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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!

  • 4
    $(\sqrt 11 + \sqrt 10)^{2002}+(\sqrt 11 - \sqrt 10)^{2002}$ is likely to be an integer. What is its last digit? – 2017-02-21
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    I don't quite know how to find it. – 2017-02-21
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    You find it by the binomial theorem. Which terms survive when combining the binomial expansions of $(\sqrt{11}+\sqrt{10})^{2002}+(\sqrt{11}-\sqrt{10})^{2002}$? Just the even numbered terms, or the odd numbered terms? Are these repeated? Which terms are integers that end in a zero? Which terms aren't? Which surviving terms are integers at all? Finally...how big is $(\sqrt{11}-\sqrt{10})^{2002}$ (*a very crude estimation here is fine*)? – 2017-02-21
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    Ok , so after I do that, I get $2[10^{1001} + ...+ 11^{1001}]$. I get that the even numbered terms cancel out because of the negative sing. And so I am just left with the odd numbered terms. I counted twice the even numbered terms. However, after that, I am not sure – 2017-02-21
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    Could someone please verify this? And give some feedback. Thank you – 2017-02-21

1 Answers 1

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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.