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## '2011 Digits' printed from http://nrich.maths.org/

$20 \div 11 = 1\frac{9}{11} = 1.81818181...$. So the first 2011 digits are 1006 '1's and 1005 '8's. Therefore the required total is $1006\times1 + 1005\times8 = 1006 + 8040 = 9046$.

*This problem is taken from the UKMT Mathematical Challenges.*