$ then $a_{2002}=11^{1001} $ and the last digit is $1$, and the first number after the decimal point is $0$.
So we are only concerned with $j $ odd.
But note $(\sqrt {11}-\sqrt {10})^{2002}=\sum (-1)^{j}a_j $.  Also $ (\sqrt {11}-\sqrt {10}) < 1/2$ so $ (\sqrt {11}-\sqrt {10})^{2002} < 1/2^{2002} $ so the first few hundred digits past the decimal point will be $0$s.
So $\sum a_j =\sum_{j\text { even}} 2 a_j - \sum (-1)^ja_j $.  $\sum a_j =\sum_{j\text { even}}  a_j$ has last digit of $1$ as we said above, so $\sum a_j =\sum_{j\text { even}} 2 a_j$ will have last digit of $2$.  And $\sum a_j =\sum_{j\text { even}} 2 a_j - \sum (-1)^ja_j$ will therefore have last digit of $1$ and the first several hundred digits past the decimal points all $9$s.
