So many nines causes irritation.

Algebra Level 4

$\large\ 9\times 99\times 9999\times ...\times \left( { 10 }^{ { 2 }^{ 2018 } } - 1 \right)$

Find the sum of the digits of the expression above, where each factor has twice as many digits as the previous one.

9\times2^{2018} - 2^{2017}

9\times2^{2018}

3\times2^{2018}

2\times9^{2018}

9\times2^{2017}

0 solutions

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